Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Consider the following procedure declaration:...
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Consider the following procedure declaration:
procedure p ;
x : integer;
procedure q;
begin x = x + 1 end;
procedure r;
x : integer;
begin x := 1; q ; write(x) end;
begin
x = 2;
r;
end;
procedure p ;
x : integer;
procedure q;
begin x = x + 1 end;
procedure r;
x : integer;
begin x := 1; q ; write(x) end;
begin
x = 2;
r;
end;
The output produced by calling P ina language with static scope will be
- a)I
- b)2
- c)3
- d)4
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Consider the following procedure declaration:procedure p ;x : integer;...
Since value output by program is the value of x which is declared in side DC function, so, output is 1 using static scoping.
Most Upvoted Answer
Consider the following procedure declaration:procedure p ;x : integer;...
Procedure Declaration
The given code snippet declares three procedures: p, q, and r. The procedure p does not have any parameters, while procedures q and r both have a local variable x of type integer.
Execution Sequence
When procedure p is called, it sets the value of x to 2 and then calls procedure r. Procedure r sets the value of x to 1, calls procedure q, and finally writes the value of x.
Static Scope
In a language with static scope, the scope of a variable is determined by its position in the source code. Variables declared within a procedure are local to that procedure and are not accessible outside of it. Therefore, the variable x declared in procedure r is different from the variable x declared in procedure p.
Procedure q
When procedure q is called from procedure r, it increments the value of x by 1. However, since it is a different x variable (local to procedure q), it does not affect the value of x in procedure r.
Procedure r
Procedure r sets x to 1, calls procedure q, and writes the value of x. The value of x in procedure r remains 1 because the increment in procedure q does not affect it. Therefore, the output of calling procedure p will be 1.
Final Answer
The correct answer is option 'A': the output produced by calling procedure P in a language with static scope will be 1.
The given code snippet declares three procedures: p, q, and r. The procedure p does not have any parameters, while procedures q and r both have a local variable x of type integer.
Execution Sequence
When procedure p is called, it sets the value of x to 2 and then calls procedure r. Procedure r sets the value of x to 1, calls procedure q, and finally writes the value of x.
Static Scope
In a language with static scope, the scope of a variable is determined by its position in the source code. Variables declared within a procedure are local to that procedure and are not accessible outside of it. Therefore, the variable x declared in procedure r is different from the variable x declared in procedure p.
Procedure q
When procedure q is called from procedure r, it increments the value of x by 1. However, since it is a different x variable (local to procedure q), it does not affect the value of x in procedure r.
Procedure r
Procedure r sets x to 1, calls procedure q, and writes the value of x. The value of x in procedure r remains 1 because the increment in procedure q does not affect it. Therefore, the output of calling procedure p will be 1.
Final Answer
The correct answer is option 'A': the output produced by calling procedure P in a language with static scope will be 1.
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Consider the following procedure declaration:procedure p ;x : integer;procedure q;begin x =x + 1 end;procedure r;x : integer;begin x := 1; q ; write(x) end;beginx = 2;r;end;The output produced by calling P ina language with static scope will bea)Ib)2c)3d)4Correct answer is option 'A'. Can you explain this answer?
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