CODEWITHSUNDEEP
cfunctionpointersparameterssubroutine
AlexT
AlexT

Reputation: 609

C - Passing pointers as parameters

Here's what I've coded:

#include <stdio.h>

void add(int *a, int i){
    while(i != 0){
        a++;
        i--;
    }
}

int main(){
    int a = 6;
    add(&a, 4);
    printf("a = %d\n", a);
    return 0;
}

And here's the result

a = 6

The result I'm expecting is 10, of course, but I don't understand why a isn't modified when I pass it as a pointer in my add function...

Upvotes: 1

Views: 821

Answers (3)

Franck Horlaville
Franck Horlaville

Reputation: 29

AntonH's comment above is the right answer.

I'll expand it:

void add(int *a, int i){
    while(i != 0){
        a++;
        i--;
    }
}

You have defined 'a' as a pointer to an int (int*). Anything you do on 'a' such as '++'ing it is done on the pointer, not on the pointed value.

You should, therefore, replace a++ with (*a)++ to increase "what is pointed by a"

Upvotes: 2

Seek Addo
Seek Addo

Reputation: 1903

You function should be like this

void add(int *a, int i){
    while(i != 0){
        (*a)++; // deference the value and increment it.
        i--;
    }
}

with this

/* move the pointer to the next location */

a++;

Upvotes: 2

Vlad from Moscow
Vlad from Moscow

Reputation: 311068

In the function there is advanced the pointer itself instead of incrementing the value pointed to by the pointer. Change the function the following way

void add(int *a, int i){
    while(i != 0){
        ( *a )++;
        i--;
    }
}

Or the expression statement ( *a )++; can be substituted for ++*a;

Also the function could be more safe if the second parameter had type unsigned int. Otherwise you have to check the sign of the parameter.

Upvotes: 4

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