C Program: Update unsigned char pointer array with function

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I have a function to update an unsigned char* and cannot find where my bug is. I'm not sure if I need to allocate memory, or if I am pointing to the wrong memory space somewhere. I tried to follow a similar structure as posted here, but have not had success with an unsigned char.

My code so far:

#include <stdio.h>

void changeArray(unsigned char **arr)
{
    unsigned char ptr[3] = {100, 101, 102};
    *arr = ptr;
    printf("%d\n", **(arr+0)); // This prints out the correct value of 100
}

int main(int argc, const char* argv[])
{
    int i = 0;
    unsigned char *blah;
    unsigned char ptr2[3] = {103, 104, 105};
    blah = ptr2;
    printf("Blah is: \n");
    for (i = 0; i < 3; i++) {
        printf("%d,",*(blah+i)); //This prints out 103,104,105
    }
    changeArray(&blah);
    printf("Blah is now: \n");
    for (i = 0; i < 3; i++) {
        printf("%d,", *(blah +i)); //This prints out 0,0,0
    }
    return 0;
}

Any help in determining how to properly access the values set in the changeArray() function would be greatly appreciated.

2

There are 2 answers

3
P.P On BEST ANSWER

With this *arr = ptr; you are storing a pointer to a variable with automatic storage duration. The behaviour undefined.

You can dynamically allocate and return a pointer that way:

void changeArray(unsigned char **arr)
{
    unsigned char ptr[3] = {100, 101, 102};
    unsigned char *p = malloc(sizeof ptr); 
    memcpy(p, ptr, sizeof ptr);
    *arr = p;
    printf("%d\n", **(arr+0)); // This prints out the correct value of 100
}

You should also do error checking if malloc failed and remember to free the allocated memory after use in main.

0
Sourav Ghosh On

The problem here is, ptr is local to changeArray() function. So once the function finishes execution, there is no existance of ptr. Hence, once you assign ptr to *arr

*arr = ptr;

and changeArray() execution gets over, accessing blah in main() now will invoke undefined behaviour.

FWIW, you don't need to pass the address of blah, you don't need a pointer-to-pointer at all. blah is already a pointer, which you can pass to changeArray() to alter the contents of the memory area it points to. You can do something like

void changeArray(unsigned char *arr)
{
     for (int i = 0; i < 3; i ++)
        arr[i] = 100+i;
}

and call it like

changeArray(blah);