pointers - when allocating memory for an array dynamically (in C), what does the (int *) cast do?

C noob here. When declaring an array during runtime, I've seen two methods for doing so. Can someone please explain the (int *) cast in the second one?

// first way
int numberElements = 5;
int *pointer = malloc(numberElements * sizeof(int));

// second way

...
int *pointer = (int *)malloc(numberElements * sizeof(int));

I just don't see what the (int *) cast is doing. With first the allocation, the array can be filled like this...

// first way cont.
... 
for (int i = 0; i < numberElements; i += 1){
    pointer[i] = 0;\

}

is this not true for the second? what would you have to do differently?

Answer

The cast does nothing. A void pointer can be assigned to any pointer without an explicit cast.

AND you shouldn't. The C99 (or C90, C11) standard does not require the cast.