GATE IT 2004 | Question: 59

Question

What is the output of the following program?

#include<stdio.h>
int funcf (int x);
int funcg (int y);
main ()
{
    int x = 5, y = 10, count;
    for (count = 1; count <= 2; ++count) {
        y += funcf(x) + funcg(x);
        printf ("%d", y);
    }
}
funcf (int x) {
    int y;
    y = funcg(x);
    return (y);
}
funcg (int x) {
    static int y = 10;
    y += 1;
    return (y + x);
}

• A. $43 \ 80$
• B. $42 \ 74$
• C. $33 \ 37$
• D. $32 \ 32$

Comments

Some random information. 

If someone is thinking they are trying to test associativity of operator '+' then notice following line in @Arjun ji's answer.

 even if the order of call is reversed, result will be same. 

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given , y+ = funcf(x) + funcg(x);

as count = 1 , count can will work upto two iterations , so two times this iteration will take place 2 times.

as addition is commutative , any function can be evaluated first and even if reverse the result will remain the same. lets suppose funcf(x) is called first.

(1) 1st  Iteration :-

for funcf(x), funcg(x) is called,

static y = 10;

y = y+1 = 10+1 = 11;

as static variable y value will be updated to 11.

return ( y+x)

y+x = 11 + 5(global variable from main) = 16

returning this value to funcf(x);

y = 16 which will be returned to funcf(x) in main .

now , funcg(x) ,

y = 11

y = y+1 = 12 . return ( y+x) = 12 + 5 = 17.

for funcg(x) value will be return 17 .

so , in main function ,

y+= funcf(x) + funcg(x);

y =  y + funcf(x) + funcg(x) = 10 ( local) + 16 + 17  = 43. so , 43 will be printed as result of 1st iteration and value of y will be updated as 43.

(II) Iteration 2:-

again , for funcf(x) , funcg(x) will  be called first

so , as y was 12 .

y = 12

y = y +1 = 12 + 1 = 13. y value will be updated as 13.

 return (y+x)

(y+x) = 13 + 5 = 18;

so funcf(x) value will return as 18 in the main function.

now, funcg(x) ,

y = 13

y = y + 1 = 13 + 1 = 14;

return ( y+x)

y+x = 14 + 5 = 19;

funcg(x) value will be returned as 19 in main function.

now , in the main function,

y = y + funcf(x) + funcg(x) = 43 + 18 + 19 = 80.

so 80 will be printed and updated as value of y.

so , answer is (a) 43 , 80

Answer 1

funcf(x) + funcg(x)

$funcf$ or $funcg$ can be executed first as whether the first operand or the second operand to '+' operator is first executed is not defined in C language and it depends on the compiler implementation. But here the order does not matter as both the operands are not affecting each other and '+' is commutative. Lets assume $funcf$ is executed first. It calls $funcg$ - so even if the order of call is reversed, result will be same. 

In first call of $funcg$, $y$ becomes $11$ and it returns $5+11 = 16$.

In second call of $funcg$, $y$ becomes $12 $ and it returns $5+12 = 17$.

So, in main $y$ is incremented by $16+17 = 33$ to become $10+33 = 43$. (Choice A)

In the second iteration $y$ will be incremented by $18+19 = 37$ to give $43+37 = 80$.

Comments

@vivekraj no, x and y are not global , all are local , but y of funcg is static so it will be intialized one one time and whenever a call to this fucnction it will remember its previous value but still its local to funcg

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why we r using x value in funcg for adding every time .

ok y is static.

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I got it @meivinay.

silly mistake

Answer 2

 The count=1 and it goes till two,so following statement will be executed twice.

 y += funcf(x) + funcg(x); 

1st call- funcg(x);   // y = 11        y+x=  16.

2nd call funcg(x); // y= 12         y+x=  17.

First iteration-> main()->y = 16+17 +10 = 43

 

Second iteration-> main() y= 18+19 +43 =80

So the Answer is A

Answer 3

So option A is correct.