What does random_ints(a,N) do and how do I use it in my code?

Question

http://developer.download.nvidia.com/CUDA/training/GTC_Express_Sarah_Tariq_June2011.pdf In the above tutorial (slide 29), they initiate 3 pointers to ints: ``` int *a, *b, *c; ``` Clearly this is of type (int *), yet they somehow make it possible for the kernel to access its indices with syntax `a[index]` They also use some (to me) unknown command to initialize their values: ``` a = (int *)malloc(size); random_ints(a, N); ``` So what does this command do? First it casts the pointer *a to point to an int (but later on `a` magically becomes a vector). I can't find any sources on what random_ints precisely does (and my compiler doesn't recognize it either because it probably requires some include). I guess it makes `a` a vector of length N with random ints (though `a` is of type `int`). I tried working around this by doing the same thing with `vector <int> * a;` etc etc but I still have trouble passing that to my kernel (it won't add the elements no matter what I try). I'm working in C++…

Accepted Answer

- You can use pointer in C/C++ in the way like normal array. See:

Example

```
int* p = new int[2];
p[1] = 2;  
*(p + 1) = 2; // same as line above - just other syntax
*(1 + p) = 2; // and other way
 1[p] = 2; // don't use it - but still valid
```

- You can allocate memory by malloc in C++ (it is derived from C way) but only for POD types, like `int`

Example

```
int* p = (int*)malloc(12 * sizeof(int)); // C way equivalent to C++: new int [12]
```

Be aware that you must free this pointer by `free(p)`, not by `delete [] p`.

- I guess the implementation of this function is just assigning N random numbers to int array represented by pointer:

Example:

```
void random_ints(int* a, int N)
{
   int i;
   for (i = 0; i < N; ++i)
    a[i] = rand();
}
```

What does random_ints(a,N) do and how do I use it in my code?

Solution

Problem