CODEWITHSUNDEEP
c
user3797829
user3797829

Reputation: 403

What will be the output of this?

I was doing this question and I have a doubt.

#include <stdio.h>
int main(void)
{
    int fun (int);
    int i=3;
    fun(i=fun(fun(i)));
    printf("%d\n",i);
    return 0;
}

int fun ( int i )
{
    i++;
    return(i);
}

I have a doubt when it gets to 

fun ( i = 5 )

What happens with this? Will the value of i go to 6 or it will be 5. According to me, it should be 6. But that is not the correct answer.

Upvotes: 0

Views: 116

Answers (3)

Serge Wautier
Serge Wautier

Reputation: 21888

The result of the call to fun() is not assigned to i. Therefore 5 is expected, not 6.

Upvotes: 1

Cahit Gungor
Cahit Gungor

Reputation: 1497

This is related to scope. In the function scope, variables defined in that scope or the parameters pass in does not any effect on outer scope variables, unless it is a pointer. So, the output of fun will be assigned to i in the fun(i = 5) but the internal operations of fun, do not effect the outer scope i. So it stays as it is before fun last call. The output is 5.

Upvotes: 1

derekerdmann
derekerdmann

Reputation: 18252

In C, parameters are passed by value. The variable i in the main function is actually different from the i inside fun(), because its value is copied when it is passed into the function.

When you call i = fun(fun(i)), 5 is isassigned into i in the main function. However, the call to fun(5) that returns 6 does not assign its result back into i, leaving it unchanged. When the output is printed, i is still 5.

Upvotes: 6

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