CSE340 Project 3: Type Checking Solved

Your goal is to finish a predictive parser and write a type checker for a given language. The input to your project will
be a program and the output will be either a) error messages if there is a type mismatch or syntax error or b) lists of
symbols with equivalent types if there is no error. Your type checker will enforce semantic checks on the input
program, and will be described in the following. First we specify the grammar of our language.
1. Grammar Description
program → decl body
decl → type_decl_section var_decl_section
type_decl_section → TYPE type_decl_list
type_decl_section → ε
type_decl_list → type_decl type_decl_list
type_decl_list → type_decl
type_decl → id_list COLON type_name SEMICOLON
type_name → REAL
type_name → INT
type_name → BOOLEAN
type_name → STRING
type_name → LONG
type_name → ID
var_decl_section → VAR var_decl_list
var_decl_section → ε
var_decl_list → var_decl var_decl_list
var_decl_list → var_decl
var_decl → id_list COLON type_name SEMICOLON
id_list → ID COMMA id_list
id_list → ID
body → LBRACE stmt_list RBRACE
stmt_list → stmt stmt_list
stmt_list → stmt
stmt → assign_stmt
stmt → while_stmt
stmt → do_stmt
stmt → switch_stmt
assign_stmt → ID EQUAL expr SEMICOLON
while_stmt → WHILE condition body
do_stmt → DO body WHILE condition SEMICOLON
switch_stmt → SWITCH ID LBRACE case_list RBRACE
case_list → case case_list
case_list → case
case → CASE NUM COLON body
expr → term PLUS expr
expr → term
term → factor MULT term
term → factor DIV term
term → factor
factor → LPAREN expr RPAREN
factor → NUM
factor → REALNUM
factor → ID
condition → ID
condition → primary relop primary
primary → ID
primary → NUM
primary → REALNUM
relop → GREATER
relop → GTEQ
relop → LESS
relop → NOTEQUAL
relop → LTEQ
The tokens used in the grammar description are:
TYPE = TYPE
VAR = VAR
REAL = REAL
INT = INT
BOOLEAN = BOOLEAN
STRING = STRING
LONG = LONG
WHILE = WHILE
DO = DO
SWITCH = SWITCH
CASE = CASE
COMMA = ,
COLON = :
SEMICOLON = ;
LBRACE = {
RBRACE = }
LPAREN = (
RPAREN = )
EQUAL = =
PLUS = +
MULT = *
DIV = /
GREATER =
GTEQ = =
LESS = <
LTEQ = <=
NOTEQUAL = <
ID = letter (letter + digit)*
NUM = 0 + (pdigit digit*)
REALNUM = NUM \. digit digit*
2. Language Semantics
As can be seen from the grammar, in this language types are first declared, then variables are declared, then the
body of the program follows.
2.1. Types
The language has five built-in types: INT , REAL , BOOLEAN , STRING , and LONG .
Programmers can declare types either explicitly or implicitly.
Explicit types are names that are not built-in types and that have their first appearance in the program as part
of the id_list of a type_decl .
Implicit types are not built-in types and not explicit programmer-declared types. Implicit types have their first
appearance as a type_name in a var_decl or a type_decl .
Example
Consider the following program written in our language:
TYPE
a : INT;
b : a;
VAR
x : b;
y : c;
{
y = x;
}
There are three types declared by the programmer in this example, a , b , and c , where a and b are explicit
types and c is an implicit type.
2.2. Variables
Programmers can declare variables either explicitly or implicitly.
Explicit variables are declared in an id_list of a var_decl .
A variable is declared implicitly if it is not declared explicitly but it appears in the program body.
Example
Consider the following program written in our language:
TYPE
a : INT;
b : a;
VAR
x : b;
y : c;
{
y = x;
z = 10;
w = z * 5;
}
This program has four variables declared: x , y , z , and w , with x and y explicitly declared and z and w
implicitly declared. Note that the implicitly declared variables z and w also have an implicitly declared type.
2.3. Declaration vs. Use
Any appearance of a name (type or variable) in the program is either a declaration or a use.
The following lists all possible declarations of a name:
1. Any appearance of a name in the id_list part of a type_decl
2. Any appearance of a name in the id_list part of a var_decl
3. The first appearance of a name in the entire program, if the name appears as type_name in a type_decl
4. The first appearance of a name in the entire program, if the name appears as type_name in a var_decl
5. The first appearance of a name in the entire program, if the name appears inside the body of the program
Any other appearance of a name is considered a use of that name.
Note that the above definitions exclude the built-in type names.
Given the following example (the line numbers are not part of the input):
01 TYPE
02 a : INT;
03 b : a;
04 VAR
05 x : b;
06 y : c;
07 {
08 y = x;
09 z = 10;
10 w = z * 5;
11 }
We can categorize all appearances of names as declaration or use:
Line 2, the appearance of name a is a declaration
Line 3, the appearance of name b is a declaration
Line 3, the appearance of name a is a use
Line 5, the appearance of name x is a declaration
Line 5, the appearance of name b is a use
Line 6, the appearance of name y is a declaration
Line 6, the appearance of name c is a declaration
Line 8, the appearance of name y is a use
Line 8, the appearance of name x is a use
Line 9, the appearance of name z is a declaration
Line 10, the appearance of name w is a declaration
Line 10, the appearance of name z is a use
2.4. Type System
Our language uses structural equivalence for checking type equivalence.
Implicit types (in variable declarations or on implicitly declared variables) will be inferred from the usage (in a
simplified form of Hindley-Milner type inference).
Here are all the type rules/constraints that your type checker will enforce (constraints are labeled from C1 to C5 for
reference):
C1: The left hand side of an assignment should have the same type as the right hand side of that assignment
C2: The operands of an operation ( PLUS , MINUS , MULT , and DIV ) should have the same type (it can be any
type, including STRING and BOOLEAN )
C3: The operands of a relational operator (see relop in grammar) should have the same type (it can be any
type, including STRING and BOOLEAN )
C4: condition should be of type BOOLEAN
C5: The variable that follows the SWITCH keyword in switch_stmt should be of type INT
The type of an expr is the same as the type of its operands
The result of p1 relop p2 is of type BOOLEAN (assuming that p1 and p2 have the same type)
NUM constants are of type INT
REALNUM constants are of type REAL
If two types cannot be determined to be the same according to the above rules, the two types are different
3. Incomplete Parser
The provided parser is incomplete, as it is missing an implementation for some of the non-terminals. You must
finish the given parser so that it can parse any valid input according to our grammar. If you detect a syntax error in
the input, you should output the following message and exit:
Syntax Error
You can start coding by finishing the parser first and then move on to implementing the type checking part. You
should make sure that your parser generates a syntax error message if the input program does not follow the
proper syntax.
We recommend that you check your code on the submission website to make sure it passes all the test cases in the
parsing category before moving on to implementing the type checking part.
Our grammar is not LL(1) i.e. it does not satisfy the conditions for predictive parser, however, it is still possible to
write a predictive parser by looking at more than one token. A notable case is when parsing condition .
4. Output
Your program will check for the following semantic errors and output the correct message when it encounters that
error. Note that there will only be at most one error per test case.
4.1. Duplication Errors
1. Errors involving programmer-defined types:
Programmer-defined type declared more than once:
Explicit type redeclared explicitly (error code 1.1)
An explicitly declared type can be declared again explicitly by appearing as part of an id_list in a type
declaration.
Implicit type redeclared explicitly (error code 1.2)
An implicitly declared type can be declared again explicitly by appearing as part of an id_list in a type
declaration.
Note that a previously declared type name (either implicit or explicit) cannot be declared again implicitly. Since
it has already been introduced, the new reference to the name (as type_name in a type_decl or var_decl )
would be a use and not a declaration.
Programmer-defined type redeclared as variable (error code 1.3)
If a previously declared type appears again in an id_list of a variable declaration, the type is redeclared as a
variable.
Programmer-defined type used as variable (error code 1.4)
If a previously declared type appears in the body of the program, the type is used as a variable.
2. Errors involving variable declarations:
Variable declared more than once (error code 2.1)
An explicitly declared variable can be declared again explicitly by appearing as part of an id_list in a
variable declaration.
Variable used as a type (error code 2.2)
If an explicitly declared variable is used as type_name in a variable declaration, the variable is used as a type.
Note that an explicitly declared variable cannot be declared again implicitly, appearances of the name in the
program body are uses. In the same way, an implicitly declared variable cannot be declared again, because all
later appearances are uses.
Also note that if a built-in type is redeclared or used in the body of the program, it should result in a syntax error.
For these errors, you should output one line in the following format:
ERROR CODE <code <symbol_name
in which <code should be replaced with the proper code (see the error codes listed above) and <symbol_name
should be replaced with the name of the type or variable related to the error.
4.2. Type Mismatch
If any of the type constraints (listed in the Type System section above) is violated in the input program, then the
output of your program should be:
TYPE MISMATCH <line_number <constraint
Where <line_number is replaced with the line number that the violation occurs and <constraint should be
replaced with the label of the violated type constraint (possible values are C1 through C5, see section on Type
System for details of each constraint). Note that you can assume that anywhere a violation can occur it will be on a
single line.
4.3. No Semantic Errors
If there are no semantic errors in the program, then your program should output lists of types and variables that are
type-equivalent. The symbols should be listed in the order they appear in the program and built-in types should be
listed first in the following order: BOOLEAN , INT , LONG , REAL , STRING . Each list must be on a single line of the
output and each symbol in the list should be separated by a single space character. Each list must be terminated by
a # character.
The following pseudo-code should explain the output format more precisely:
for each built-in type T:
{
output T
output all names that are type-equivalent with T in order of their appearance
mark outputted names to avoid re-printing them later
output "#\n"
}i
f there are unprinted names left:
{
for each unprinted name N in order of appearance:
{
output N
output all other names that are type-equivalent with N in order of their appearance
output "#\n"
}
}
The phrase in order of appearance in the above pseudo-code means that names that appear before others in
the program should be printed first. This order should be easy to maintain since it is the natural order of storing
names in your symbol table.
5. Examples
Given the following:
TYPE
a, b, c, b : INT;
VAR
x : a;
{
x = 10;
}
The output will be the following:
ERROR CODE 1.1 b
Given the following:
TYPE
a : INT;
VAR
x : INT;
b, a : STRING;
{
x = 10;
}
The output should be the following:
ERROR CODE 1.3 a
Given the following:
VAR
x1 : INT;
x2, x3, x1 : a;
{
x1 = 0;
}
The output should be the following:
ERROR CODE 2.1 x1
Given the following:
VAR
x, y : STRING;
z : x;
{
y = x;
}
The output should be the following:
ERROR CODE 2.2 x
Given the following:
VAR
x100 : INT;
y : STRING;
{
x100 = y;
}
The output should be the following:
TYPE MISMATCH 5 C1
Given the following:
VAR
x : INT;
{
x = 100;
y = 20.10;
y = x;
}
The output should be the following:
TYPE MISMATCH 6 C1
Given the following:
VAR
x, y : a1;
{
WHILE x < 10
{
x = x + y;
y = y * 1.0;
}
}
The output should be the following:
TYPE MISMATCH 7 C2
Given the following:
TYPE
a, b : INT;
c : a;
d : STRING;
VAR
x : e;
y : c;
test : d;
{
a1 = 100;
b1 = a1 + (10 * 50);
foo = b1 / 50;
SWITCH foo
{
CASE 1:
{
foo = 0;
}
CASE 2:
{
test = test * test;
}
}
h = x;
}
The output should be the following:
BOOLEAN #
INT a b c y a1 b1 foo #
LONG #
REAL #
STRING d test #
x e h #
6. Evaluation
Your submission will be graded on passing the automated test cases.
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 105 points):
Parsing: 37 points
Errors involving programmer-defined types (error codes 1.x): 18 points
Errors involving variable declarations (error codes 2.x): 10 points
Type mismatch errors and no semantic error cases: 40 points
The parsing category is not partially graded, you need to pass all test cases in that category to get the 37 points. All
other categories are partially graded.