Consider the following MIPS assembly language code

Question

s3a · Answer

Problem:
Consider the following MIPS assembly language code: 

1000 lw $5, 0($3) # does some work
1004 beq $5, $0, done # does some work
1008 jal foo1
1012 ...

foo1:
1100 addi $29, $29, -4 # $29 is the stack pointer
1104 sw $31, 0($29) # $31 is the return address
1108 jal foo2
1112 ...

After executing the instruction on line 1104, the contents of register $31 is: 
(a) 1008 
(b) 1112
(c) The content of $29
(d) 1108
(e) 1012

Answer:
The answer is (e).

I can see that (c) is wrong but, could someone please explain to me why (e) is the correct answer?

If more information is needed, just ask.

Any input would be greatly appreciated!

ganeshie8 · Answer

jal instruction at line 1008 sets the register $31 to return address of 1012

s3a · Answer

But, on line 1100, the stack pointer is decremented by 4 and then that updated stack pointer is stored into the return register which seems to clash with what you said ... could you elaborate, please?

ganeshie8 · Answer

stack pointer reg is $29 right, wat it got to do with reg $31 ?

ganeshie8 · Answer

the value in $29 reg is decremented. how it wil change the value in $31 ?

s3a · Answer

What exactly does line 1104 do?

ganeshie8 · Answer

Jumps to the calculated address and stores the return address in $31

ganeshie8 · Answer

return adress = immediate adress after jal instruction

s3a · Answer

So, does $29 get incremented by 4 and then decremented by 4 on line 1100?

ganeshie8 · Answer

we adding -4 so it gets decremented by 4. thats all to it

s3a · Answer

Is the program counter the same thing as $29? I ask because http://www.mrc.uidaho.edu/mrc/people/jff/digital/MIPSir.html says $t = $s + imm; advance_pc (4);

ganeshie8 · Answer

PC counter gets incremented every instruction right

s3a · Answer

But, is the stack pointer the same thing as the program counter?

If so, that would explain the +4-4 to me.

ganeshie8 · Answer

no ways. stackpointer and pc are two different registers

ganeshie8 · Answer

i dint get the +4-4 thingy.. can u explain pls

s3a · Answer

If they were the same thing (which you said, they are not), I was thinking that adding 4 to the program counter and subtracting 4 from the stack pointer would "neutralize" each other.

ganeshie8 · Answer

OK. as i said, PC gets incremented automatically after completing each instruction. we dont have to do anything explicit.

where as stackpointer we need to program

s3a · Answer

Okay, I think I got it. Tell me if the following is correct.:

jal foo1 stores the address of line 1012 in $31 and then runs the code in foo1 but, foo1 does not change the contents of $31 so the final answer is just 1012.

Also, line 1104 goes to the address of $29 and stores the contents of $31 in $29 and not the contents of $29 into $31, right?

ganeshie8 · Answer

thats perfect ! both statements are correct.

s3a · Answer

Thanks a lot, again! :)

ganeshie8 · Answer

yw :)