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Official PYQ · 1987Q1-xxRuntime EnvironmentsMCQ1 mark

GATE CS 1987 Q1-xx — Runtime Environments

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For the program given below what will be printed by the write statements marked (1)(1) and (2)(2) in the program if the variables are dynamically scoped?

Var x, y:interger; 
procedure P(n:interger);
begin
     x := (n+2)/(n-3);
end;

procedure Q 
Var x, y:interger;
begin   
    x:=3;
    y:=4; 
    P(y);
    Write(x);                                __(1)
end;
 
begin
    x:=7;
    y:=8;
    Q; 
    Write(x);                                   __(2)
end.
Choose one option
View worked solution

Answer: B

Under dynamic scoping, a free variable in a procedure refers to the most recently declared active variable with that name in the execution call stack.

Step-by-Step Execution Trace:

  1. Main Block:
    • Global variables x and y are allocated.
    • x := 7 and y := 8.
    • Procedure Q is called.
  2. Inside Procedure Q:
    • Local variables x and y are allocated.
    • x := 3 and y := 4.
    • Procedure P(y) is called with argument y = 4 (so parameter n = 4).
  3. Inside Procedure P:
    • The active call sequence is: Main → Q → P.
    • The variable x inside P is free. Under dynamic scoping, it refers to the most recently declared active x, which is the local variable x of Q.
    • The assignment is computed as: x:=n+2n−3=4+24−3=6x := \frac{n + 2}{n - 3} = \frac{4 + 2}{4 - 3} = 6
    • Thus, the local variable x of Q is updated to 66.
  4. Back in Q:
    • The statement Write(x) marked as (1)(1) is executed.
    • Since the local x of Q was updated to 66, it prints 66.
  5. Back in Main Block:
    • Procedure Q finishes execution, and its local variables are deallocated.
    • The statement Write(x) marked as (2)(2) is executed.
    • Here, x refers to the global variable x, which remains unchanged at 77.

Therefore, the printed values are 6,76, 7, which corresponds to Option B.

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