unexpected output in function calls

Question

i have a program in which i have implemented main function as below and finally i getting a unexpected value of i.

 int main()
    {
        int fun (int);
        int i=3;
        fun(i=fun(fun(i)));
        printf("%d",i);
        return 0;
    }

and my function implementation is

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

}

my output is:

5

what i expected was:

6

Answer 1

i = fun(i=fun(fun(i)));

This would give you 6

fun(i=fun(fun(i)));

This gives 5 because the last call does not assign the value to i.

Also as mentioned by Tom below i is being passed by value not reference, if you did pass by reference then it would be 6 if you did fun(fun(fun(&i))); (depending on what parameter type the function takes/returns).

Answer 2

You are passing the argument by value, and returning that value. The variable i is only modified during initialization (set to 3) and in the call to the outer fun where it takes the value 5 returned from fun(fun(3))

EDIT C++ only (before @awoodland removed that tag from the question):

If you want to modify the external variable in the function you can do so by using references:

int& fun( int & x ) {
  return ++x;
}
int main() {
   int i = 3;
   fun( fun( fun( i ) ) );
// fun( i = fun( fun( i ) ) ); // alternatively
}

Answer 3

i = 3;

fun(i=fun(fun(i)));
        +
        |
        V
fun(i=fun(fun(3)));
        +
        |
        V
fun(i=fun(4));       /* fun (3) returns 4 */
        +
        |
        V
   fun(i=5);         /* fun (4) returns 5 and it is stored in i */
        +
        |
        V
    fun(5);         /* fun (5) returns 6, but it is NOWHERE stored, i is still 5 */

print i results in 5 .

Answer 4

My guess is that your expectation comes from the i++ being convention for i = i + 1 and while that is true your issue here is scope. When you call a function in C like this add:

int add( int a, int b ) {
     a = a + b;
     return a;
}

you are passing by value. Which is to say that C is generating a duplicate of the values, thus you have a difference scope and thus things that happen inside of add only affect the things inside of add.

The other method you can pass data around in C is by reference like this

int * add( int * a, int b ) {
   (*a) = (*a) + b;
   return a;
}

that function will mutate the memory pointed to by a* and thus is "violating" it's scope. Using pass by reference you could have your function act in the manner you expected.