The purpose of function call prologue value $ra in MIPS

Question

The following was the way I was taught to call functions in MIPS in school. I was simply told to "memorize" the "prologue" and the "epilogue", but never explained its concept. Here it is:

First, initialize $sp, the stack pointer. Then call jal f:

li      $sp, 0x7ffffffc
jal     f                # call function f

When inside the function, first do the "prologue":

sub     $sp, $sp, 8      # set new stack pointer
sw      $ra, 8($sp)      # save return address in stack
sw      $fp, 4($sp)      # save old frame pointer in stack
add     $fp, $sp, 8      # set new frame pointer

When returning, do the "epilogue":

lw      $ra, 8($sp)      # load return address from stack
lw      $fp, 4($sp)      # restore old frame pointer from stack
add     $sp, $sp, 8      # reset stack pointer
jr      $ra              # return to caller using saved return address

Question:

The values stored inside $ra never changed (unless my functions used it, but I shouldn't be using $ra in the first place). The prologue just saved it inside the stack, and restored the same value back into $ra when returning. jr is acting upon the same thing, so why is storing $ra in the stack necessary in the first place? And why do I need to restore it back if it never changed?

It makes sense to store $fp inside the stack because $fp was changed. But $ra is just the same returning address. Is the processor looking through the stack to match the values inside $ra with ones on the stack?

m0skit0 m0skit0 · Accepted Answer · 2016-12-11T17:28:33

why is storing $ra in the stack necessary in the first place? And why do I need to restore it back if it never changed?

In your particular case, $ra is never changed. But what would happen if a function calls another function? In that case, $ra would indeed change. In fact your function prologue is missing saving the $sX registers which also (based on MIPS ABI) should be saved by the callee.

Is the processor looking through the stack to match the values inside $ra with ones on the stack?

No, in MIPS the processor has no understanding of what "stack" is. The stack is a software construct, hardware doesn't care about it.

The purpose of function call prologue value $ra in MIPS

1 Answers