Thread: Assigning pointers to array in function

Hello!

I have this code:

Code:

#include <stdio.h>

void assignPoint(int [], int*, int*);
void printPoint(int*, int*);

int main(void){
    int array[3];
    int* p1;
    int* p2;
    assignPoint(array, p1, p2);
    printPoint(p1, p2);
}

void assignPoint(int array[], int* p1, int* p2){
    p1 = &array[1];
    p2 = &array[2];
}

void printPoint(int* p1, int* p2){
    printf("%d %d", *p1, *p2);
}

I am getting segmentation here. My question is: Do pointers keep their values after function returns or do I have to return them?

[SPOILER]

Originally Posted by jimblumberg

By default everything is passed by value (including pointers), so in your example your assignPoint() function is working with copies of the pointer so you can't change where the pointer is pointing.

By the way your compiler should be able to warn you about the problems:


If you want to change where the pointer is pointing you need something like:

Code:

#include <stdio.h>

void assignPoint(int [], int**, int**);
void printPoint(int*, int*);

int main(void){
    int array[3];
    int* p1;
    int* p2;
    assignPoint(array, &p1, &p2);
    printPoint(p1, p2);
}

void assignPoint(int array[], int** p1, int** p2){
    *p1 = &array[1];
    *p2 = &array[2];
}

void printPoint(int* p1, int* p2){
    printf("%d %d", *p1, *p2);
}

And don't forget you need to initialize the elements of the array before you get some meaningful information.


Jim

[/SPOILER]
Thanks!

Yes my compiler warn me but I wrote the code on another computer (which didnt have internet access) so I had to rewrite the code in chrome