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 (function f1) it does not decay. In the other function f it decays. Why is the type of T in function f not const char (buff&)[3] but rather const char* (if I understand it correctly)?

#include <iostream>

template <class T>
void f(T buff) {
    std::cout << "f:buff size:" << sizeof(buff) << std::endl;       //prints 4
}

template <class T>
void f1(T& buff) {
    std::cout << "f:buff size:" << sizeof(buff) << std::endl;       //prints 3
}

int main(int argc, char *argv[]) {
    const char buff[3] = {0,0,0};
    std::cout << "buff size:" << sizeof(buff) << std::endl;         //prints 3
    f(buff);
    f1(buff);
    return 0;
}

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 <class T>
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 <class T>
void f1(T& buff) {

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