AAPCS32 2026-10-24
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The 32-bit calling convention for the ARM architecture family. Its base variant passes initial arguments in r0–r3, returns a word in r0, preserves r4–r8, r10, r11, and sp, and requires eight-byte stack pointer alignment at public calls.[ref]
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Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Digital systems, linear algebra, introduction to formal proof / 2 / c / Solution 2026-10-25
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Keep total in the callee-saved register r4, and the first returned sum in the callee-saved register r5. Suppose the surrounding function has made an aligned stack frame with the local seven-halfword array at the current sp, occupying fourteen bytes in a sixteen-byte slot. One possible frame setup is push {r4,r5,r6,lr} followed by sub sp,sp,#16; the earlier code initializes total, regular, and special before this fragment.
    ldr  r0, =regular       @ assembler literal load of the global array address
    movs r1, #7
    bl   sum16
    movs r5, r0             @ first sum must survive the next call

    add  r0, sp, #0         @ current frame's local special array
    movs r1, #7
    bl   sum16
    muls r0, r5, r0         @ second sum times first sum
    adds r4, r4, r0         @ update total
The AAPCS32 permits each call to overwrite r0–r3, r12, and condition flags, but requires preservation of r4–r8, r10, r11, and sp. Thus both r4 and r5 remain valid even if sum16 is replaced by another conforming implementation. bl changes the link processor register lr, so the enclosing function saves its own return address in its prologue. Its epilogue can use add sp,sp,#16 and pop {r4,r5,r6,pc}. The frame uses thirty-two bytes in total, preserving the required eight-byte stack pointer alignment at both calls.[ref]
The two calls may be evaluated in this order. Assume the product and final addition fit the signed 32-bit result expected by the source program; unlike each individual seven-element sum, their product need not fit. The assembly language operations retain their low 32 bits if this assumption fails, whereas signed integer overflow in the C programming language has different semantics.
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Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Digital systems, linear algebra, introduction to formal proof resit / 1 / a / Solution 2026-10-25
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Use r0 for the argument and result. With 32-bit integers and pointers, the three fields have byte offsets 0, 4 and 8. Assume an acyclic finite binary tree, no concurrent mutation, a valid aligned call stack, and no signed integer overflow in any evaluated sum. These are representation and arithmetic assumptions, rather than properties of arbitrary C programming language implementations.
.syntax unified
.cpu cortex-m0
.thumb
.global sum_rec
.thumb_func
sum_rec:
    cmp  r0, #0
    beq  .Lempty
    push {r4, r5, r6, lr}    @ 16-byte frame; r6 is alignment padding
    movs r4, r0              @ retain this invocation's node
    ldr  r5, [r4, #0]        @ running value begins with its label
    ldr  r0, [r4, #4]
    bl   sum_rec             @ r0 = left subtree sum
    adds r5, r5, r0
    ldr  r0, [r4, #8]
    bl   sum_rec             @ r0 = right subtree sum
    adds r0, r0, r5
    pop  {r4, r5, r6, pc}    @ restore caller state and return
.Lempty:
    movs r0, #0
    bx   lr
The node and partial sum are held in callee-saved registers r4,r5. Each recursive invocation saves its caller's values before replacing them, so a child returns with the parent's values intact. Saving lr preserves the return address across the two bl instructions. Caller-saved registers may be overwritten by a child; no parent value is left live only in them. Pushing four words maintains eight-byte stack alignment at recursive calls.[ref]
Correctness follows by structural induction. An empty subtree returns zero. For a nonempty subtree, assume both recursive calls return their respective label sums. The final addition then combines the node's label with exactly those two sums, so r0 returns the sum of all labels. The instruction forms and field offsets fit the short load and arithmetic encodings.[ref]
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Stack alignment 2026-10-24
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Stack alignment constrains the stack pointer to a multiple of a specified byte boundary. AAPCS32 requires word alignment generally and eight-byte alignment at public interfaces. Saving four 32-bit words preserves eight-byte alignment; saving three would change it by four bytes and may require padding before another call. Frame size and the alignment of local objects must be considered together.[ref]
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