When a function has a specific-size array parameter, why is it replaced with a pointer?

Question

Given the following program,

#include 

using namespace std;

void foo( char a[100] )
{
    cout << \"foo() \" << sizeof( a ) <         


        

Answer 1

Yes. In C and C++ you cannot pass arrays to functions. That's just the way it is.

Why are you doing plain arrays anyway? Have you looked at boost/std::tr1::array/std::array or std::vector?

Note that you can, however, pass a reference to an array of arbitrary length to a function template. Off the top of my head:

template< std::size_t N >
void f(char (&arr)[N])
{
  std::cout << sizeof(arr) << '\n';
}