Question

Consider the following pseudocode, assuming nested subroutines, lexical scope, and that local variables (including formal parameters) are stored in the stack. procedure main() g: integer procedure B(a: integer) x: integer procedure A(n: integer) g := n procedure R(m: integer) write_integer (x) x := x/2 -- integer division if x > 1 R(m + 1) else A (m) --body of B x := a * a R(1) --body of main B(3) write_integer(g) (a) What does this program print? (b) Show the frames on the stack when A has just been called. For each frame, show the static and dynamic links as well as the values of the local variables. (c) Explain how A finds g. Use RISC code as you did for homework 4. Use the stack frame layout as defined on page 42 in lecture 9.

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Answer 1

Sel: (a) This program will print 9 4 2 3. (6) The language lises nested Subroutines and static scoping → The frames are to be shown on the stack when A has been called. A(3) * 3 R13) m 2 Dynamic R(2) m2 static links links R(1) ma B(3) 2 Kall connected to b] Main of gok @ A finds of by first looking to see if it is defined locally. Since it is not, A troverses its static link to Bi . g is not defined in B, it traverses B's static dink to Main, where g is defined. As per HomeworkLib sule, I have written the answer properly.