Question

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Declaring a C function to return an array

I am new to C, and need to your thoughts to help me to return the result array from the following function:

void getBase(int n, int b)
{
    const size_t SIZE = 32;
    char arr[32+1]={0}; int digits=SIZE, i;
    char* ptr = arr;
    while (n > 0)
    {
        int t = n%b;
        n/=b;
        arr[--digits] = numbers[t];
    }
    while ( *ptr == '\0') ptr++;
    // NEED To return a ref to `ptr`
}

My solution:

void getBase(int n, int b, /*send some  array as a parameter*/ char* str)
{
    const size_t SIZE = 32;
    char arr[32+1]={0}; int digits=SIZE, i;
    char* ptr = arr;
    while (n > 0)
    {
        int t = n%b;
        n/=b;
        arr[--digits] = numbers[t];
    }
    while ( *ptr == '\0') ptr++;

    /* and use strcpy ... perhaps memcpy if non-string )*/
    strcpy(str, ptr);
}

I need further ideas….

Thanks.

Answer 1

Your solution looks fine.

Instead, you don’t even need the local arr array at all. You can just write directly into str:

EDIT : Cleaned up and working version.

const char numbers[] = "0123456789abcdef";

void getBase(int n, int b, char* str)
{
    const size_t SIZE = 32;
    int digits=SIZE;
    while (n > 0)
    {
        int t = n%b;
        n/=b;
        str[--digits] = numbers[t];
    }

    int length = SIZE - digits;

    memmove(str,str + digits,length);
    str[length] = '\0';
}

You just have to make sure that your str is large enough to avoid an array-overrun.

---

int main(){

    char str[33];

    getBase(684719851,10,str);

    printf(str);

    return 0;
}

Output:

684719851