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CSE 340 Project 4 Solution
The goal of this project is to give you some hands-on experience with implementing a small compiler. You will write a compiler for a simple language. You will not be generating assembly code. Instead, you will generate an intermediate representation (a data structure that represents the program). The execution of the program will be done after compilation by interpreting the generated intermediate representation.
1 Introduction
You will write a small compiler that will read an input program and represent it in an internal data structure. The data structure consists of two parts: (1) a representation of instructions to be executed and (2) a representation of the memory of the program (locations for variables).
Instructions are represented by a data structure that includes the operand(s) of the instruction (if any) and that specify the next instruction to be executed. After the data structures are generated by your compiler, your compiler will execute the generated instructions representation by interpreting it. This means that the program will traverse the data structure and at every node it visits, it will execute the node by changing the content of memory locations corresponding to operands and deciding what is the next instruction to execute (program counter). The output of your compiler is the output that the input program should produce. These steps are illustrated in the following gure
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The remainder of this document is organized as follows:
1. Grammar De nes the programming language syntax including grammar.
2. Execution Semantics Describe statement semantics for if, while, switch and print state-ments.
3. How to generate the intermediate representation Explains step by step how to generate the intermediate representation (data structure). You should read this sequentially and not skip around.
4. Executing the intermediate representation Basically, you have two options. If you are using C/C++, you should use the code we provide to execute the intermediate representation.
If you are using Java, it describes the strict rules to follow in executing the intermediate representation. Those rules will be enforced.
5. Requirements Lists the programming languages you are allowed to use in your solution (C/C++ or Java) and other requirements.
6. Grading Describes the grading scheme.
7. Bonus Project Describes the requirements for the bonus project.
2 Grammar
The grammar for this project is a simpli ed form of the grammar from the previous project, but there are a couple extensions.
program ! var section body
var section ! id list SEMICOLON
id list ! ID COMMA id list j ID
body ! LBRACE stmt list RBRACE
stmt list ! stmt stmt list j stmt
stmt ! assign stmt j print stmt j while stmt j if stmt j switch stmt
assign stmt ! ID EQUAL primary SEMICOLON
assign stmt ! ID EQUAL expr SEMICOLON
expr ! primary op primary
primary ! ID j NUM
op ! PLUS j MINUS j MULT j DIV
print stmt ! print ID SEMICOLON
while stmt ! WHILE condition body
if stmt ! IF condition body
condition ! primary relop primary
relop ! GREATER j LESS j NOTEQUAL
switch stmt ! SWITCH ID LBRACE case list RBRACE
switch stmt ! SWITCH ID LBRACE case list default case RBRACE
for stmt ! FOR LPAREN assign stmt SEMICOLON condition SEMICOLON RPAREN body
case list ! case case list j case
case ! CASE NUM COLON body
default case ! DEFAULT COLON body
Some highlights of the grammar:
1. Expressions are greatly simpli ed and are not recursive.
2. There is no type declaration section.
3. Division is integer division and the result of the division of two integers is an integer.
4. if statement is introduced. Note that if stmt does not have else.
5. for statement is introduced. Note that for has a very general syntax similar to that of the Clanguage for loop.
6. A print statement is introduced. Note that the print keyword is in lower case, but other keywords are all upper-case letters.
7. There is no variable declaration list. There is only one id list in the global scope and that contains all the variables.
8. There is no type speci ed for variables. All variables are int by default.
3 Execution Semantics
All statements in a statement list are executed sequentially according to the order in which they appear. Exception is made for body of if stmt, while stmt and switch stmt as explained below. In what follows, I will assume that all values of variables as well as constants are stored in locations.
This assumption is used by the execution procedure that we provide. This is not a restrictive assumption. For variables, you will have locations associated with them. For constants, you can reserve a location in which you store the constant (this is like having an unnamed immutable variable).
3.1 Assignment Statement
To execute an assignment statement, the expression on the lefthand side of the equal sign is evalu-ated and the result is stored in the location associated with the righthand side of the expression.
3.2 Expression
To evaluate an expression, the values in the locations associated with the two operands are obtained and the expression operator is applied to these values resulting in a value for the expression.
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3.3 Boolean Condition
A boolean condition takes two operands as parameters and returns a boolean value. It is used to
control the execution of while and if statements. To evaluate a condition, the values in the locations
associated with the operands are obtained and the relational operator is applied to these values
resulting in a true or false value. For example, if the values of the two operands a and b are 3 and
4 respectively, a < b evaluates to true.
3.4 If statement
if stmt has the standard semantics:
1. The condition is evaluated.
2. If the condition evaluates to true, the body of the if stmt is executed, then the next statement
following the if in the stmt list is executed.
3. If the condition evaluates to false, the statement following the if in the stmt list is executed
These semantics apply recursively to nested if stmt.
3.5 While statement
while stmt has the standard semantics.
1. The condition is evaluated.
2. If the condition evaluates to true, the body of the while stmt is executed. The next statement
to execute is the while stmt itself.
3. If the condition evaluates to false, the body of the while stmt is not executed. The next
statement to execute is the next statement following the while stmt in the stmt list.
These semantics apply recursively to nested while stmt. The code block:
WHILE condition
f
stmt list
g
is equivalent to:
label : IF condition
f
stmt list
goto label
g
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Note that goto statements do not appear in the input program, but our intermediate represen-
tation includes GotoStatement which is used in conjunction with IfStatement to represent while
and switch statements.
3.6 For statement
The for stmt is very similar to the for statement in the C language. The semantics are de ned by
giving an equivalent construct.
FOR ( assign stmt 1 ; condition ; assign stmt 2 )
f
stmt list
g
is equivalent to:
assign stmt 1
WHILE condition
f
stmt list
assign stmt 2
g
For example, the following snippet of code:
FOR ( a = 0; a < 10; a = a + 1; )
{
print a;
}
is equivalent to:
a = 0;
WHILE a < 10
{
print a;
a = a + 1;
}
3.7 Switch statement
switch stmt has the following semantics1:
1The switch statement in the C language has di erent syntax and semantics. It is also more dangerous!
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1. The value of the switch variable is checked against each case number in order.
2. If the value matches the number, the body of the case is executed, then the statement following
the switch stmt in the stmt list is executed.
3. If the value does not match the number, next case is evaluated.
4. If a default case is provided and the value does not match any of the case numbers, then the
body of the default case is executed and then the statement following the switch stmt in the
stmt list is executed.
5. If there is no default case and the value does not match any of the case numbers, then the
statement following the switch stmt in the stmt list is executed.
These semantics apply recursively to nested switch stmt. The code block:
SWITCH var f
CASE n1 : f stmt list 1 g
...
CASE nk : f stmt list k g
g
is equivalent to:
IF var == n1 f
stmt list 1
goto label
g
...
IF var == nk f
stmt list k
goto label
g
label :
And for switch statements with default case, the code block:
SWITCH var f
CASE n1 : f stmt list 1 g
...
CASE nk : f stmt list k g
DEFAULT : f stmt list default g
g
is equivalent to:
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IF var == n1 f
stmt list 1
goto label
g
...
IF var == nk f
stmt list k
goto label
g
stmt list default
label :
3.8 Print statement
The statement
print a;
prints the value of variable a at the time of the execution of the print statement.
4 How to generate the code
The intermediate code will be a data structure (a graph) that is easy to interpret and execute. I
will start by describing how this graph looks for simple assignments then I will explain how to deal
with while statements.
Note that in the explanation below I start with incomplete data structures then I ex-
plain what is missing and make them more complete. You should read the whole explanation.
4.1 Handling simple assignments
A simple assignment is fully determined by: the operator (if any), the id on the left-hand side, and
the operand(s). A simple assignment can be represented as a node:
struct AssignmentStatement {
struct ValueNode* left_hand_side;
struct ValueNode* operand1;
struct ValueNode* operand2;
ArithmeticOperatorType op; // operator
}
For assignment without an expression on the right-hand side, the operator is set to OPERATOR NONE
and there is only one operand. To execute an assignment, you need the values of the operand(s),
apply the operator, if any, to the operands and assign the resulting value of the right-hand side
to the left hand side. For literals (NUM), the value is the value of the number. For variables, the
value is the last value stored in the location associated with the variable. Initially, all variables
are initialized to 0. In this representation, the locations associated with variables as well as the
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locations in which constants are stored are value nodes ( struct valueNode), a node that contains
a value. The assignment statement contains pointers to the value nodes of the left-hand side and
the operand(s).
Multiple assignments are executed one after another. So, we need to allow multiple assignment
nodes to be linked to each other. This can be achieved as follows:
struct AssignmentStatement {
struct ValueNode* left_hand_side;
struct ValueNode* operand1;
struct ValueNode* operand2;
ArithmeticOperatorType op; // operator
struct AssignmentStatement* next;
}
This structure only accepts ValueNode as operands. To handle literal constants (NUM), you need
to create ValueNode for them and initialize the value stored in these value nodes to the constant
value. This initialization, as well as the initialization of values in locations associated with variable
is done while parsing.
This will now allow us to execute a sequence of assignment statements represented in a linked-
list: we start with the head of the list, then we execute every assignment in the list one after the
other.
Begin Note It is important to distinguish between compile-time initialization and runtime
execution. For example, consider the program
a b;
{
a = 3;
b = 5;
}
The intermediate representation for this program will have have two assignment statements:
one to copy the value in the location that contains the value 3 to the location associated with a
and one to copy the value in the location that contains the value 5 to the location associated with
b. The values 3 and 5 will not be copied to the locations of a and b at compile-time. End Note
This is simple enough, but does not help with executing other kinds of statements. We consider
them one at a time.
4.2 Handling print statements
The print statement is straightforward. It can be represented as
struct PrintStatement
{
struct ValueNode* id;
}
Now, we ask: how can we execute a sequence of statements that are either assignment or print
statement (or other types of statements)? We need to put both kinds of statements in a list and not
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just the assignment statements as we did above. So, we introduce a new kind of node: a statement
node. The statement node has a eld that indicates which type of statement it is. It also has elds
to accommodate the remaining types of statements. It looks like this:
struct StatementNode {
StatementType type; // NOOP_STMT, GOTO_STMT, ASSIGN_STMT, IF_STMT, PRINT_STMT
union {
struct AssignmentStatement* assign_stmt;
struct PrintStatement* print_stmt;
struct IfStatement* if_stmt;
struct GotoStatement* goto_stmt;
};
struct StatementNode* next;
}
This way we can go through a list of statements and execute one after the other. To execute
a particular node, we check its type. If it is PRINT STMT, we execute the print stmt eld, if it is
ASSIGN STMT, we execute the assign stmt eld and so on. With this modi cation, we do not need
a next eld in the AssignmentStatement structure (as we explained above), instead, we put the
next eld in the statement node.
This is all ne, but we do not yet know how to generate the list to execute later. The idea is to
have the functions that parses non-terminals return the code for the non-terminals. For example
for a statement list, we have the following pseudecode (missing many checks):
struct StatementNode* parse_stmt_list()
{
struct StatementNode* st; // statement
struct StatementNode* stl; // statement list
st = parse_stmt();
if (nextToken == start of a statement list)
{
stl = parse_stmt_list();
append stl to st; // this is pseudecode
return st;
}
else
{
ungetToken();
return st;
}
}
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And to parse body we have the following pseudecode:
struct StatementNode* parse_body()
{
struct StatementNode* stl;
match LBRACE
stl = parse_stmt_list();
match RBRACE
return stl;
}
4.3 Handling if and while statements
More complications occur with if and while statements. The structure for an if statement can be
as follows:
struct IfStatement {
ConditionalOperatorType condition_op;
struct ValueNode* condition_operand1;
struct ValueNode* condition_operand2;
struct StatementNode* true_branch;
struct StatementNode* false_branch;
}
The condition op, condition operand1 and condition operand2 elds are the operator and
operands of the condition of the if statement. To generate the node for an if statement, we need to
put together the condition, and stmt list that are generated in the parsing of the if statement.
The true branch and false branch elds are crucial to the execution of the if statement. If
the condition evaluates to true then the statement speci ed in true branch is executed otherwise
the one speci ed in false branch is executed. We need one more type of node to allow loop back
for while statements. This is a GotoStatement.
struct GotoStatement {
struct StatementNode* target;
}
To generate code for the while and if statements, we need to put a few things together. The outline
given above for stmt list needs to be modi ed as follows (this is missing details and shows only the
main steps).
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struct StatementNode* parse_stmt()
{
...
create statement node st
if next token is IF
{
st-type = IF_STMT;
create if-node; // note that if-node is pseudecode and is not
// a valid identifier in C, C++ or Java
st-if_stmt = if-node;
parse the condition and set if-node-condition_op, if-node-condition_operand1 and if-node-condition_operand2
if-node-true_branch = parse_body(); // parse_body returns a pointer to a list of statements
create no-op node // this is a node that does not result
// in any action being taken
append no-op node to the body of the if // this requires a loop to get to the end of
// if-node-true_branch by following the next field
// you know you reached the end when next is NULL
// it is very important that you always appropriately
// initialize fields of any data structures
// do not use uninitialized pointers
set if-node-false_branch to point to no-op node
set st-next to point to no-op node
...
} else ...
}
The following diagram shows the desired structure for the if statement:
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The stmt list code should be modi ed because of the extra no-op node:
struct StatementNode* parse_stmt_list()
{
struct StatementNode* st; // statement
struct StatementNode* stl; // statement list
st = parse_stmt();
if (nextToken == start of a statement list)
{
stl = parse_stmt_list();
if st-type == IF_STMT
{
append stl to the no-op node that follows st
// st
// |
// V
// no-op
// |
// V
// stl
}
else
{
append stl to st;
// st
// |
// V
// stl
}
return st;
}
else
{
ungetToken();
return st;
}
}
Handling while statement is similar. Here is the outline for parsing a while statement and creating
the data structure for it:
...
create statement node st
if next token is WHILE
{
st-type = IF_STMT; // handling WHILE using if and goto nodes
create if-node // if-node is not a valid identifier see
// corresponding comment above
st-if_stmt = if-node
parse the condition and set if-node-condition_op, if-node-condition_operand1 and if-node-condition_operand2
if-node-true_branch = parse_body();
create a new statement node gt // This is of type StatementNode
gt-type = GOTO_STMT;
create goto-node // This is of type GotoStatement
gt-goto_stmt = goto-node;
goto-node-target = st; // to jump to the if statement after
// executing the body
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append gt to the body of the while // append gt to the body of the while
// this requires a loop. check the comment
// for the if above.
create no-op node
set if-node-false_branch to point to no-op node
set st-next to point to no-op node
}
...
The following diagram shows the desired structure for the while statement:
4.4 Handling switch and for statements
You can handle the switch and for statements similarly, but you should gure that yourself. Use
a combination of IfStatement and GotoStatement to support the semantics of the switch and for
statements. See sections 3.7 and 3.6 for the semantics of the switch and for statements.
5 Executing the intermediate representation
After the graph data structure is built, it needs to be executed. Execution starts with the rst
node in the list. Depending on the type of the node, the next node to execute is determined. The
general form for execution is illustrated in the following pseudo-code.
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pc = first node
while (pc != NULL)
{
switch (pc-type)
{
case ASSIGN_STMT: // code to execute pc-assign_stmt ...
pc = pc-next
case IF_STMT: // code to evaluate condition ...
// depending on the result
// pc = pc-if_stmt-true_branch
// or
// pc = pc-if_stmt-false_branch
case NOOP_STMT: pc = pc-next
case GOTO_STMT: pc = pc-goto_stmt-target
case PRINT_STMT: // code to execute pc-print_stmt ...
pc = pc-next
}
}
C/C++ implementations If you are developing in C/C++, we have provided you with the data
structures and the code to execute the graph and you must use it. There are two les compiler.h
and compiler.cc, you need to write your code in separate le(s) and #include compiler.h. The
entry point of your code is a function declared in compiler.h:
struct StatementNode* parse_generate_intermediate_representation();
You need to implement this function.
The main() function is provided in compiler.cc:
int main()
{
struct StatementNode * program;
program = parse_generate_intermediate_representation();
execute_program(program);
return 0;
}
It calls the function that you will implement which is supposed to parse the program and
generate the intermediate representation, then it calls the execute program function to execute
the program. You should not modify any of the given code. In fact if you write your program in
C/C++, you should only submit the le(s) that contain your own code and we will add the given
part and compile the code before testing. If you write your program in Java, you should strictly
follow the guidelines for executing the intermediate representation.
JAVA Implementation: If you write your solution in Java, executing the graph should
be done non-recursively and without any function calls. Even helper functions are not
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allowed for the execution of the graph. This is a requirement that will be checked by
inspecting your code. Little credit will be assigned if this requirement is not met.
6 Requirements
1. Write a compiler that generates intermediate representation for the code and write an inter-
preter to execute the intermediate representation. The interpreter is provided for C/C++
implementations.
2. Language: You can use Java, C, or C++ for this assignment.
3. Any language other than Java, C or C++ is not allowed for this project.
4. If you use C/C++ for this project, you should use the provided code and only implement the
required functions.
5. If you use Java, you will need to write everything yourself but the requirements on how to
execute the intermediate representation will be checked manually when grading.
6. Platform: As previous projects, the reference platform is CentOS 6.7
7. You can assume that there are no syntax or semantic errors in the input program.
7 Grading
The test cases provided with the assignment, do not contain any test case for switch and for
statements. However, test cases with switch and for statements will be added for grading the
project. Make sure you test your code extensively with input programs that contain switch and for
statements.
The test cases (there will be multiple test cases in each category, each with equal weight) will
be broken down in the following way (out of 100 points):
Assignment statements: 20
If statements: 25
While statements: 25
Switch and For statements: 20
All statements: 10
Note that all test cases depend on successful implementation of print statements (otherwise
your program cannot generate correct output). Also, assignment statements are needed for if,
while, switch, statement test cases all depend on assignment statements.
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8 Bonus Project
8.1 Bonus Project Options
You have two options for the bonus project:
1. Resubmit project 2. For this option you are given another chance to submit project 2. The
grade you obtain will be reduced by 30% and replaces the grade you already obtained on
project 2. So, if your grade for project 2 was 20 and your grade for the replacement is 90, the
grade for project 2 will be changed from 20 to 63 (63 = 90 reduced by 30%)
2. Do a new project that replaces the lowest grade between project 2 and project 3. The grade
you obtain under this option will replace the lower of the two grades that you obtained on
project 2 or 3 (if the grade you obtain on the bonus is lower than what you already got on
projects 2 and 3, no replacement is made).
If you make multiple submissions, only the last submission will count.
8.2 New Bonus Project (option 2)
Support the following grammar:
program ! var section body
var section ! VAR int var decl array var decl
int var decl ! id list SEMICOLON
array var decl ! id list COLON ARRAY LBRAC NUM RBRAC SEMICOLON
id list ! ID COMMA id list j ID
body ! LBRACE stmt list RBRACE
stmt list ! stmt stmt list j stmt
stmt ! assign stmt j print stmt j while stmt j if stmt j switch stmt
assign stmt ! var access EQUAL expr SEMICOLON
var access ! ID j ID LBRAC expr RBRAC
expr ! term PLUS expr
expr ! term
term ! factor MULT term
term ! factor
factor ! LPAREN expr RPAREN
factor ! NUM
factor ! var access
print stmt ! print var access SEMICOLON
while stmt ! WHILE condition body
if stmt ! IF condition body
condition ! expr relop expr
relop ! GREATER j LESS j NOTEQUAL
switch stmt ! SWITCH var access LBRACE case list RBRACE
switch stmt ! SWITCH var access LBRACE case list default case RBRACE
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case list ! case case list j case
case ! CASE NUM COLON body
default case ! DEFAULT COLON body
Note that LBRAC is "[" and LBRACE is "f". The former is used for arrays and the latter is
used for body. Assume that all arrays are integer arrays and are indexed from 0 to size - 1, where
size is the size of the array speci ed in the var section after the ARRAY keyword and between "["
and "]".
The data structures and code that we have provided for the regular assignment will not be
enough for the bonus, you will need to modify those to support arrays. Submit all code les for
the bonus project (including the modi ed compiler.h and compiler.cc).
All restrictions imposed on the execution of the intermediate representation for the
regular project apply to the bonus project as well. You are not allowed to call any
functions while executing the intermediate representation. You are not allowed to
execute the program recursively.
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1 Introduction
You will write a small compiler that will read an input program and represent it in an internal data structure. The data structure consists of two parts: (1) a representation of instructions to be executed and (2) a representation of the memory of the program (locations for variables).
Instructions are represented by a data structure that includes the operand(s) of the instruction (if any) and that specify the next instruction to be executed. After the data structures are generated by your compiler, your compiler will execute the generated instructions representation by interpreting it. This means that the program will traverse the data structure and at every node it visits, it will execute the node by changing the content of memory locations corresponding to operands and deciding what is the next instruction to execute (program counter). The output of your compiler is the output that the input program should produce. These steps are illustrated in the following gure
1
The remainder of this document is organized as follows:
1. Grammar De nes the programming language syntax including grammar.
2. Execution Semantics Describe statement semantics for if, while, switch and print state-ments.
3. How to generate the intermediate representation Explains step by step how to generate the intermediate representation (data structure). You should read this sequentially and not skip around.
4. Executing the intermediate representation Basically, you have two options. If you are using C/C++, you should use the code we provide to execute the intermediate representation.
If you are using Java, it describes the strict rules to follow in executing the intermediate representation. Those rules will be enforced.
5. Requirements Lists the programming languages you are allowed to use in your solution (C/C++ or Java) and other requirements.
6. Grading Describes the grading scheme.
7. Bonus Project Describes the requirements for the bonus project.
2 Grammar
The grammar for this project is a simpli ed form of the grammar from the previous project, but there are a couple extensions.
program ! var section body
var section ! id list SEMICOLON
id list ! ID COMMA id list j ID
body ! LBRACE stmt list RBRACE
stmt list ! stmt stmt list j stmt
stmt ! assign stmt j print stmt j while stmt j if stmt j switch stmt
assign stmt ! ID EQUAL primary SEMICOLON
assign stmt ! ID EQUAL expr SEMICOLON
expr ! primary op primary
primary ! ID j NUM
op ! PLUS j MINUS j MULT j DIV
print stmt ! print ID SEMICOLON
while stmt ! WHILE condition body
if stmt ! IF condition body
condition ! primary relop primary
relop ! GREATER j LESS j NOTEQUAL
switch stmt ! SWITCH ID LBRACE case list RBRACE
switch stmt ! SWITCH ID LBRACE case list default case RBRACE
for stmt ! FOR LPAREN assign stmt SEMICOLON condition SEMICOLON RPAREN body
case list ! case case list j case
case ! CASE NUM COLON body
default case ! DEFAULT COLON body
Some highlights of the grammar:
1. Expressions are greatly simpli ed and are not recursive.
2. There is no type declaration section.
3. Division is integer division and the result of the division of two integers is an integer.
4. if statement is introduced. Note that if stmt does not have else.
5. for statement is introduced. Note that for has a very general syntax similar to that of the Clanguage for loop.
6. A print statement is introduced. Note that the print keyword is in lower case, but other keywords are all upper-case letters.
7. There is no variable declaration list. There is only one id list in the global scope and that contains all the variables.
8. There is no type speci ed for variables. All variables are int by default.
3 Execution Semantics
All statements in a statement list are executed sequentially according to the order in which they appear. Exception is made for body of if stmt, while stmt and switch stmt as explained below. In what follows, I will assume that all values of variables as well as constants are stored in locations.
This assumption is used by the execution procedure that we provide. This is not a restrictive assumption. For variables, you will have locations associated with them. For constants, you can reserve a location in which you store the constant (this is like having an unnamed immutable variable).
3.1 Assignment Statement
To execute an assignment statement, the expression on the lefthand side of the equal sign is evalu-ated and the result is stored in the location associated with the righthand side of the expression.
3.2 Expression
To evaluate an expression, the values in the locations associated with the two operands are obtained and the expression operator is applied to these values resulting in a value for the expression.
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3.3 Boolean Condition
A boolean condition takes two operands as parameters and returns a boolean value. It is used to
control the execution of while and if statements. To evaluate a condition, the values in the locations
associated with the operands are obtained and the relational operator is applied to these values
resulting in a true or false value. For example, if the values of the two operands a and b are 3 and
4 respectively, a < b evaluates to true.
3.4 If statement
if stmt has the standard semantics:
1. The condition is evaluated.
2. If the condition evaluates to true, the body of the if stmt is executed, then the next statement
following the if in the stmt list is executed.
3. If the condition evaluates to false, the statement following the if in the stmt list is executed
These semantics apply recursively to nested if stmt.
3.5 While statement
while stmt has the standard semantics.
1. The condition is evaluated.
2. If the condition evaluates to true, the body of the while stmt is executed. The next statement
to execute is the while stmt itself.
3. If the condition evaluates to false, the body of the while stmt is not executed. The next
statement to execute is the next statement following the while stmt in the stmt list.
These semantics apply recursively to nested while stmt. The code block:
WHILE condition
f
stmt list
g
is equivalent to:
label : IF condition
f
stmt list
goto label
g
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Note that goto statements do not appear in the input program, but our intermediate represen-
tation includes GotoStatement which is used in conjunction with IfStatement to represent while
and switch statements.
3.6 For statement
The for stmt is very similar to the for statement in the C language. The semantics are de ned by
giving an equivalent construct.
FOR ( assign stmt 1 ; condition ; assign stmt 2 )
f
stmt list
g
is equivalent to:
assign stmt 1
WHILE condition
f
stmt list
assign stmt 2
g
For example, the following snippet of code:
FOR ( a = 0; a < 10; a = a + 1; )
{
print a;
}
is equivalent to:
a = 0;
WHILE a < 10
{
print a;
a = a + 1;
}
3.7 Switch statement
switch stmt has the following semantics1:
1The switch statement in the C language has di erent syntax and semantics. It is also more dangerous!
5
1. The value of the switch variable is checked against each case number in order.
2. If the value matches the number, the body of the case is executed, then the statement following
the switch stmt in the stmt list is executed.
3. If the value does not match the number, next case is evaluated.
4. If a default case is provided and the value does not match any of the case numbers, then the
body of the default case is executed and then the statement following the switch stmt in the
stmt list is executed.
5. If there is no default case and the value does not match any of the case numbers, then the
statement following the switch stmt in the stmt list is executed.
These semantics apply recursively to nested switch stmt. The code block:
SWITCH var f
CASE n1 : f stmt list 1 g
...
CASE nk : f stmt list k g
g
is equivalent to:
IF var == n1 f
stmt list 1
goto label
g
...
IF var == nk f
stmt list k
goto label
g
label :
And for switch statements with default case, the code block:
SWITCH var f
CASE n1 : f stmt list 1 g
...
CASE nk : f stmt list k g
DEFAULT : f stmt list default g
g
is equivalent to:
6
IF var == n1 f
stmt list 1
goto label
g
...
IF var == nk f
stmt list k
goto label
g
stmt list default
label :
3.8 Print statement
The statement
print a;
prints the value of variable a at the time of the execution of the print statement.
4 How to generate the code
The intermediate code will be a data structure (a graph) that is easy to interpret and execute. I
will start by describing how this graph looks for simple assignments then I will explain how to deal
with while statements.
Note that in the explanation below I start with incomplete data structures then I ex-
plain what is missing and make them more complete. You should read the whole explanation.
4.1 Handling simple assignments
A simple assignment is fully determined by: the operator (if any), the id on the left-hand side, and
the operand(s). A simple assignment can be represented as a node:
struct AssignmentStatement {
struct ValueNode* left_hand_side;
struct ValueNode* operand1;
struct ValueNode* operand2;
ArithmeticOperatorType op; // operator
}
For assignment without an expression on the right-hand side, the operator is set to OPERATOR NONE
and there is only one operand. To execute an assignment, you need the values of the operand(s),
apply the operator, if any, to the operands and assign the resulting value of the right-hand side
to the left hand side. For literals (NUM), the value is the value of the number. For variables, the
value is the last value stored in the location associated with the variable. Initially, all variables
are initialized to 0. In this representation, the locations associated with variables as well as the
7
locations in which constants are stored are value nodes ( struct valueNode), a node that contains
a value. The assignment statement contains pointers to the value nodes of the left-hand side and
the operand(s).
Multiple assignments are executed one after another. So, we need to allow multiple assignment
nodes to be linked to each other. This can be achieved as follows:
struct AssignmentStatement {
struct ValueNode* left_hand_side;
struct ValueNode* operand1;
struct ValueNode* operand2;
ArithmeticOperatorType op; // operator
struct AssignmentStatement* next;
}
This structure only accepts ValueNode as operands. To handle literal constants (NUM), you need
to create ValueNode for them and initialize the value stored in these value nodes to the constant
value. This initialization, as well as the initialization of values in locations associated with variable
is done while parsing.
This will now allow us to execute a sequence of assignment statements represented in a linked-
list: we start with the head of the list, then we execute every assignment in the list one after the
other.
Begin Note It is important to distinguish between compile-time initialization and runtime
execution. For example, consider the program
a b;
{
a = 3;
b = 5;
}
The intermediate representation for this program will have have two assignment statements:
one to copy the value in the location that contains the value 3 to the location associated with a
and one to copy the value in the location that contains the value 5 to the location associated with
b. The values 3 and 5 will not be copied to the locations of a and b at compile-time. End Note
This is simple enough, but does not help with executing other kinds of statements. We consider
them one at a time.
4.2 Handling print statements
The print statement is straightforward. It can be represented as
struct PrintStatement
{
struct ValueNode* id;
}
Now, we ask: how can we execute a sequence of statements that are either assignment or print
statement (or other types of statements)? We need to put both kinds of statements in a list and not
8
just the assignment statements as we did above. So, we introduce a new kind of node: a statement
node. The statement node has a eld that indicates which type of statement it is. It also has elds
to accommodate the remaining types of statements. It looks like this:
struct StatementNode {
StatementType type; // NOOP_STMT, GOTO_STMT, ASSIGN_STMT, IF_STMT, PRINT_STMT
union {
struct AssignmentStatement* assign_stmt;
struct PrintStatement* print_stmt;
struct IfStatement* if_stmt;
struct GotoStatement* goto_stmt;
};
struct StatementNode* next;
}
This way we can go through a list of statements and execute one after the other. To execute
a particular node, we check its type. If it is PRINT STMT, we execute the print stmt eld, if it is
ASSIGN STMT, we execute the assign stmt eld and so on. With this modi cation, we do not need
a next eld in the AssignmentStatement structure (as we explained above), instead, we put the
next eld in the statement node.
This is all ne, but we do not yet know how to generate the list to execute later. The idea is to
have the functions that parses non-terminals return the code for the non-terminals. For example
for a statement list, we have the following pseudecode (missing many checks):
struct StatementNode* parse_stmt_list()
{
struct StatementNode* st; // statement
struct StatementNode* stl; // statement list
st = parse_stmt();
if (nextToken == start of a statement list)
{
stl = parse_stmt_list();
append stl to st; // this is pseudecode
return st;
}
else
{
ungetToken();
return st;
}
}
9
And to parse body we have the following pseudecode:
struct StatementNode* parse_body()
{
struct StatementNode* stl;
match LBRACE
stl = parse_stmt_list();
match RBRACE
return stl;
}
4.3 Handling if and while statements
More complications occur with if and while statements. The structure for an if statement can be
as follows:
struct IfStatement {
ConditionalOperatorType condition_op;
struct ValueNode* condition_operand1;
struct ValueNode* condition_operand2;
struct StatementNode* true_branch;
struct StatementNode* false_branch;
}
The condition op, condition operand1 and condition operand2 elds are the operator and
operands of the condition of the if statement. To generate the node for an if statement, we need to
put together the condition, and stmt list that are generated in the parsing of the if statement.
The true branch and false branch elds are crucial to the execution of the if statement. If
the condition evaluates to true then the statement speci ed in true branch is executed otherwise
the one speci ed in false branch is executed. We need one more type of node to allow loop back
for while statements. This is a GotoStatement.
struct GotoStatement {
struct StatementNode* target;
}
To generate code for the while and if statements, we need to put a few things together. The outline
given above for stmt list needs to be modi ed as follows (this is missing details and shows only the
main steps).
10
struct StatementNode* parse_stmt()
{
...
create statement node st
if next token is IF
{
st-type = IF_STMT;
create if-node; // note that if-node is pseudecode and is not
// a valid identifier in C, C++ or Java
st-if_stmt = if-node;
parse the condition and set if-node-condition_op, if-node-condition_operand1 and if-node-condition_operand2
if-node-true_branch = parse_body(); // parse_body returns a pointer to a list of statements
create no-op node // this is a node that does not result
// in any action being taken
append no-op node to the body of the if // this requires a loop to get to the end of
// if-node-true_branch by following the next field
// you know you reached the end when next is NULL
// it is very important that you always appropriately
// initialize fields of any data structures
// do not use uninitialized pointers
set if-node-false_branch to point to no-op node
set st-next to point to no-op node
...
} else ...
}
The following diagram shows the desired structure for the if statement:
11
The stmt list code should be modi ed because of the extra no-op node:
struct StatementNode* parse_stmt_list()
{
struct StatementNode* st; // statement
struct StatementNode* stl; // statement list
st = parse_stmt();
if (nextToken == start of a statement list)
{
stl = parse_stmt_list();
if st-type == IF_STMT
{
append stl to the no-op node that follows st
// st
// |
// V
// no-op
// |
// V
// stl
}
else
{
append stl to st;
// st
// |
// V
// stl
}
return st;
}
else
{
ungetToken();
return st;
}
}
Handling while statement is similar. Here is the outline for parsing a while statement and creating
the data structure for it:
...
create statement node st
if next token is WHILE
{
st-type = IF_STMT; // handling WHILE using if and goto nodes
create if-node // if-node is not a valid identifier see
// corresponding comment above
st-if_stmt = if-node
parse the condition and set if-node-condition_op, if-node-condition_operand1 and if-node-condition_operand2
if-node-true_branch = parse_body();
create a new statement node gt // This is of type StatementNode
gt-type = GOTO_STMT;
create goto-node // This is of type GotoStatement
gt-goto_stmt = goto-node;
goto-node-target = st; // to jump to the if statement after
// executing the body
12
append gt to the body of the while // append gt to the body of the while
// this requires a loop. check the comment
// for the if above.
create no-op node
set if-node-false_branch to point to no-op node
set st-next to point to no-op node
}
...
The following diagram shows the desired structure for the while statement:
4.4 Handling switch and for statements
You can handle the switch and for statements similarly, but you should gure that yourself. Use
a combination of IfStatement and GotoStatement to support the semantics of the switch and for
statements. See sections 3.7 and 3.6 for the semantics of the switch and for statements.
5 Executing the intermediate representation
After the graph data structure is built, it needs to be executed. Execution starts with the rst
node in the list. Depending on the type of the node, the next node to execute is determined. The
general form for execution is illustrated in the following pseudo-code.
13
pc = first node
while (pc != NULL)
{
switch (pc-type)
{
case ASSIGN_STMT: // code to execute pc-assign_stmt ...
pc = pc-next
case IF_STMT: // code to evaluate condition ...
// depending on the result
// pc = pc-if_stmt-true_branch
// or
// pc = pc-if_stmt-false_branch
case NOOP_STMT: pc = pc-next
case GOTO_STMT: pc = pc-goto_stmt-target
case PRINT_STMT: // code to execute pc-print_stmt ...
pc = pc-next
}
}
C/C++ implementations If you are developing in C/C++, we have provided you with the data
structures and the code to execute the graph and you must use it. There are two les compiler.h
and compiler.cc, you need to write your code in separate le(s) and #include compiler.h. The
entry point of your code is a function declared in compiler.h:
struct StatementNode* parse_generate_intermediate_representation();
You need to implement this function.
The main() function is provided in compiler.cc:
int main()
{
struct StatementNode * program;
program = parse_generate_intermediate_representation();
execute_program(program);
return 0;
}
It calls the function that you will implement which is supposed to parse the program and
generate the intermediate representation, then it calls the execute program function to execute
the program. You should not modify any of the given code. In fact if you write your program in
C/C++, you should only submit the le(s) that contain your own code and we will add the given
part and compile the code before testing. If you write your program in Java, you should strictly
follow the guidelines for executing the intermediate representation.
JAVA Implementation: If you write your solution in Java, executing the graph should
be done non-recursively and without any function calls. Even helper functions are not
14
allowed for the execution of the graph. This is a requirement that will be checked by
inspecting your code. Little credit will be assigned if this requirement is not met.
6 Requirements
1. Write a compiler that generates intermediate representation for the code and write an inter-
preter to execute the intermediate representation. The interpreter is provided for C/C++
implementations.
2. Language: You can use Java, C, or C++ for this assignment.
3. Any language other than Java, C or C++ is not allowed for this project.
4. If you use C/C++ for this project, you should use the provided code and only implement the
required functions.
5. If you use Java, you will need to write everything yourself but the requirements on how to
execute the intermediate representation will be checked manually when grading.
6. Platform: As previous projects, the reference platform is CentOS 6.7
7. You can assume that there are no syntax or semantic errors in the input program.
7 Grading
The test cases provided with the assignment, do not contain any test case for switch and for
statements. However, test cases with switch and for statements will be added for grading the
project. Make sure you test your code extensively with input programs that contain switch and for
statements.
The test cases (there will be multiple test cases in each category, each with equal weight) will
be broken down in the following way (out of 100 points):
Assignment statements: 20
If statements: 25
While statements: 25
Switch and For statements: 20
All statements: 10
Note that all test cases depend on successful implementation of print statements (otherwise
your program cannot generate correct output). Also, assignment statements are needed for if,
while, switch, statement test cases all depend on assignment statements.
15
8 Bonus Project
8.1 Bonus Project Options
You have two options for the bonus project:
1. Resubmit project 2. For this option you are given another chance to submit project 2. The
grade you obtain will be reduced by 30% and replaces the grade you already obtained on
project 2. So, if your grade for project 2 was 20 and your grade for the replacement is 90, the
grade for project 2 will be changed from 20 to 63 (63 = 90 reduced by 30%)
2. Do a new project that replaces the lowest grade between project 2 and project 3. The grade
you obtain under this option will replace the lower of the two grades that you obtained on
project 2 or 3 (if the grade you obtain on the bonus is lower than what you already got on
projects 2 and 3, no replacement is made).
If you make multiple submissions, only the last submission will count.
8.2 New Bonus Project (option 2)
Support the following grammar:
program ! var section body
var section ! VAR int var decl array var decl
int var decl ! id list SEMICOLON
array var decl ! id list COLON ARRAY LBRAC NUM RBRAC SEMICOLON
id list ! ID COMMA id list j ID
body ! LBRACE stmt list RBRACE
stmt list ! stmt stmt list j stmt
stmt ! assign stmt j print stmt j while stmt j if stmt j switch stmt
assign stmt ! var access EQUAL expr SEMICOLON
var access ! ID j ID LBRAC expr RBRAC
expr ! term PLUS expr
expr ! term
term ! factor MULT term
term ! factor
factor ! LPAREN expr RPAREN
factor ! NUM
factor ! var access
print stmt ! print var access SEMICOLON
while stmt ! WHILE condition body
if stmt ! IF condition body
condition ! expr relop expr
relop ! GREATER j LESS j NOTEQUAL
switch stmt ! SWITCH var access LBRACE case list RBRACE
switch stmt ! SWITCH var access LBRACE case list default case RBRACE
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case list ! case case list j case
case ! CASE NUM COLON body
default case ! DEFAULT COLON body
Note that LBRAC is "[" and LBRACE is "f". The former is used for arrays and the latter is
used for body. Assume that all arrays are integer arrays and are indexed from 0 to size - 1, where
size is the size of the array speci ed in the var section after the ARRAY keyword and between "["
and "]".
The data structures and code that we have provided for the regular assignment will not be
enough for the bonus, you will need to modify those to support arrays. Submit all code les for
the bonus project (including the modi ed compiler.h and compiler.cc).
All restrictions imposed on the execution of the intermediate representation for the
regular project apply to the bonus project as well. You are not allowed to call any
functions while executing the intermediate representation. You are not allowed to
execute the program recursively.
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