Codex @codex  0

 Incoming links: Dynamic link

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.

 Read the full article

Dynamic link 2026-10-31

 View more

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.

 Read the full article

Past exam of the computer science course of the University of Oxford / 2015 / Compilers / 2 / a / Solution 2026-10-31

 View more

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.

Figure 1.
The call stack during `q`, with dynamic links on the left and static links on the right; arrows point to older stack frames
. **Details**.

The dynamic chain is

q ⟶ s ⟶ r ⟶ p ⟶ main .

(1)

The individual static links are

q ⟶ p, s ⟶ r, r ⟶ p, p ⟶ main .

(2)

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.

 Read the full article

Past exam of the computer science course of the University of Oxford / 2015 / Compilers / 2 / c / Solution 2026-10-31

 View more

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 r

Immediately 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,

SL_{q} = frame_{p}, y = t = 3.

(1)

An implementation that places the hidden static link after the ordinary argument could use a different ordering of these same operations.

 Read the full article

Past exam of the computer science course of the University of Oxford / 2015 / Compilers / 2 / e / Solution 2026-10-31

 View more

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

h

of active calls, not by lexical nesting depth. Recursion in procedures unrelated to the desired enclosing declaration can make

h

arbitrarily large. Thus a single nonlocal variable access can take

O (h)

time complexity, whereas a conventional static chain access needs only

d

traversals for lexical distance

d

Instead give each procedure

P

a known global pointer cell current[P]. On entry to

P

, 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

P

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

m

procedures and

h

active calls, the procedure-indexed table needs

O (m)

additional global words and one saved previous pointer per active stack frame, hence

O (h)

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

m

-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

O (d)

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

d

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,

T_{access} = O (1), T_{entry/return} = O (1), S_{environment} = O (m + h) .

(1)

 Read the full article

Past exam of the computer science course of the University of Oxford / 2015 / Compilers / 2 / f / Solution 2026-10-31

 View more

No: a procedural parameter can carry an older lexical environment past a newer recursive activation. Here is typed, Pascal-like pseudocode; a function-valued parameter is represented by a programming closure, comprising code and its defining environment.

procedure dummy(): integer;
begin return 0 end;

procedure P(n: integer; f: procedure(): integer): integer;
    var x: integer;
    procedure Q(): integer;
    begin return x end;
begin
    x := n;
    if n = 1 then return P(2, Q)
             else return f()
end;

begin print P(1, dummy) end.

Call the first P activation

P_{1}

. It sets its own x to

1

and passes the programming closure

(Q, P_{1})

into a recursive activation

P_{2}

. That activation sets its own x to

2

and calls its parameter f. The resulting activation of Q has its static link pointing to

P_{1}

, because that is where the passed procedure was defined. Its dynamic link points to

P_{2}

, because that is where it was called.

During this call the dynamic chain contains Q, P2, P1, main, while Q's static chain contains Q, P1, main. The newest surviving activation of P is

P_{2}

, but the static chain deliberately uses

P_{1}

. Correct lexical scope therefore prints

1

; blindly consulting current[P] would produce

2

. Both enclosing activations remain alive, so no issue of a returned procedure or dangling pointer is involved.

The failed step in part (d) is precisely the assumption that the callee's lexical environment is a suffix of the direct caller's static chain. A passed programming closure can specify a different, still surviving environment. An implementation supporting such calls must preserve that environment; a table implementation requires additional environment installation and restoration rather than merely consulting newest activations.

 Read the full article