c - The use of sizeof in malloc

I am trying to wrap the creation of a matrix into a function, but I am having problems trying to understand the following code snippet extracted from a book:

  // An error checked malloc() wrapper function
  void *ec_malloc(unsigned int size) {
     void *ptr;
     ptr = malloc(size);
     if(ptr == NULL)
        fatal("in ec_malloc() on memory allocation");
     return ptr;
  }

I've already check this question:

Do I cast the result of malloc?

And now I now it is not necessary to cast the result. But what I don't understand is the use of malloc(size) without a sizeof operator. For example, to create a matrix, lets say int **matrix I've also created this function:

  // An error checked malloc() wrapper function
  void **double_ec_malloc(unsigned int size) {
     void **ptr;

     ptr = malloc(size);
     if(ptr == NULL)
        fatal("in ec_malloc() on memory allocation");
     return ptr;
  }

And then I do:

  int **matrixA = double_ec_malloc(size);


  int i = 0;
  for (i = 0; i < size; i++){
    matrixA[i] = ec_malloc(size);

The man of malloc says:

The malloc() function allocates size bytes and returns a pointer to the allocated memory.

Let size be 4, then in ptr = malloc(size) I am allocating 4 bytes, but if matrix is of type int. Wouldn't I need sizeof int * 4? Because right now I think I am not allocating enough memory for an integer matrix.

Answer

Since ec_malloc() doesn't take a datatype argument, it assumes you will do sizeof(datatype) * size yourself. Thus the argument unsigned int size should be in bytes.

Notice how that it is exactly how malloc() itself behaves.