Why has this function no effect when called?

Question

I'm a newcomer C learner and following code is my problem:

I wrote it to change the value of the passed variable, but in the output notihng has happened.

Here is my code:

#inlcude <stdio.h>

void change(int i, int j) {
    i = i + j;
}

int main() {
    int a = 50;

    printf("a before = %d\n", a);

    change(a, 10);

    printf("a after = %d\n", a);
    return 0;
}

What is the reason for this error?

Answer 1

This is a very simple case in C.

In C, It uses a method called Call-By-Value to evaluate function parameters. It means when you called a function with your arguments the called function receives the values of those arguments not the arguments.

Which in this program, function "change" receives integer 50 (value the 'a' is holding) and 10 instead of a and 10. So i is 50 and j is 10 when you called the function. In the function i is changed to 60. But a remains the same. If you add a printf() to print the value of i in "change" you can see the change happened.

So to make "change" effective you have to use pointers, a special "tool" in C.

you have to change your "change" function as following:

void change(int *pi, int *pj)
{
    *pi += *pj;
}

and you should change the call also.

int main()
{
    int a = 50, b = 10;
    /* printf line */
    change(&a, &b);
    /* printf line */
}

in the function "change"

   void change(int i, int j)

is changed into

   void change(int *pi, int *pj)

this means that pi, pj are pointers to int. pi & pj are variables to hold an address of a variable.

*pi & *pj are the variable that resides in the addresses that pi & pj have. Which means that *pi refers to (points to) a, *pj points to b

   *pi += *pj

is equal to

   a = a + b;

and you can see that I have changed way "change" called

it is now

   change(&a, &b);

instead of

   change(a, 10);

the ampersand (&) gives the address of a variable So &a and &b are the addresses of a & b, Which are meant to be hold in pi and pj

I think this lengthy answer would help you. However you should read some lessons about pointers.

Answer 2

Function parameters are its local variables. Function parameters are initialized by copies of the values of the supplied arguments. So any changing of a parameter does not influence on the corresponding argument. After exiting the function its parameters (local variables) are destroyed.

For example you can imagine your function

void change(int i, int j) {
    i = i + j;
}

that is called like

change(a, 10);

the following way

void change(/*int i, int j*/) {
    int i = a;
    int j = 10;

    i = i + j;
}

As you can see variable a itself will not be changed.

There are two approaches. Either the first parameter will be passed by reference or the function will return its result. For example

#inlcude <stdio.h>

void change( int *i, int j) {
    *i = *i + j;
}

int main() {
    int a = 50;

    printf("a before = %d\n", a);

    change( &a, 10);

    printf("a after = %d\n", a);
    return 0;
}

Or

#inlcude <stdio.h>

int change(int i, int j) {
    i = i + j;

    return i;
}

int main() {
    int a = 50;

    printf("a before = %d\n", a);

    a = change(a, 10);

    printf("a after = %d\n", a);
    return 0;
}

Answer 3

The variable i and j in change() are copies from the variables inside main(). You have to learn about pointers to achieve what you are trying to do. Basically it look like this:

#include <stdio.h>

void change(int *i, int j) {
    *i = *i + j;
}

int main() {
    int a = 50;

    printf("a before = %d\n", a);

    change(&a, 10);

    printf("a after = %d\n", a);
    return 0;
}