Past exam of the computer science course of the University of Oxford / 2015 / Compilers / 2 / f

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 $\boxed{1}$; 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.