C++ passing an array pointer as a function argument

Question

I'm trying to use pointers of arrays to use as arguments for a function which generates an array.

void generateArray(int *a[],  int *si){
  srand(time(0));
  for (int j=0;j<*si;j++)
       *a[j]=(0+rand()%9);
} //end generateArray;

int main() {
  const int size=5;
  int a[size];

  generateArray(&a, &size);

  return 0;
} //end main

But when I compile this this message appears:

cannot convert `int (*)[5]' to `int**' for argument `1' to `void generateArray(int**, int*)'

Answer 1

You're over-complicating it - it just needs to be:

void generateArray(int *a, int si)
{
    for (int j = 0; j < si; j++)
        a[j] = rand() % 9;
}

int main()
{
    const int size=5;
    int a[size];

    generateArray(a, size);

    return 0;
}

When you pass an array as a parameter to a function it decays to a pointer to the first element of the array. So there is normally never a need to pass a pointer to an array.

Answer 2

int *a[], when used as a function parameter (but not in normal declarations), is a pointer to a pointer, not a pointer to an array (in normal declarations, it is an array of pointers). A pointer to an array looks like this:

int (*aptr)[N]

Where N is a particular positive integer (not a variable).

If you make your function a template, you can do it and you don't even need to pass the size of the array (because it is automatically deduced):

template<size_t SZ>
void generateArray(int (*aptr)[SZ])
{
    for (size_t i=0; i<SZ; ++i)
        (*aptr)[i] = rand() % 9;
}

int main()
{    
    int a[5];    
    generateArray(&a);
}

You could also take a reference:

template<size_t SZ>
void generateArray(int (&arr)[SZ])
{
    for (size_t i=0; i<SZ; ++i)
        arr[i] = rand() % 9;
}

int main()
{    
    int a[5];    
    generateArray(a);
}

Answer 3

You do not need to take a pointer to the array in order to pass it to an array-generating function, because arrays already decay to pointers when you pass them to functions. Simply make the parameter int a[], and use it as a regular array inside the function, the changes will be made to the array that you have passed in.

void generateArray(int a[],  int si) {
    srand(time(0));
    for (int j=0;j<*si;j++)
        a[j]=(0+rand()%9);
}

int main(){
    const int size=5;
    int a[size];
    generateArray(a, size);
    return 0;
}

As a side note, you do not need to pass the size by pointer, because you are not changing it inside the function. Moreover, it is not a good idea to pass a pointer to constant to a parameter that expects a pointer to non-constant.