I am trying to add jal functionality to the following but I am stuck with how does it work. I know that it stores the old PC+4 value in the $ra register and then transfers the control to the function which transfers back the control by return $ra but how do I implement it in the hardware?
There are two things you need to do.
Add a mux at the input of the Registers so that the PC+4 value can be selected as the data to be written. With the appropriate control signal this will allow you to write PC+4 as an additional effect of the "jal $ra" instruction.
Implement the return "jr $ra" instruction. You will need to add a mux to the chain of logic that selects the next PC so that "read data 1" from the register file can be selected as the next PC when the instruction is "jr xxx".
We ad a new line that takes the PC+4 and sends it to the "Write Data" input of the register file. We need to add a multiplexer. We also need to make sure the "Write Register" field is set to the address of $RA. Register $RA is register number 31 most of the time.
Related
Here's the datapath:
So this seems like a pretty common question but I can't seem to find any answers on how to extend the datapath to implement SLL and SRL.
This is how I would think to do it but I'm not entirely sure:
It would need another mux right next to Read data 1 next to the register file. This mux would take Read data 1 (rs) and Read data 2 (rt) as inputs. It would select Read data 1 if we're not doing a shift operation, and it would select rt if we ARE doing a shift operation (since sll and srl use rt, not rs). This would then be fed into the ALU.
Next, we would need to branch Instruction[10:6] (the shift amount) off of Instruction[15:0], and Instruction[10:6] would then be fed into the other port of the ALU. Is this correct thinking?
This is sll on single cycle datapath, but i am not sure if the ALU now gets 5 instead of 4 bits control input.
If u make sll then the first ALU input would be shamt and the second is the register to be shifted, ALU know if it must make shift because of instruction field, because it is a R-Type instruction. Then the shifted data will be saved in rd register.
SLL SC datapath
You need to modify the datapath for the SLL instruction, adding a input line to the ALU with the "shamt" field in order to determine de shift amount. The ALU will identify the SLL operation by the ALUop field.
Modiffied datapath
You are going in the correct direction. As stated in one of the answers, there can be one additional port added to the ALU which will consider the shamt amount (bits [10:6]). There can be some internal hardware such as a MUX in the ALU which takes care of selecting either the shamt field or Read Data 2 from the output of register file.
I am asked to add the shift left logical instruction to a single cycle datapath. I know I need to feed the SHAMT field to the ALU, but I'm not sure how to do this. I understand the basics of single cycle data paths for R-format, branch, load word, and store word, but I'm not sure how the SLL plays in... Can anyone help explain how a single cycle SLL datapath works?
If this is the wrong form for this post, I'd be much obliged if someone would direct me to the correct site.
You simply need to decode the opcode of the SLL instruction and use it to set the ALUOp input of the ALU to 11. You also need to set the multiplexers to put the source register and the shift amount at the appropriate inputs to the ALU.
So I'm going over some old quizzes for my Computer Organization final and I must have missed this lecture or something. I'm decently proficient in programming MIPS, but this problem has me completely stumped. Could someone help me understand this?
The diagram is missing lines connecting the various parts of the processor as well as a multiplexor for determining if the next instruction address is coming from the PC+4 or from a register as in a jr ra instruction.
There needs to be a line from the ALU to the write data portion of the register file. This is for R-Type instructions, as their result will need to be written back to the destination register. Going into the ALU needs to be lines from Read data 1 and Read data 2, this is how the values from the registers make their way into the ALU for R-type instructions.
A couple lines have been added from the instruction memory to the registers, you're missing the one to the Write Register though (this specifies the destination register for R-type instructions).
For the PC, the line going into the adder goes into the other input (the above one). The 4 for the adder is constant, as each instruction address is 4 bytes after the previous one, so unless we're jumping to an address we will be executing the instruction immediately after the current instruction. The line from the PC to the read address is also necessary as the PC specifies which instruction address to find the current instruction at. The line going into the PC comes from either the result of the PC+4 addition or from the register specified in a jr ra instruction.
To handle this decision a multiplexor is needed. Multiplexors have two inputs and one output, so this one will have one input from the PC+4 adder (for regular R-type instructions) and another from Read Data 1 (for jr ra instructions). The input from Read Data 1 should be visualized as a split from the line between Read Data 1 and the ALU. The output will go right back to the PC, as it determines the next instruction to execute.
I think that's everything that's wanted for that question, as the prof specifies that control signals are already generated (the multiplexor is a type of control unit, but I think it's necessary nonetheless). Hope that helps!
I'm having trouble understanding how the instruction jal works in the MIPS processor.
My two questions are:
a) What is the value stored in R31 after "jal": PC+4 or PC+8?
b) If it's really PC+8, what happens to the instruction at PC+4? Is it executed before the jump or is it never executed?
In Patterson and Hennessy (fourth edition), pg 113:
"jump-and-link instruction: An instruction that jumps to and address and simultaneously saves the address of the following instruction in a register ($ra in MIPS)"
"program counter (PC): The register containing the address of the instruction in the program being executed"
After reading those two statements, it follows that the value saved in $ra should be (PC+4).
However, in the MIPS reference data (green card) that comes with the book, the jal instruction's algorithm is defined like this:
"Jump and Link : jal : J : R[31]=PC+8;PC=JumpAddr"
This website also states that "it's really PC+8", but strangely, after that it says that since pipelining is an advanced topic "we'll assume the return address is PC+4".
I come from 8086 assembly, so I'm aware that there's a big difference between returning to an address and to the one following it, because programs won't work if I just assume something that's not true. Thanks.
The address in $ra is really PC+8. The instruction immediately following the jal instruction is in the "branch delay slot". It is executed before the function is entered, so it shouldn't be re-executed when the function returns.
Other branching instructions on the Mips also have branch delay slots.
The delay slot is used to do something useful in the time it takes to execute the jal instruction.
I got the same question. Googled this excellent answer of Richard and also another link I wish to add here.
The link is http://chortle.ccsu.edu/AssemblyTutorial/Chapter-26/ass26_4.html
with this wonderful explanation of double adding 4 to the PC.
So the actual execution has two additions: 1) newPC=PC+4 by pipelining and 2) another addition $ra=newPC+4 by the jal instruction resulting the effective $ra = (address of the jal instruction)+8.
The code for this is too long to post so Ill just describe it. I've created a 5 stage mips pipe that almost works. The catch is that EVERY lw instruction that reaches the instruction decode stage overwrites the control signal values in the execution stage. Not only that it causes the PC to skip can instruction, i.e from 300 -> 308. I just need some idea on where to look for bugs since this is a class assignment. If we take out all the LW instructions the CPU works fine.
Example:
The adder in the EX stage is going to sub $4 $1 $2 which should be 1
Once LW enters the ID stage ALUsrc is asserted AND ALUop is changed from subtract to add
This forces the adder in the EX stage to add $4 $1 $2 resulting in 5 being stored in $4
http://en.wikipedia.org/wiki/File:MIPS_Architecture_%28Pipelined%29.svg
The MIPS 5 Stage Pipeline (annotated to show Write Reg Select and enable)
The bottom line through the pipeline stages represents the register file write (back) port address and write enable and WB is the data from memory.
http://www.mrc.uidaho.edu/mrc/people/jff/digital/MIPSir.html
Load Word Instruction
Description:
A word is loaded into a register from the specified address.
Operation: $t = MEM[$s + offset]; advance_pc (4);
Syntax: lw $t, offset($s)
Encoding:
1000 11ss ssst tttt iiii iiii iiii iiii
Where the write register address ($t) input is read from data memory address comprised of register file register $s offset with the immediate value i which gets sign extended. Your $4 is $t above, $1 or $2 is $s while the remaining register file output lane sounds to be suborned for the sign extended immediate.
From your description it sounds like you aren't using a three port register file with one port a write only port.
With a three port register file the only time you run into conflicts is when you attempt to use the new register file value from memory before it is read from memory and written to the register file. That can be managed by a compiler scheduling NOOPs until the outstanding register file write is retired when a following instruction is trying to use it, or stalling the IF/ID in hardware when it's output contains a reference to an outstanding register file write.
There are three instructions that can be in flight to the right of IF/ID, each with a write to register file address and a write enable. You'd need to compare both instruction decode register file addresses to all three of those and stall IF/ID until those clear out. The write enable stored in each of those three pipeline stages are used to determine whether the write register address in those pipeilne stages should be compared.
Because the ID/EX, EX/MEM and MEM/WB write register file addresses are not used anywhere else the circuitry for doing the comparison can be collocated with IF/ID and the Register File, preventing unnecessary layout delays affecting the minimum clock cycle.
Using a two port register file is much simpler and infers IF/ID stalling until the write enable comes back from MEM/WB, effectively turning any memory reading instructions into 3 cycle instructions (or more, data memory can stall if it's a cache or slow). It makes a three port register file more or less necessary for performance reasons. There's an implied multiplexer to source for at least one of the two register file port controls (write enable, write address) from the MEM/WB stage when IF/ID is stalled (for memory->regfile).
Data memory access can stall MEM/WB, just like instruction memory access can also stall IF/ID. A stalled IF/ID doesn't issue a write enable for the register file to ID/EX nor does a stalled MEM/WB.