Consider the following program

Question

int i = 1;
int main()
{
    int a[]= { 0,1, 2} ;
    f(a[i], i);
    printf("%d", a[i]);
}

void f(int x, int y)
{
    y++;
    x=5*i;
}

In above function f() uses " call by name" technique, what is the output printed?

a) 2        b) 10           c) 5         d) 1

Comments

@Isha-karn which place are these parameter passing questions taken from?

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@shyam you are just taking this call by text l to be same as call by value.both are different!!!

http://www.cs.rit.edu/~rpj/courses/plc/lab4/lab47.html#Q12

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@Arjun Sir these questions are taken from test papers of a coaching institute.

Answer 1

Using a C style creates confusion for call by name. The scope of variables comes from the function where it is called (dynamic scoping) in call by name.

y++; i becomes 2

x = 5*i; a[2] becomes 10.

So, 10 should be printed.

Answer 2

The output printed is 1 i.e., option d). The function f() doesn't return any value neither does it make changes in the original value of a[i] and hence no matter what goes inside the body of f(), the output will be the initial value of a[i] i.e., 1.

Answer 3

option b .

as per concept of macro implementation in c,

http://www.cs.rit.edu/~rpj/courses/plc/lab4/lab47.html#Q12

Comments

Can you please explain how is above code snippet related to Macros in C?

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i have given a link regarding this.

macros also just substitutes the name with it's value wherever it is used in the program.

call by text/name does the same .

in c++ inline functions does the same by just putting  the function body at every place where it is called.

Answer 4

D should be the answer. this is call by value so a[i] is unchanged.

Comments

In question it says call by name.