Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Digital systems, linear algebra, introduction to formal proof / 3 / b / Solution 2026-10-31
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The entry sequence creates one 12-byte stack frame, then preserves the return address and both used callee-saved registers:
subi $sp, $sp, 12    # allocate three four-byte words
sw   $ra, 0($sp)     # save return address of this activation
sw   $s0, 4($sp)     # save caller's s0
sw   $s1, 8($sp)     # save caller's s1
The closing sequence reverses these effects:
lw   $ra, 0($sp)     # recover return address of this activation
lw   $s0, 4($sp)     # restore caller's s0
lw   $s1, 8($sp)     # restore caller's s1
addi $sp, $sp, 12    # deallocate the frame
jr   $ra            # return to the caller
Loads must precede deallocation because their offsets refer to the current stack frame. The return value in $v0 is deliberately retained. Saving $ra is essential even though it is not an $s processor register: otherwise the second nested call would destroy the address to which this activation must return.
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