Question

Can you explain this:

void foo(const char data[10])
{
   char copy[10];
   const char copy1[10] = {};

   printf("%i", sizeof(copy)); //prints 10, as expected
   printf("%i", sizeof(copy1)); //prints 10, as expected
   printf("%i", sizeof(data)); //prints 4, WTF?

}

Looks like function parameters are treated as simple pointers for sizeof.
But WHY does this happen? Is it documented anywhere?

Answer 1

I never saw that syntax before in C++ but maybe you meant

void foo(const char data[10])

Anyway, in C++ arrays decay to pointers when passed to a function. So the function has no way of knowing how big the passed arrays are. In that sense, what I wrote above is completely equivalent to:

void foo(const char data[])

There is also a C FAQ about this subject.