why does the array decay to a pointer in a template function

Question

I don\'t understand why the array decays to a pointer in a template function.

If you look at the following code: When the parameter is forced to be a reference (fun

Answer 1

Because arrays can not be passed by value as a function parameter.
When you pass them by value they decay into a pointer.

In this function:

template 
void f(T buff) {

T can not be char (&buff)[3] as this is a reference. The compiler would have tried char (buff)[3] to pass by value but that is not allowed. So to make it work arrays decay to pointers.

Your second function works because here the array is passed by reference:

template 
void f1(T& buff) {

// Here T& => char (&buff)[3]