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Incoming links: Frame pointer
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.
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 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 a Solution 2026-10-31
Use the following explicit tree notation for the intermediate representation.
C(k) is a constant, R(r) a processor register, M(e) a word load from memory address e, ADD addition, and SHL(e,2) a left shift by two bits. MOVE(destination,source) assigns either a processor register or the computer memory word designated by M. Parentheses show complete tree structure, so these displays specify trees rather than already shared Directed acyclic graphs.Let be the constant base memory address of the global array
a, and the constant offset of local i from the frame pointer fp. Each integer occupies four bytes, and array indices are zero-based; a fixed nonzero lower bound can instead be absorbed into . Local t is represented by R(t). After replacing multiplication by four with a left shift, the three trees are:MOVE(R(t),
M(ADD(C(A),
SHL(M(ADD(C(A),
SHL(M(ADD(R(fp),C(I))),2))),2))))
MOVE(M(ADD(C(A),
SHL(M(ADD(C(A),
SHL(M(ADD(R(fp),C(I))),2))),2))),
M(ADD(C(A),SHL(M(ADD(R(fp),C(I))),2))))
MOVE(M(ADD(C(A),SHL(M(ADD(R(fp),C(I))),2))),
R(t))Every
M(ADD(R(fp),C(I))) loads i; the next enclosing M loads a[i]; the outermost M in the first tree loads a[a[i]]. Each array index is converted to a byte offset by a left shift of two bits. The address rule used in all three trees is The source language's array bounds checks are omitted as requested. 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