Problem how to calculate size of "Array"

[h4ever]

From different languages I am used to arrays, here in C/C++ it is a bit harder. Can you help? I am writing C code now. I have file which uses openCL to generate program for GPU. Now, there is a part of code which detects some informations about how many devices there is in the machine (I dont know how this exactly works. The complete code is long so I paste the code here:
http://paste.ofcode.org/U5g2Z3Cj5ZuRBnJf2b5cqv
and here the key part - this is how they use it:

 size_t nDevices;
 size_t * binary_sizes;
 char ** binaries;
 ...
    err = clGetProgramInfo( program, CL_PROGRAM_NUM_DEVICES, sizeof(nDevices), &nDevices, NULL );
 binary_sizes = (size_t *)malloc( sizeof(size_t)*nDevices );
 ...
binaries = (char **)malloc( sizeof(char *)*nDevices );
 ...

 int i;
    for( i = 0; i < nDevices; i++ )
    {
        if( binary_sizes[i] != 0 )
        {
            binaries[i] = (char *)malloc( sizeof(char)*binary_sizes[i] );
        }
        else
        {
            binaries[i] = NULL;
        }
    }

Now I am writing the code to write the binaries data, and this is where I need to find out how to calculate the count or "how many times the loop should be repeated". So if you have two GPU devices in machine (as far as I understand openCL) the binaries data should contain two sets of data for each GPU device. So I have this:

void write(char ** binaries, size_t nDevices){
    char str[1000];
    int n;
    for( n = 0; n < nDevices; n++ )
    {
    FILE* f = fopen("input.bin", "wb");
    fseek(f, 0, SEEK_SET);
    int i;
    for (i = 0; i < jel_size; i++)
            fwrite(&foo[i], sizeof(struct foo), 1, f);
    fclose(f);
    }
}

Also I need advice how to turn the n to valid filename for fopen.

[Duthomhas]

http://www.cplusplus.com/faq/sequences/arrays/sizeof-array/

[h4ever]

Thanks for article, very good, but can you explain me why in my code, there is:

char ** binaries; // Why two pointers here? Why pointer to pointer?
binaries = (char **)malloc( sizeof(char *)*nDevices );

[Duthomhas]

Any time you see type* you are looking at a pointer to type, which can be treated as if it were an array of type.

#include <iostream> 

void print5( int* xs )  // xs is a pointer to an integer (which we'll treat like an array)
{
  for (int n = 0; n < 5; n++)
    std::cout << xs[ n ] << "  ";
  std::cout << "\n";
}

int main()
{
  int a[] = { 1, 2, 3, 4, 5 };

  print5( &(a[0]) );  // pass as argument the address to the first element of a[]
}

Any time you see type** you are looking at a pointer to a pointer to type.

Now, the array thing again: you may have an array of pointers to type, or a pointer to an array of type, or an array of arrays of type.

void print2( int** xs )  // xs here is a pointer to pointer to int, or an array of pointers to int...
{
  for (int n = 0; n < 2; n++)
    print5( xs[ n ] );
}

int main()
{
  int a[] = { 1, 2, 3, 4, 5 };
  int b[] = { 6, 7, 8, 9, 10 };

  int* xs[] = { &(a[0]), &(a[1]) };

  print2( &(xs[0]) );  // the address of the first element of a[] is a pointer to a pointer to int.
}

Hope this helps.