C to the MIPS problem
So, I worked on this for a few days and I managed to get through it all, except one part got me sick.
int mystery(int a0)
{
if (a0 == 0)
{
return 0;
}
else
{
return mystery(a0 - 1) + a0;
}
}
I have this recursive function and I have MIPS code. The C code works, but I have a problem somewhere in my MIPS code that makes it wrong without inserting 2.
.text
main:
li $a0, 2
jal mystery
move $a0, $v0
jal putDec
li $a0, '\n'
li $v0, 11
syscall
li $a0, 3
jal mystery
move $a0, $v0
jal putDec
li $a0, '\n'
li $v0, 11
syscall
li $v0, 10
syscall
putDec:
li $v0, 1
syscall
jr $ra
mystery:
bne $0, $a0, recur
li $v0, 0
jr $ra
recur:
sub $sp, $sp, 8
sw $ra, 4($sp)
sub $a0, $a0, 1
jal mystery
sw $v0, 0($sp)
jal mystery
lw $t0, 0($sp)
addu $v0, $v0, $t0
addu $v0, $v0, 1
add $a0, $a0, 1
lw $ra, 4($sp)
add $sp, $sp, 8
jr $ra
Anything about the "riddle" of the label is fine where there is formality to actually enter arguments and print after. The problem I'm running into is getting values above 3 to print out the correct numbers. Help would be greatly appreciated if someone could see where I am doing my mistake. Thanks you
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When after a lot of work and tracking the code one step at a time, I think I finally got it, I had to change it a bit, but in fact it seems to me easier now. Here is the code in case you want to review it again if you see an issue that could occur with any specific values that I have not tested. Thanks for the help.
New recursive mystery function:
mystery:
bne $0, $a0, recur
li $v0, 0
jr $ra
recur:
sub $sp, $sp, 8
sw $ra, 4($sp)
sw $a0, 0($sp)
sub $a0, $a0, 1
jal mystery
lw $t0, 0($sp) #
addu $v0, $v0, $t0
lw $ra, 4($sp)
add $sp, $sp, 8
jr $ra
Thanks again.:)
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Try to take a step back and compare the structure of your C code and assembly without worrying about the details.
There is a conditional expression in C code that leads to either the base case (which just returns a value) or the recursive case. The recursive case performs subtraction, recursive invocation, mystery
and addition.
Now let's look at the assembly version: this too has a conditional expression leading either to the base case or to the recursive case. But look at the structure of the recursive case: there are two recursive calls in mystery
there! This is a strong hint that he is unlikely to do the same ...
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