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Binary array sum by recursion 2026-10-18
Keep an array unchanged, and sum entries in the half-open integer interval of array indices . The precondition requires integers and exact, terminating integer addition. An empty range returns the empty sum zero; a one-entry range returns that entry. Every longer range splits at its middle array index:For the original array , call . The left half returns , the right half returns , and the original invocation returns . The right half's left endpoint increases from zero to two: what decreases is range length , not every argument. Negative entries affect the answer but do not affect algorithm termination.
S(A, l, r):
require integer indices with 0 <= l <= r <= length(A)
size = r - l
if size == 0:
return 0
if size == 1:
return A[l]
middle = l + floor(size / 2)
left_sum = S(A, l, middle)
right_sum = S(A, middle, r)
return left_sum + right_sum
Seven calls compute an array sum with only three frames active at once
. Original binary tree of invocations for , with integer intervals labeling ranges and upward arrows labeling returned sums. Small numbers give the order in which invocations begin when the left child runs first. The highlighted graph path contains the three active stack frames while the entry is read; the result from the completed left sibling is retained by its parent. Details.
For range length , the children have lengths and , both positive and strictly less than . The floor function chooses the split; the ceiling function gives the other length. Algorithm termination follows by this natural number measure. For algorithm correctness, apply mathematical induction on : the two base cases return the specified sum; for a larger range, the two disjoint child ranges cover every original array index exactly once, so addition of their correct sums gives the required sum.
For , the invocation tree is a full binary tree with leaves, one per array entry. If is its number of nonleaf vertices, counting edges gives , so . Thus the total invocation count is , and there are combining additions. Here gives seven calls and three combining additions. For , there is one call and no combining addition; for , there is also one call and no combining addition, but no entry leaf, so the formula does not apply.
The call stack holds the active branch, rather than the entire invocation tree. Execute the left child completely before the right child, retaining the left result in the parent stack frame. The maximum number of active stack frames is for , and one for . Balanced splitting therefore needs frames for , compared with the frames of the factorial recursive process. Assuming constant-cost array access, index operations and addition, its time complexity is and its auxiliary space complexity is machine words for , excluding the stored input. Pass the same array with different endpoints, rather than copying slices. These bounds require all stored indices and partial sums to fit in a constant number of machine words; exact unbounded integer addition and storage require a separate bit cost analysis.
Dynamic chain 2026-10-31
The chain of active callers obtained by following dynamic links from the current stack frame. It records call history, whereas a static chain records lexical environments.
Dynamic link 2026-10-31
A pointer from an activation's stack frame to its caller's stack frame, used to restore the caller when the activation returns. The dynamic link is independent of a static link to the lexically enclosing environment.
Explicit-stack depth-first tree traversal 2026-10-24
An explicit-stack depth-first tree traversal stores pending subtree roots instead of suspended recursive calls. In a binary tree, visit the current node, defer its nonempty right child on a stack, and continue with its left child; pop when the current subtree is empty. The unvisited subtrees remain disjoint and, together with the visited nodes, partition the original tree. This loop invariant proves that accumulating each visited label gives the tree sum. At most one right sibling per ancestor is pending, so auxiliary storage is for depth and running time is for nodes. Unlike a full recursive stack frame, a pending entry need contain only a subtree pointer.
Factorial recursion and returning frames 2026-10-18
For a natural number , the factorial satisfies and for . The following description includes the zero base case explicitly; a negative or nonintegral input is outside its precondition:For , the arguments on descent are . Each positive-argument invocation postpones one multiplication until its recursive call returns. At the deepest point, the call stack has five stack frames: four suspended callers and the active zero base case. The invocation at zero returns to the invocation at one, rather than finishing the entire computation. Returns then giveA stack frame retains the caller's argument and where to resume. When the invocation at three receives , it uses its own , obtains , and returns that value to the invocation at four. The recursive calls share a definition, but their local variables belong to separate invocations.
F(n):
require n is an integer and n >= 0
if n == 0:
return 1
child_result = F(n - 1)
return n * child_result
The natural number measure decreases by one on every recursive call, so algorithm termination follows. For algorithm correctness, use mathematical induction: the zero base case returns ; assuming the child's answer is , the caller returns . This also explains why keeping the caller's matters.
There are exactly invocations and multiplications, including one invocation and no multiplications when . Ordinary execution of this recursive process uses at most active stack frames, hence frames for in Big O notation. If each test, call, subtraction and multiplication is counted as constant cost, its time complexity is . That is an arithmetic-operation model: exact arbitrary-length integer multiplication has growing cost, and neither the factorial output nor every stack frame's numeric contents occupies a universally fixed number of bits.
Frame pointer 2026-10-31
A processor register or equivalent value designating a stable reference location within the current stack frame. Parameters, local variables and saved links can be accessed at fixed offsets from the frame pointer even if the evaluation stack changes height.
Java virtual machine 2026-11-05
The Java virtual machine is the abstract execution machine specified for Java class files and bytecode. It defines runtime data areas, stack frames, operand stacks, local variables and instruction behavior, while leaving implementation choices such as interpretation and just-in-time compilation open.[ref]
For example, two integer loads followed by an integer-add instruction obtain operands from a stack frame and leave their sum on its operand stack. A method call establishes a new stack frame and later returns a result to its caller. These specified effects remain the contract even if a compiler keeps the values in native registers and emits one machine instruction. The virtual-machine interface therefore supports portable compiled code without requiring every implementation to execute bytecode one instruction at a time.
Operand stack 2026-11-05
An operand stack holds intermediate values used by bytecode operations in a Java virtual machine method invocation. Each stack frame has its own last-in-first-out operand stack, initially empty; its maximum depth is recorded in the class file. Values can be loaded from local variables, consumed by operations and replaced by results.[ref]
Starting from an empty operand stack,
iconst_2; iconst_3; iadd produces the successive contents 2, 2,3 and 5, with the top shown at the right. The addition consumes both operands. This abstract execution rule does not require a physical memory push for each operation: just-in-time compilation may retain values in processor registers while preserving the specified behavior. Past exam of the computer science course of the University of Oxford 2015 Compilers 2 a Solution 2026-10-31
The surviving call stack, from oldest to newest, is
main, p, r, s, q. A dynamic link identifies the caller's stack frame, whereas a static link identifies the current activation of the lexically enclosing procedure. These need not identify the same stack frame.
The call stack during
. Details. q, with dynamic links on the left and static links on the right; arrows point to older stack frames
The dynamic chain isThe individual static links are
main has no enclosing procedure; its links can be null. The p stack frame contains t=3 and x=3, and the q stack frame contains y=3. Return addresses are code addresses, not links to another stack frame. In particular, q returns dynamically to s while obtaining x statically from p. Past exam of the computer science course of the University of Oxford 2015 Compilers 2 b Solution 2026-10-31
We specify a small stack machine instruction convention because the paper does not prescribe instruction spellings or numeric stack frame offsets.
PUSHFP pushes the current frame pointer. LOAD k replaces a stack-top frame address by the value at fixed offset k in that frame. Let SL be the static link offset, X the offset of p's x, and Y the offset of q's parameter y. ADD pops two integers and pushes their sum. RETVAL returns the stack-top value, removes the current activation and restores the caller through the usual calling convention.PUSHFP ; address of q's frame
LOAD SL ; q's static link: address of p's frame
LOAD X ; value of p.x
PUSHFP ; address of q's frame, above x
LOAD Y ; q's local parameter y
ADD ; x + y
RETVAL ; return the sum to sOne static link traversal suffices for
x, despite three intervening calls in the dynamic chain. y is local to q. In this execution the two loads produce and , so the returned integer is . Past exam of the computer science course of the University of Oxford 2015 Compilers 2 c Solution 2026-10-31
Continue the instruction convention above. The current stack frame is now
s's. Its static chain passes through r to p; q is declared directly inside p, so the new q activation must receive a static link to that same p stack frame. Let T be the offset of p's parameter t.Define
CALL q,1 to expect, above the pre-existing evaluation stack, the static link for q followed by its one argument. It consumes these, creates q's stack frame with its dynamic link pointing to the current s activation, and leaves one return value when q returns. This explicit convention avoids dependence on a particular real instruction set architecture.PUSHFP ; s
LOAD SL ; r
LOAD SL ; p, the required static parent of q
DUP ; preserve p for CALL while also using it to read t
LOAD T ; replace the duplicate p address by p.t
CALL q,1 ; q(p.t), with static parent p and dynamic caller s
RETVAL ; return q's result from s to rImmediately before
CALL, the evaluation stack ends in p-frame, 3. Two static link traversals find both the argument's environment and the environment to pass to q. In terms of the hidden argument and ordinary argument, An implementation that places the hidden static link after the ordinary argument could use a different ordering of these same operations. Past exam of the computer science course of the University of Oxford 2015 Compilers 2 d Solution 2026-10-31
Use mathematical induction on the sequence of call and return events. The invariant concerns the static chain of the currently executing activation: every stack frame on that chain is the most recent surviving activation of the procedure it represents. It need not hold for the saved chain of an arbitrary suspended caller while a recursive callee is running.
Initially only
main is active, so the claim holds. Suppose it holds just before a direct call to a procedure , whose lexical parent is . Without procedural parameters, visibility of means that its declaration lies in the current procedure or in a lexically enclosing procedure. Therefore the activation of needed for the call is on the caller's static chain. By the induction hypothesis, it is the newest surviving activation of .Create a new activation of whose static link points to this activation. The inherited chain is a suffix of the caller's chain, so its stack frames still represent the newest surviving activations of their procedures. Adding does not make any of those inherited activations out of date: cannot be its own lexical ancestor. The new activation itself is plainly the newest activation of . This covers recursive calls too: an older activation of may survive on the dynamic chain, but is absent from the new activation's static chain.
On return, the callee and every call it made have finished. The caller's saved stack frame and static chain are restored. Any temporarily newer activation of a procedure represented on that chain has returned, so each restored activation is again the most recent surviving one. Thus return preserves the invariant as well.
The claim holds for the executing procedure's static chain throughout every execution using direct calls under lexical scope. Equivalently, writing for the procedure represented by a stack frame ,
Past exam of the computer science course of the University of Oxford 2015 Compilers 2 e Solution 2026-10-31
Searching the dynamic chain would require identifying the procedure associated with each visited stack frame. More seriously, the search length is bounded by the number of active calls, not by lexical nesting depth. Recursion in procedures unrelated to the desired enclosing declaration can make arbitrarily large. Thus a single nonlocal variable access can take time complexity, whereas a conventional static chain access needs only traversals for lexical distance .
Instead give each procedure a known global pointer cell
current[P]. On entry to , save its previous value in the new stack frame and set current[P] to the new frame pointer. On return, restore the saved value. A nonlocal variable declared in is then loaded using current[P] and its known frame offset. The proof in part (d) justifies that lookup in the absence of procedural parameters. This is a procedure-indexed version of a display for lexical scope; a conventional display for lexical scope can instead use lexical levels as indices.With procedures and active calls, the procedure-indexed table needs additional global words and one saved previous pointer per active stack frame, hence activation storage. The saved word can replace the conventional per-frame static link if that link is no longer needed. A conventional static chain has one pointer per active stack frame and no such -entry table. Both schemes retain ordinary dynamic links and return information.
The table gives constant-time nonlocal access and constant-time save/restore work on each entry/return, independently of lexical distance. Conventional static chain access costs per access, and establishing the callee's static link may also require following an appropriate portion of the caller's chain. For a fixed program is statically bounded, but it still affects generated instruction counts. These costs assume constant-cost pointer operations and omit shared ordinary call overhead. For the procedure-indexed table,
Past exam of the computer science course of the University of Oxford 2015 Compilers 3 b Solution 2026-10-31
The useful repeated program expressions, nested inside one another, are the memory address of
a[i], its value, and the memory address of a[a[i]]. Materialize each maximal reusable tree at the level at which it is shared, using fresh temporary variables u, v, and w:MOVE(R(u), ADD(C(A),SHL(M(ADD(R(fp),C(I))),2)))
MOVE(R(v), M(R(u)))
MOVE(R(w), ADD(C(A),SHL(R(v),2)))
MOVE(R(t), M(R(w)))
MOVE(M(R(w)), R(v))
MOVE(M(R(u)), R(t))Thus
u is a saved memory address, v the old integer value of a[i], w the second saved memory address, and t the old value at that second location. The two large address trees occur both in loads and as store destinations. After extracting them, the repeated load of a[i] inside the second address and as the middle assignment's source becomes v. Extracting every smaller nested tree again would add no benefit.The relevant alias analysis facts are that local
i does not occupy any location written through global a, that frame pointer fp and base stay fixed, and that t,u,v,w are distinct processor registers or otherwise nonoverlapping temporary variables. There are no intervening calls, volatile memory accesses, or concurrent changes. Under these assumptions, loading i once remains valid even across the array stores. Before the middle store, no array location has yet changed, so both its target memory address and its source value can be computed and saved earlier.No assumption that the two array locations are distinct is required. The middle store may change
a[i] when its target coincides with it, but v is not loaded again afterwards: only the saved address u is needed for the final store. It would be unsound to retain an arbitrary array load across an aliasing store and then treat it as the new contents; that is not what these trees do.For a language with a true global array, a distinct local variable, and no arbitrary means of fabricating pointers to unrelated storage, these facts follow simply from declarations and the stack frame layout. Accesses must denote valid array elements. In an unrestricted low-level language with overlapping storage, arbitrary pointer arithmetic or unchecked out-of-range accesses, that conclusion requires extra guarantees. Omitting emitted array bounds checks does not itself prove that all memory addresses are valid. If denotes a word load, the saved quantities are
Make the code a leaf procedure called
array_diff, copy the four argument processor registers into the temporaries used above, and replace the final label by a return. The copies are real addu instructions, avoiding an unexplained move pseudo-instruction:array_diff:
addu $t0,$a0,$zero
addu $t1,$a1,$zero
addu $t2,$a2,$zero
addu $t3,$a3,$zero
# insert the code of (i), including its initial zero-count test
# use the local label array_diff_done in place of done
array_diff_done:
jr $ra
sll $zero,$zero,0In the literal insertion, the
done: line from (i) is the array_diff_done: line displayed here; it is not a second return block. Under the usual calling convention, temporary processor registers are caller-saved. This procedure calls no other procedure and changes no saved processor registers, so it needs no stack frame or saved return address.For a 32-bit constant , define and using its unsigned binary representation. Load it with
lui followed by ori; both are single MIPS instructions. For base addresses and count , the main program is:main:
lui $a0,H_A
ori $a0,$a0,L_A
lui $a1,H_B
ori $a1,$a1,L_B
lui $a2,H_C
ori $a2,$a2,L_C
lui $a3,H_N
ori $a3,$a3,L_N
jal array_diff
sll $zero,$zero,0
main_after_call:
# c now contains the differences; continue the main program hereHere each
H_... and L_... is an assembler-time numeric constant, not an executed operation. lui places the high half and clears the low half; ori inserts the low half without sign extension. ThusThe jal instruction saves the return address after its branch delay slot in $ra; jr $ra resumes at main_after_call. A main routine that itself needs to return to a caller must save its incoming $ra before jal and restore it afterwards.The calling convention passes the integer argument in
$a0, returns the result in $v0, and uses jal with $ra for a call and its return address. The callee returns using jr $ra. Its stack frame grows downwards from $sp and must be removed before returning, restoring the caller's stack pointer.The
$s processor registers are callee-saved registers: a callee that uses $s0 or $s1 must restore their incoming values. The $a, $v, and $t processor registers are caller-saved registers; a caller must preserve any of their values that it needs after a call. Consequently foo keeps its own argument in $s0 and the first call's result in $s1, where nested calls preserve them. Because a nested jal overwrites $ra, foo also saves its incoming return address. Every recursive activation has its own stack frame.The supplied MIPS instructions use the paper's simplified convention: pseudoinstructions such as
subi are accepted, and control transfers have no branch delay slot. The completed code below follows that convention. A hardware implementation with delayed control transfers requires suitable scheduling or nop instructions throughout the supplied code as well.The entry sequence creates one 12-byte stack frame, then preserves the return address and both used callee-saved registers:The closing sequence reverses these effects:Loads must precede deallocation because their offsets refer to the current stack frame. The return value in
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 s1lw $ra, 0($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 $ra # return to the caller$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.Use the conventional integer calling convention:
$a0 holds the argument and $v0 holds the result. The function is recursive, so every call must preserve its own return address and original argument. The first recursive result must also survive the second call; a caller-saved register alone would not protect it.This implementation reserves a 32-byte stack frame, keeping the stack pointer eight-byte aligned. Offsets 0–15 are available as the outgoing argument area, 16 stores the original argument, 20 stores the first result, 24 is padding, and 28 stores the return address. It changes no callee-saved register. Assemble in classic MIPS delayed-branch mode with instruction reordering disabled, such as
.set noreorder.foo:
addiu $sp,$sp,-32 # allocate 32-byte aligned frame
sw $ra,28($sp) # save caller return address
sw $a0,16($sp) # save original n
beq $a0,$zero,foo_base
sll $zero,$zero,0
addiu $t0,$zero,1
beq $a0,$t0,foo_base
sll $zero,$zero,0
addiu $t0,$zero,2
beq $a0,$t0,foo_base
sll $zero,$zero,0
addiu $a0,$a0,-1
jal foo
sll $zero,$zero,0
sw $v0,20($sp) # preserve foo(n-1) across next call
lw $a0,16($sp) # recover original n
sll $zero,$zero,0
addiu $a0,$a0,-3
jal foo
sll $zero,$zero,0
lw $t0,20($sp)
sll $zero,$zero,0
addu $v0,$v0,$t0 # combine the two recursive results
beq $zero,$zero,foo_return
sll $zero,$zero,0
foo_base:
addu $v0,$a0,$zero # base case result is n
foo_return:
lw $ra,28($sp)
sll $zero,$zero,0
addiu $sp,$sp,32 # release the frame
jr $ra
sll $zero,$zero,0All displayed operation mnemonics denote single machine instructions. In particular,
sll $zero,$zero,0 is the actual harmless shift encoding often abbreviated by the pseudoinstruction nop; addu $v0,$a0,$zero performs a copy without the move pseudoinstruction. Each branch or jump has an explicit harmless branch delay slot, and harmless shifts also separate loads from their subsequent uses. The base cases return the original argument; other calls return the sum of the two recursive values. Every return restores the saved return address and exactly undoes the stack allocation.[ref]MIPS32 Architecture For Programmers Volume II by MIPS Technologies. Revision 2.00, ADDIU and ADDU, pp. 37–38; JAL and JR, pp. 139 and 144
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 totalThe 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.
Stack frame Created 2026-10-18 Updated 2026-10-24
The record for one active invocation in a call stack, containing information such as local variables and the place to resume its caller. Returning removes that invocation's stack frame, allowing the caller's computation to continue.