Question

Having an issue with pointers and functions.

#include <stdio.h>

int f(int *a, int *b);

int main()

{
int a = 2, b = 7;
b = f(&b, &a);
printf("a = %d,\n", a);
printf("b = %d\n", b);
return 0;
}
int f(int* a, int* b) {
    (*a) = (*a) + 3;
    (*b) = 2*(*a) - (*b)+5;
    printf("a = %d b = %d\n", *a, *b);
    return(*a)+(*b);

}

can someone explain to me why the output is

a = 10 b =23

a = 23

b = 33

I understand that (&b, &a) flips the numbers in the function but why does b = 23 and not 13 ?

Answer 1

Ans :

int main()

{
int a = 2, b = 7;
b = f(&b, &a);
printf("a = %d,\n", a);
printf("b = %d\n", b);
return 0;
}

in that function initailly value of a=2 and b=7

and when we calling the function f(&b, &a); we are passing the reference of the variable. It is call by reference type function. inside that function we are passing the address of the variable a and b.

so in the first arguement we are passing reference(address) of b and in the second arguement, address of a is passed.

  Inside the function f

int f(int* a, int* b) {
    (*a) = (*a) + 3;
    (*b) = 2*(*a) - (*b)+5;
    printf("a = %d b = %d\n", *a, *b);
    return(*a)+(*b);

}

we are getting the address of b into a and address of a into b.

So here address of variable in the memory are flipped.

In that function currently a is holding value 7 and b is holding value 2.

in the expression     (*a) = (*a) + 3; will be evaluated as (*a)=7+3=10;

so value of (*a) is 10.

In the next expession (*b) = 2*(*a) - (*b)+5; here (*b)=2

=> (*b) = 2 * 10 - 2 +5 =23

so    printf("a = %d b = %d\n", *a, *b); will print    a =10 b =23. inside the function these are local variable for the function.

This statement return(*a)+(*b) will return 10 +23=33

So  return(*a)+(*b); will return 33

and that value is hold by b in the main function b = f(&b, &a); so here value of b will became 33.

and value of a will became 33.

Handwritten detailed solution :