Test by Bikram | Mock GATE | Test 1 | Question: 9

Question

Consider the following program fragment :

int foo(int ,int );
int a;
int ar[4] = { 1,0, 2,3};
main()
{
a = 0;
foo(ar[a], ar[ar[a]]);
printf("%d,%d ,%d,%d ", ar[0] , ar[1] ,ar[2] , ar[3]);
}


foo(int x, int y)
{
x= x+1;
y=y+1;
x=x+1;
y=y+1;
ar[1] =50;
}

What value will be printed by the program if parameter passed by call by name?

• A. $1, 50, 2, 3$
• B. $3, 50, 2, 3$
• C. $3, 50, 3, 4$
• D. $3, 2, 2, 3$

Comments

Parameter passing is not in syllabus right? Wasnt it removed from 2016? I mean its not wrong to know, just confirming...

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@Bikram sir, i got A as answer , also this showing  , please verify 

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@shivam001 C does not use call by name.

Answer 1

Call by name is same as macro

In this question, ar[a] corresponds to x & ar[ar[a]] corresponds to y.

Now, x=x+1;

That means ar[a]=ar[a]+1;

                    ar[0]=ar[0]+1;

                    ar[0]=1+1;

                    ar[0]=2;

            Now y= y+1;

                   ar[ar[a]]=ar[ar[a]]+1;

                   ar[ar[0]]=ar[ar[0]]+1;

                   ar[2]=2+1;

                    ar[2]=3;

              Now, again x=x+1;

                        ar[0]=ar[0]+1;

                    ar[0]=2+1;

                    ar[0]=3;

Now y= y+1;

                   ar[ar[a]]=ar[ar[a]]+1;

                   ar[ar[0]]=ar[ar[0]]+1;

                   ar[3]=ar[3]+1;

                    ar[3]=3+1;

                     ar[3]=4;

               ar[1]=50; (given in the query)

Therefore ,  option c matches.

Comments

Bcoz a is initialed to 0.

See

a=0;

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ok!!

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good Q

Sir when we called f(ar[a],ar[ar[a]])

at compile time will it not be resolved to f(ar[0],ar[1]) [Code Optimization Folding]

Answer 2

Ans should be option A . 
under the function call nothing will get affected other that value of ar[1] , which is made to 50 . 

so final ans should be 1, 50 , 2, 3
check the output here : http://ideone.com/P3LKiW

Comments

function foo(index, value-increment)
    sum = 0
    loop index = 1 to 3
        sum = sum + value-increment
    return sum

x = 3
foo(x, 1 / x)

@Bikram sir, @Arjun sir --- plz tell me the outpit for call by need??

I think its 3.

And when parmeters are called by reference how this 1/x thing works??

According to me in call by reference, a new value 1/x=1/3 is created and stored at some memory location whose address is passed to the calling function, because in call by reference parameters are evaluated before calling the function. Hence it will be 1/3+1/3+1/3

In call by name 1/x will be evaluated inside the function, so no new variable for 1/x is created. so it will be 1+1/2+1/3.

Am i right???

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@sushmita Exactly. You won't be getting any question wrong in this area in GATE..

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thanx a lot sir :)