How do you get the size of array that is passed into the function?

Question

I am trying to write a function that prints out the elements in an array. However when I work with the arrays that are passed, I don't know how to iterate over the array.

void
print_array(int* b)
{
  int sizeof_b = sizeof(b) / sizeof(b[0]);
  int i;
  for (i = 0; i < sizeof_b; i++)
    {
      printf("%d", b[i]);
    }
}

What is the best way to do iterate over the passed array?

Answer 1

You need to also pass the size of the array to the function.
When you pass in the array to your function, you are really passing in the address of the first element in that array. So the pointer is only pointing to the first element once inside your function.

Since memory in the array is continuous though, you can still use pointer arithmetic such as (b+1) to point to the second element or equivalently b[1]

void print_array(int* b, int num_elements)
{
  for (int i = 0; i < num_elements; i++)
    {
      printf("%d", b[i]);
    }
}

This trick only works with arrays not pointers:

sizeof(b) / sizeof(b[0])

... and arrays are not the same as pointers.

Answer 2

Why don't you use function templates for this (C++)?

template<class T, int N> void f(T (&r)[N]){
}

int main(){
    int buf[10];
    f(buf);
}

EDIT 2:

The qn now appears to have C tag and the C++ tag is removed.