Question

Need to have an array full of random numbers within range. Code:

void fillArray(int *arr){
        for(int i=0;i<(sizeof arr);i++){
            arr[i] = rand() % 2 - 2;
            }
        }

int *arrPoint(int *arr, int max){
    for (int i=0;i<max;i++){
        printf("%d is %d\n",i,arr[i]);
    }
    return arr;
}

int main(int argc, char *argv[]){
    srand ( time(NULL) );
    int arr_f[15];
    fillArray(arr_s);
    arrPoint(arr_s, 15);
    system("PAUSE");
    return EXIT_SUCCESS;
}    

Output:

    0 is -2
1 is -1
2 is -2
3 is -1
4 is 0
5 is 0
6 is 4633240
7 is 2686652
8 is 1973724226
9 is 1974338216
10 is 2686716
11 is 1973744850
12 is 8
13 is 1973752206
14 is 1973752162
Press any key to continue . . .

What the hell? Putting rand() % 2 into brackets doesn’t help either. What these numbers are and how do I get rid of them?

P.S. Tried this in crappy Dev-C++ and Code::Blocks with the same result. Need the program to be small sized (putting it to dropbox), so no, I can’t use 100mb boost lib.

Answer 1

void fillArray(int *arr){
    for(int i=0;i<(sizeof arr);i++){

Since arr is a pointer to an integer, sizeof arr is equivalent to sizeof (int *), which is apparently 4 (32-bits) on your platform. That’s clearly not what you want. You only pass fillArray a pointer to the first element of the array.

If you need the number of elements in the array a pointer points to, you need to pass that information. The C and C++ languages provides no way to tell how many bytes a pointer points to given just the pointer.

You do it correctly in arrPoint. Do it that way in fillArray.