C Program: Update unsigned char pointer array with function

Question

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.

P.P · Accepted Answer · 2015-06-11 15:15:22Z

2

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.

edited Jun 11, 2015 at 15:15

answered Jun 11, 2015 at 15:09

P.P

122k20 gold badges193 silver badges251 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Sourav Ghosh Over a year ago

Sir, if i'm not wrong, passing blah itself will do the job, pointer to pointer is not necessary. Can you please review my answer once for correctness?

P.P Over a year ago

It depends. I did the way I did because OP passes a &blah and updating the pointer with *arr=ptr;, which is typically how pointer is updated. The way you did will modify the array ptr2. So that's also fine if OP wanted to modify the array ptr2.

Sourav Ghosh Over a year ago

Sir, based on the usage of the modified array, I came to my conclusion that it's only the values that matter to OP, that's why I suggested so. Your approach is also absolutely correct also, no doubts. :-)

Sourav Ghosh · Accepted Answer · 2015-06-11 15:19:34Z

1

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);

edited Jun 11, 2015 at 15:19

answered Jun 11, 2015 at 15:09

Sourav Ghosh

135k17 gold badges192 silver badges271 bronze badges

Collectives™ on Stack Overflow

C Program: Update unsigned char pointer array with function

2 Answers 2

3 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related