GATE CSE 1987 | Question: 1-xix

Question

Study the following program written in a block-structured language:

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.

What will be printed by the write statements marked $(1)$ and $(2)$ in the program if the variables are statically scoped?

• A. $3, 6$
• B. $6, 7$
• C. $3, 7$
• D. None of the above.

Answer 1

Using Static Scoping:
First, procedure Q is called from the main procedure. Q has local variables x and y with values 3 and 4 respectively. This local variable y (value 4) is being passed to procedure P during call, and received in local variable n inside procedure P. Now, as P does not have any local definition for variable x, it will assign the evaluated value of (n+2)/(n-3) i.e. (4+2)/(4-3)=6 to the global variable x, which was previously 7. After the call of procedure P, procedure Q writes the value of local variable x which is still 3. Lastly, the main procedure writes the value of global variable x which has been changed to 6 inside procedure P. So, the output will be 3, 6.

Using Dynamic Scoping:
The same sequence of statements will be executed using dynamic scoping. However, as there is no local definition of variable x in procedure P, it will consider the recent definition in the calling sequence; as P is being called from procedure Q, definition of x from Q will be used, and value of x will be changed to 6 from 3. Now, when Q writes local variable x, 6 will be printed. The write global variable x from main procedure will print 7 (as value of the global variable x has not been changed). So, the output will be 6, 7.

Correct Answer: $A$

Comments

nice explanation ,deserves to be best answer

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nice explanation@sutanay3

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Nice answer

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P does not have any local definition for variable x,----it should be 

P does not have any local declaration for variable x.

Answer 2

 

In static scoping, the free variable is replaced by the global variable, which means if a variable is not defined locally it will take the value from a global variable.

So option (A) is correct.

Comments

@Vinil
Is the answer to your image 3,3 ?
Also for the above question my doubt is we can easily find x=7 in the same block, isn't it ?
So why are we printing x=6, because if we consider the memory model diagram, we can clearly see that, x = 6 is lost at line 1.
So why at lne 2 x is not considered at 7, why the answer is x=6 at line 2, can you explain with a memory model diagram, considering the memory stack.

PS : Thanks for replying

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@Hira thakur
begin
    x:=7;
    y:=8;
    Q; 
Write(x);                                   __(2) 
end.

here we have the local value x=7 and local value should be given preference than global value .So, why Write(x) is printing 6 .

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 dragonball
$x=7$ is updated to $x=6$ because in the $P$ function call when an expression is evaluated and the value is store in $x$, so $x$ is not defined there that’s why that value is updated here. Write(x) value can’t be 7 because before print there is a $Q$ function call which changes the value of $x$.