Question

Consider the following program. program main int x; proc A; int x; proc B; { x:= 3; call C; x := x + 1; print x; } (end B) proc C; int x; { x:= 1; call D; print x; call F; print x; } (end C) { x:= 2; call B; x :=x+1; print x; } (end A) proc D; proc E; int x; { x:= 5; call F; print x; } (end E) { x:= x +1; print x; call E; x := x + 1; print x; } (end D) proc F; { print x; x:=x+2; } (end F) { x:= 1; call A; print x; } (end main) (a) What is the program output assuming static scoping rules are employed? (b) What is the corresponding output when dynamic scoping rules are employed?

          Consider the following program.
program main
int x;
proc A;
int x; proc B; { x:= 3; call C; x := x + 1; print x; } (end B)
proc C; int x; { x:= 1; call D; print x; call F; print x; } (end C)
{ x:= 2; call B; x :=x+1; print x; } (end A)
proc D;
proc E;
int x;
{ x:= 5; call F; print x; } (end E)
{ x:= x +1; print x; call E; x := x + 1; print x; } (end D)
proc F; { print x; x:=x+2; } (end F)
{ x:= 1; call A; print x; } (end main)
(a) What is the program output assuming static scoping rules are
employed?
(b) What is the corresponding output when dynamic scoping rules are
employed?

Consider the following program.
program main
int x;
proc A;
int x; proc B; x:= 3; call C; x := x + 1; print x; (end B)
proc C; int x; x:= 1; call D; print x; call F; print x; (end C)
x:= 2; call B; x :=x+1; print x; (end A)
proc D;
proc E;
int x;
x:= 5; call F; print x; (end E)
x:= x +1; print x; call E; x := x + 1; print x; (end D)
proc F; print x; x:=x+2; (end F)
x:= 1; call A; print x; (end main)
(a) What is the program output assuming static scoping rules are
employed?
(b) What is the corresponding output when dynamic scoping rules are
employed?

Added by Gloria E.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

03/16/2023

Step 1

(a) For static scoping: Show more…

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Consider the following program: program main int x; proc A; int x; proc B; { x := 3; call C; x := x + 1; print x; } (end B) proc C; int x; { x := 1; call D; print x; call F; print x; } (end C) { x := 2; call B; x := x + 1; print x; } (end A) proc D; proc E; int x; { x := 5; call F; print x; } (end E) { x := x + 1; print x; call E; x := x + 1; print x; } (end D) proc F; { print x; x := x + 2; } (end F) { x := 1; call A; print x; } (end main)