Starting from:
$24.99
CS2200 Extra Credit Project: Pipeline Solution
1 Why Pipelining?
The datapath design that we implemented for Project 1 was, in fact, grossly inefficient. By focusing on increasing throughput, a pipelined processor can get more instructions done per clock cycle. In the real world, that means higher performance, lower power draw, and most importantly, happy customers!
2 Project Requirements
In this extra credit project, you will make a pipelined processor that implements the LC-22 ISA. There will be five stages in your pipeline:
1. IF - Instruction Fetch
2. ID/RR - Instruction Decode/Register Read
3. EX - Execute (ALU operations)
4. MEM - Memory (both reads and writes with memory)
5. WB - Writeback (writing to registers)
Before you move on, read Appendix A: LC-22 Instruction Set Architecture to understand the ISA that you will be implementing. Understanding the instructions supported by your ISA will make designing your pipeline much easier. We provide you with a CircuitSim file with the some of the structure laid out.
3 Building the Pipeline
First, you will have to build the hardware to support all of your instructions. You will have to make each stage such that it can accommodate the actions of all instructions passing through it. Use the book (Ch. 5) to get an idea of what the pipeline looks like and to understand the function of each stage before you start building your circuits.
1. IF Stage
The IF stage is responsible for:
• Getting the instruction from I-MEM at location PC
• Updating the PC
2. ID/RR Stage
The ID/RR stage is responsible for:
• Decoding the instruction
1
• Reading the appropriate registers
Please look at Appendix A: LC-22 Instruction Set Architecture in order to understand the instruction formats! You will be provided a dual ported register file (DPRF), which allows you to read from two registers and write one register all at the same time.
3. EX Stage
The EX stage is responsible for:
• Performing all necessary arithmetic and logic calculations
• Resolving any branch or JALR instructions
In the Execute (EX) stage, you will perform any arithmetic computations required by the instruction. This stage should host a complete ALU to perform the actual adding or NANDing as required by the instruction. For memory access instructions, this stage will perform the Base + Offset computation required to determine the memory address to access.
4. MEM Stage
The MEM stage is responsible for:
• Reading from or writing a result to memory
All you need to do is to use the value calculated in the EX stage as the address for the RAM. Note that you must use the maximum address length for the RAM block - this is 16 bits. To accomplish this, simply take the lower 16 bits of the calculated address. Depending on the instruction, this stage will need to pass either the value read from memory or the value computed in EX to the WB stage.
5. WB Stage
The WB stage is responsible for:
• Writing results back to the DPRF (dual-ported register file)
4 General Advice
Subcircuits
For this project, we highly encourage using modular design and creating subcircuits when necessary. We strongly recommend using subcircuits when building your pipeline buffers, stages, and forwarding unit. A modular design will make it easier to debug and test your circuit during development.
Pipeline Buffers
For deciding what to pass through buffers, remember that we need to support the requirements of every possible instruction. Think of what each instruction needs to fulfill its duty, and pass a union of all those requirements. (By union we mean the mathematical union, for example say I1 needs PC and Rx, while I2 needs Rx and Ry, then you should pass PC, Rx and Ry through the buffer). You can also feel free to implement your hardware such that you re-use space in the buffer for different purposes depending on the instruction, but this is not required.
Control Signals
In the Project 1 datapath, recall that we had one main ROM that was the single source of all the control signals on the datapath. Now that we are spreading out our work across different stages of the pipeline, you have a choice of how to implement your signals!
The first thing to note is that in a pipelined processor, each stage is like a simple one-cycle processor that can do exactly ONE thing intended for that stage in a single cycle. In this sense, there is really no need for a control ROM anymore! Therefore in real processors, each stage of the pipeline is implemented using hardwired control as discussed in Chapter 3 of the textbook. However, to keep your design simple for debugging and getting it working, we are going to suggest using a control ROM to generate the needed control signals for the different independent stages of the pipelined processor.
There are two options:
1. You can either have a single large main ROM in ID/RR which calculates all the control signals for every stage.
OR
2. you can have a small(er) ROM in each stage which takes in the opcode and assert the proper signals for that operation.
Note that if you choose the first method, you will need to pass all the signals needed for later stages through the earlier stages, and in the second method, you will need to pass the instruction opcode though all the stages so that you know which signals to assert during that stage.
Stalling and Data Forwarding
One must stall the pipeline when an instruction cannot proceed to the next stage because a value is not yet available to an instruction. This usually happens because of a data hazard. For example, consider two instructions in the following program:
1. LW $t0, 5($t1)
2. ADDI $t0, $t0, 1
To stall the pipeline, the stages preceding the stalled stage should disable writes into their buffers, i.e. they should continue to output the previous value into the next stage. The stalled stage itself will output NOOP (example, ADD $zero, $zero, $zero) instructions down the pipeline until the cause of the stall finishes.
It is recommended that you make a forwarding unit that implements various stock rules. The forwarding unit should take in the two register values you are reading, the output value from the EX stage, the output value from the MEM stage, and the output value from the WB stage. To forward a value from a future stage back to ID/RR, you must check to see if the destination register number from a particular stage is equal to your source register numbers in the ID/RR stage. If so, you must forward the value from that stage to your ID/RR stage.
Note, forwarding cannot save you from one situation: when the destination register of a LW instruction is the source register of an instruction immediately after it. In this case, sometimes called “load-to-use”, you must stall the instruction in the ID/RR stage. It is your job to flesh out all of the stall and forwarding rules.
Keep in mind: the zero register can never change, therefore it should not be considered for forwarding and stalling situations.
Branch Prediction
For this project we will be predicting the branch is not taken, and so the pipeline will continue fetching sequentially. Upon resolving the branch, the pipeline should continue normally in the case of a correct prediction, or flush the incorrectly fetched instructions in the case of an incorrect prediction.
Flushing the Pipeline
5 Testing
When you have constructed your pipeline, you should test it instruction by instruction to see if you have all the necessary components to ensure proper execution.
Be careful to only use the instructions listed in the appendix - there are some subtle points in having a separate instruction and data memory. Load the assembled program into both the instruction memory and the data memory and let your processor execute it. Any writes to memory will only affect the data memory.
6 Grading
One point will be given for each of the following:
• Your pipeline produces the correct output when running our test program.
• All stages of the pipeline have been implemented with all required components and connections visible.
We will not accept regrades for the extra credit project.
7 Deliverables
Please submit all of the required files in a .tar.gz archive generated by one of the following:
The generated archive should contain at a minimum the following files:
• LC22-Pipeline.sim
Always re-download your assignment from Canvas after submitting to ensure that all necessary files were properly uploaded. If what we download does not work, you will get a 0 regardless of what is on your machine.
8 Appendix A: LC-22 Instruction Set Architecture
The LC-22 is a simple, yet capable computer architecture. The LC-22 combines attributes of both ARM and the LC-2200 ISA defined in the Ramachandran & Leahy textbook for CS 2200.
The LC-22 is a word-addressable, 32-bit computer. All addresses refer to words, i.e. the first word (four bytes) in memory occupies address 0x0, the second word, 0x1, etc.
All memory addresses are truncated to 16 bits on access, discarding the 16 most significant bits if the address was stored in a 32-bit register. This provides 256 KB of addressable memory.
8.1 Registers
The LC-22 has 16 general-purpose registers. While there are no hardware-enforced restraints on the uses of these registers, your code is expected to follow the conventions outlined below.
Table 1: Registers and their Uses
Register Number Name Use Callee Save?
0 $zero Always Zero NA
1 $at Assembler/Target Address NA
2 $v0 Return Value No
3 $a0 Argument 1 No
4 $a1 Argument 2 No
5 $a2 Argument 3 No
6 $t0 Temporary Variable No
7 $t1 Temporary Variable No
8 $t2 Temporary Variable No
9 $s0 Saved Register Yes
10 $s1 Saved Register Yes
11 $s2 Saved Register Yes
12 $k0 Reserved for OS and Traps NA
13 $sp Stack Pointer No
14 $fp Frame Pointer Yes
15 $ra Return Address No
1. Register 0 is always read as zero. Any values written to it are discarded. Note: for the purposes of this project, you must implement the zero register. Regardless of what is written to this register, it should always output zero.
3. Register 2 is where you should store any returned value from a subroutine call.
4. Registers 3 - 5 are used to store function/subroutine arguments. Note: registers 2 through 8 should be placed on the stack if the caller wants to retain those values. These registers are fair game for the callee (subroutine) to trash.
5. Registers 6 - 8 are designated for temporary variables. The caller must save these registers if they want these values to be retained.
7. Register 12 is reserved for handling interrupts. While it should be implemented, it otherwise will not have any special use on this assignment.
8. Register 13 is the everchanging top of the stack; it keeps track of the top of the activation record for a subroutine.
9. Register 14 is the anchor point of the activation frame. It is used to point to the first address on the activation record for the currently executing process.
10. Register 15 is used to store the address a subroutine should return to when it is finished executing.
8.2 Instruction Overview
The LC-22 supports a variety of instruction forms, only a few of which we will use for this project. The instructions we will implement in this project are summarized below.
Table 2: LC-22 Instruction Set
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0000 DR SR1 unused SR2
0001 DR SR1 unused SR2
0010 DR SR1 immval20
0011 DR BaseR offset20
0100 SR BaseR offset20
0101 unused offset20
0110 RA AT unused
0111 unused
1000 SR1 SR2 offset20
1001 SR1 SR2 offset20
1010 DR unused PCoffset20
ADD
NAND
ADDI
LW
SW
BR
JALR
HALT
BLT
BGT
LEA
8.2.1 Conditional Branching
Branching in the LC-22 ISA is a bit different than usual. We have a set of branching instructions including BR, an unconditional branch, as well as BLT and BGT, which offer the ability to branch upon a certain condition being met. The BLT and BGT instructions use comparison operators, comparing the values of two source registers. If the comparisons are true (for example, with the BGT instruction, if SR1 > SR2), then we will branch to the target destination of incrementedPC + offset20.
8.3 Detailed Instruction Reference
8.3.1 ADD
Assembler Syntax
ADD DR, SR1, SR2
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0000 DR SR1 unused SR2
Operation
DR = SR1 + SR2;
Description
The ADD instruction obtains the first source operand from the SR1 register. The second source operand is obtained from the SR2 register. The second operand is added to the first source operand, and the result is stored in DR.
8.3.2 NAND
Assembler Syntax
NAND DR, SR1, SR2
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0001 DR SR1 unused SR2
Operation
DR = ~(SR1 & SR2);
Description
The NAND instruction performs a logical NAND (AND NOT) on the source operands obtained from SR1 and SR2. The result is stored in DR.
HINT: A logical NOT can be achieved by performing a NAND with both source operands the same.
For instance,
NAND DR, SR1, SR1
...achieves the following logical operation: DR←SR1.
8.3.3 ADDI
Assembler Syntax
ADDI DR, SR1, immval20
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0010 DR SR1 immval20
Operation
DR = SR1 + SEXT(immval20);
Description
The ADDI instruction obtains the first source operand from the SR1 register. The second source operand is obtained by sign-extending the immval20 field to 32 bits. The resulting operand is added to the first source operand, and the result is stored in DR.
8.3.4 LW
Assembler Syntax
LW DR, offset20(BaseR)
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0011 DR BaseR offset20
Operation
DR = MEM[BaseR + SEXT(offset20)];
Description
An address is computed by sign-extending bits [19:0] to 32 bits and then adding this result to the contents of the register specified by bits [23:20]. The 32-bit word at this address is loaded into DR.
8.3.5 SW
Assembler Syntax
SW SR, offset20(BaseR)
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0100 SR BaseR offset20
Operation
MEM[BaseR + SEXT(offset20)] = SR;
Description
An address is computed by sign-extending bits [19:0] to 32 bits and then adding this result to the contents of the register specified by bits [23:20]. The 32-bit word obtained from register SR is then stored at this address.
8.3.6 BR
Assembler Syntax
BR offset20
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0101 unused offset20
Operation
PC = incrementedPC + offset20
Description
A branch is unconditionally taken. The PC will be set to the sum of the incremented PC (since we have already undergone fetch) and the sign-extended offset[19:0].
8.3.7 JALR
Assembler Syntax
JALR RA, AT
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0110 RA AT unused
Operation
RA = PC;
PC = AT;
Description
First, the incremented PC (address of the instruction + 1) is stored into register RA. Next, the PC is loaded with the value of register AT, and the computer resumes execution at the new PC.
8.3.8 HALT
Assembler Syntax
HALT
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0111 unused
Description
The machine is brought to a halt and executes no further instructions.
8.3.9 BLT
Assembler Syntax
BLT SR1, SR2, offset20
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1000 SR1 SR2 offset20
Operation
if (SR1 < SR2) {
PC = incrementedPC + offset20
}
Description
A branch is taken if SR1 is less than SR2. If this is the case, the PC will be set to the sum of the incremented PC (since we have already undergone fetch) and the sign-extended offset[19:0].
8.3.10 BGT
Assembler Syntax
BGT SR1, SR2, offset20
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1001 SR1 SR2 offset20
Operation
if (SR1 > SR2) {
PC = incrementedPC + offset20
}
Description
A branch is taken if SR1 is greater than SR2. If this is the case, the PC will be set to the sum of the incremented PC (since we have already undergone fetch) and the sign-extended offset[19:0].
8.3.11 LEA
Assembler Syntax
LEA DR, label
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1010 DR unused PCoffset20
Operation
DR = PC + SEXT(PCoffset20);
Description
An address is computed by sign-extending bits [19:0] to 32 bits and adding this result to the incremented PC (address of instruction + 1). It then stores the computed address into register DR.
The datapath design that we implemented for Project 1 was, in fact, grossly inefficient. By focusing on increasing throughput, a pipelined processor can get more instructions done per clock cycle. In the real world, that means higher performance, lower power draw, and most importantly, happy customers!
2 Project Requirements
In this extra credit project, you will make a pipelined processor that implements the LC-22 ISA. There will be five stages in your pipeline:
1. IF - Instruction Fetch
2. ID/RR - Instruction Decode/Register Read
3. EX - Execute (ALU operations)
4. MEM - Memory (both reads and writes with memory)
5. WB - Writeback (writing to registers)
Before you move on, read Appendix A: LC-22 Instruction Set Architecture to understand the ISA that you will be implementing. Understanding the instructions supported by your ISA will make designing your pipeline much easier. We provide you with a CircuitSim file with the some of the structure laid out.
3 Building the Pipeline
First, you will have to build the hardware to support all of your instructions. You will have to make each stage such that it can accommodate the actions of all instructions passing through it. Use the book (Ch. 5) to get an idea of what the pipeline looks like and to understand the function of each stage before you start building your circuits.
1. IF Stage
The IF stage is responsible for:
• Getting the instruction from I-MEM at location PC
• Updating the PC
2. ID/RR Stage
The ID/RR stage is responsible for:
• Decoding the instruction
1
• Reading the appropriate registers
Please look at Appendix A: LC-22 Instruction Set Architecture in order to understand the instruction formats! You will be provided a dual ported register file (DPRF), which allows you to read from two registers and write one register all at the same time.
3. EX Stage
The EX stage is responsible for:
• Performing all necessary arithmetic and logic calculations
• Resolving any branch or JALR instructions
In the Execute (EX) stage, you will perform any arithmetic computations required by the instruction. This stage should host a complete ALU to perform the actual adding or NANDing as required by the instruction. For memory access instructions, this stage will perform the Base + Offset computation required to determine the memory address to access.
4. MEM Stage
The MEM stage is responsible for:
• Reading from or writing a result to memory
All you need to do is to use the value calculated in the EX stage as the address for the RAM. Note that you must use the maximum address length for the RAM block - this is 16 bits. To accomplish this, simply take the lower 16 bits of the calculated address. Depending on the instruction, this stage will need to pass either the value read from memory or the value computed in EX to the WB stage.
5. WB Stage
The WB stage is responsible for:
• Writing results back to the DPRF (dual-ported register file)
4 General Advice
Subcircuits
For this project, we highly encourage using modular design and creating subcircuits when necessary. We strongly recommend using subcircuits when building your pipeline buffers, stages, and forwarding unit. A modular design will make it easier to debug and test your circuit during development.
Pipeline Buffers
For deciding what to pass through buffers, remember that we need to support the requirements of every possible instruction. Think of what each instruction needs to fulfill its duty, and pass a union of all those requirements. (By union we mean the mathematical union, for example say I1 needs PC and Rx, while I2 needs Rx and Ry, then you should pass PC, Rx and Ry through the buffer). You can also feel free to implement your hardware such that you re-use space in the buffer for different purposes depending on the instruction, but this is not required.
Control Signals
In the Project 1 datapath, recall that we had one main ROM that was the single source of all the control signals on the datapath. Now that we are spreading out our work across different stages of the pipeline, you have a choice of how to implement your signals!
The first thing to note is that in a pipelined processor, each stage is like a simple one-cycle processor that can do exactly ONE thing intended for that stage in a single cycle. In this sense, there is really no need for a control ROM anymore! Therefore in real processors, each stage of the pipeline is implemented using hardwired control as discussed in Chapter 3 of the textbook. However, to keep your design simple for debugging and getting it working, we are going to suggest using a control ROM to generate the needed control signals for the different independent stages of the pipelined processor.
There are two options:
1. You can either have a single large main ROM in ID/RR which calculates all the control signals for every stage.
OR
2. you can have a small(er) ROM in each stage which takes in the opcode and assert the proper signals for that operation.
Note that if you choose the first method, you will need to pass all the signals needed for later stages through the earlier stages, and in the second method, you will need to pass the instruction opcode though all the stages so that you know which signals to assert during that stage.
Stalling and Data Forwarding
One must stall the pipeline when an instruction cannot proceed to the next stage because a value is not yet available to an instruction. This usually happens because of a data hazard. For example, consider two instructions in the following program:
1. LW $t0, 5($t1)
2. ADDI $t0, $t0, 1
To stall the pipeline, the stages preceding the stalled stage should disable writes into their buffers, i.e. they should continue to output the previous value into the next stage. The stalled stage itself will output NOOP (example, ADD $zero, $zero, $zero) instructions down the pipeline until the cause of the stall finishes.
It is recommended that you make a forwarding unit that implements various stock rules. The forwarding unit should take in the two register values you are reading, the output value from the EX stage, the output value from the MEM stage, and the output value from the WB stage. To forward a value from a future stage back to ID/RR, you must check to see if the destination register number from a particular stage is equal to your source register numbers in the ID/RR stage. If so, you must forward the value from that stage to your ID/RR stage.
Note, forwarding cannot save you from one situation: when the destination register of a LW instruction is the source register of an instruction immediately after it. In this case, sometimes called “load-to-use”, you must stall the instruction in the ID/RR stage. It is your job to flesh out all of the stall and forwarding rules.
Keep in mind: the zero register can never change, therefore it should not be considered for forwarding and stalling situations.
Branch Prediction
For this project we will be predicting the branch is not taken, and so the pipeline will continue fetching sequentially. Upon resolving the branch, the pipeline should continue normally in the case of a correct prediction, or flush the incorrectly fetched instructions in the case of an incorrect prediction.
Flushing the Pipeline
5 Testing
When you have constructed your pipeline, you should test it instruction by instruction to see if you have all the necessary components to ensure proper execution.
Be careful to only use the instructions listed in the appendix - there are some subtle points in having a separate instruction and data memory. Load the assembled program into both the instruction memory and the data memory and let your processor execute it. Any writes to memory will only affect the data memory.
6 Grading
One point will be given for each of the following:
• Your pipeline produces the correct output when running our test program.
• All stages of the pipeline have been implemented with all required components and connections visible.
We will not accept regrades for the extra credit project.
7 Deliverables
Please submit all of the required files in a .tar.gz archive generated by one of the following:
The generated archive should contain at a minimum the following files:
• LC22-Pipeline.sim
Always re-download your assignment from Canvas after submitting to ensure that all necessary files were properly uploaded. If what we download does not work, you will get a 0 regardless of what is on your machine.
8 Appendix A: LC-22 Instruction Set Architecture
The LC-22 is a simple, yet capable computer architecture. The LC-22 combines attributes of both ARM and the LC-2200 ISA defined in the Ramachandran & Leahy textbook for CS 2200.
The LC-22 is a word-addressable, 32-bit computer. All addresses refer to words, i.e. the first word (four bytes) in memory occupies address 0x0, the second word, 0x1, etc.
All memory addresses are truncated to 16 bits on access, discarding the 16 most significant bits if the address was stored in a 32-bit register. This provides 256 KB of addressable memory.
8.1 Registers
The LC-22 has 16 general-purpose registers. While there are no hardware-enforced restraints on the uses of these registers, your code is expected to follow the conventions outlined below.
Table 1: Registers and their Uses
Register Number Name Use Callee Save?
0 $zero Always Zero NA
1 $at Assembler/Target Address NA
2 $v0 Return Value No
3 $a0 Argument 1 No
4 $a1 Argument 2 No
5 $a2 Argument 3 No
6 $t0 Temporary Variable No
7 $t1 Temporary Variable No
8 $t2 Temporary Variable No
9 $s0 Saved Register Yes
10 $s1 Saved Register Yes
11 $s2 Saved Register Yes
12 $k0 Reserved for OS and Traps NA
13 $sp Stack Pointer No
14 $fp Frame Pointer Yes
15 $ra Return Address No
1. Register 0 is always read as zero. Any values written to it are discarded. Note: for the purposes of this project, you must implement the zero register. Regardless of what is written to this register, it should always output zero.
3. Register 2 is where you should store any returned value from a subroutine call.
4. Registers 3 - 5 are used to store function/subroutine arguments. Note: registers 2 through 8 should be placed on the stack if the caller wants to retain those values. These registers are fair game for the callee (subroutine) to trash.
5. Registers 6 - 8 are designated for temporary variables. The caller must save these registers if they want these values to be retained.
7. Register 12 is reserved for handling interrupts. While it should be implemented, it otherwise will not have any special use on this assignment.
8. Register 13 is the everchanging top of the stack; it keeps track of the top of the activation record for a subroutine.
9. Register 14 is the anchor point of the activation frame. It is used to point to the first address on the activation record for the currently executing process.
10. Register 15 is used to store the address a subroutine should return to when it is finished executing.
8.2 Instruction Overview
The LC-22 supports a variety of instruction forms, only a few of which we will use for this project. The instructions we will implement in this project are summarized below.
Table 2: LC-22 Instruction Set
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0000 DR SR1 unused SR2
0001 DR SR1 unused SR2
0010 DR SR1 immval20
0011 DR BaseR offset20
0100 SR BaseR offset20
0101 unused offset20
0110 RA AT unused
0111 unused
1000 SR1 SR2 offset20
1001 SR1 SR2 offset20
1010 DR unused PCoffset20
ADD
NAND
ADDI
LW
SW
BR
JALR
HALT
BLT
BGT
LEA
8.2.1 Conditional Branching
Branching in the LC-22 ISA is a bit different than usual. We have a set of branching instructions including BR, an unconditional branch, as well as BLT and BGT, which offer the ability to branch upon a certain condition being met. The BLT and BGT instructions use comparison operators, comparing the values of two source registers. If the comparisons are true (for example, with the BGT instruction, if SR1 > SR2), then we will branch to the target destination of incrementedPC + offset20.
8.3 Detailed Instruction Reference
8.3.1 ADD
Assembler Syntax
ADD DR, SR1, SR2
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0000 DR SR1 unused SR2
Operation
DR = SR1 + SR2;
Description
The ADD instruction obtains the first source operand from the SR1 register. The second source operand is obtained from the SR2 register. The second operand is added to the first source operand, and the result is stored in DR.
8.3.2 NAND
Assembler Syntax
NAND DR, SR1, SR2
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0001 DR SR1 unused SR2
Operation
DR = ~(SR1 & SR2);
Description
The NAND instruction performs a logical NAND (AND NOT) on the source operands obtained from SR1 and SR2. The result is stored in DR.
HINT: A logical NOT can be achieved by performing a NAND with both source operands the same.
For instance,
NAND DR, SR1, SR1
...achieves the following logical operation: DR←SR1.
8.3.3 ADDI
Assembler Syntax
ADDI DR, SR1, immval20
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0010 DR SR1 immval20
Operation
DR = SR1 + SEXT(immval20);
Description
The ADDI instruction obtains the first source operand from the SR1 register. The second source operand is obtained by sign-extending the immval20 field to 32 bits. The resulting operand is added to the first source operand, and the result is stored in DR.
8.3.4 LW
Assembler Syntax
LW DR, offset20(BaseR)
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0011 DR BaseR offset20
Operation
DR = MEM[BaseR + SEXT(offset20)];
Description
An address is computed by sign-extending bits [19:0] to 32 bits and then adding this result to the contents of the register specified by bits [23:20]. The 32-bit word at this address is loaded into DR.
8.3.5 SW
Assembler Syntax
SW SR, offset20(BaseR)
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0100 SR BaseR offset20
Operation
MEM[BaseR + SEXT(offset20)] = SR;
Description
An address is computed by sign-extending bits [19:0] to 32 bits and then adding this result to the contents of the register specified by bits [23:20]. The 32-bit word obtained from register SR is then stored at this address.
8.3.6 BR
Assembler Syntax
BR offset20
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0101 unused offset20
Operation
PC = incrementedPC + offset20
Description
A branch is unconditionally taken. The PC will be set to the sum of the incremented PC (since we have already undergone fetch) and the sign-extended offset[19:0].
8.3.7 JALR
Assembler Syntax
JALR RA, AT
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0110 RA AT unused
Operation
RA = PC;
PC = AT;
Description
First, the incremented PC (address of the instruction + 1) is stored into register RA. Next, the PC is loaded with the value of register AT, and the computer resumes execution at the new PC.
8.3.8 HALT
Assembler Syntax
HALT
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
0111 unused
Description
The machine is brought to a halt and executes no further instructions.
8.3.9 BLT
Assembler Syntax
BLT SR1, SR2, offset20
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1000 SR1 SR2 offset20
Operation
if (SR1 < SR2) {
PC = incrementedPC + offset20
}
Description
A branch is taken if SR1 is less than SR2. If this is the case, the PC will be set to the sum of the incremented PC (since we have already undergone fetch) and the sign-extended offset[19:0].
8.3.10 BGT
Assembler Syntax
BGT SR1, SR2, offset20
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1001 SR1 SR2 offset20
Operation
if (SR1 > SR2) {
PC = incrementedPC + offset20
}
Description
A branch is taken if SR1 is greater than SR2. If this is the case, the PC will be set to the sum of the incremented PC (since we have already undergone fetch) and the sign-extended offset[19:0].
8.3.11 LEA
Assembler Syntax
LEA DR, label
Encoding
31302928272625242322212019181716151413121110 9 8 7 6 5 4 3 2 1 0
1010 DR unused PCoffset20
Operation
DR = PC + SEXT(PCoffset20);
Description
An address is computed by sign-extending bits [19:0] to 32 bits and adding this result to the incremented PC (address of instruction + 1). It then stores the computed address into register DR.
1 file (188.9KB)
