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 (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

It is because arrays cannot be passed by value to a function. So in order to make it work, the array decays into a pointer which then gets passed to the function by value.

In other words, passing an array by value is akin to initializing an array with another array, but in C++ an array cannot be initialized with another array:

char buff[3] = {0,0,0};
char x[3] = buff; //error 

So if an array appears on the right hand side of =, the left hand side has to be either pointer or reference type:

char *y = buff; //ok - pointer
char (&z)[3] = buff; //ok - reference

Demo : http://www.ideone.com/BlfSv

It is exactly for the same reason auto is inferred differently in each case below (note that auto comes with C++11):

auto a = buff; //a is a pointer - a is same as y (above)
std::cout << sizeof(a) << std::endl; //sizeof(a) == sizeof(char*)

auto & b = buff; //b is a reference to the array - b is same as z (above)
std::cout << sizeof(b) << std::endl; //sizeof(b) == sizeof(char[3])

Output:

4 //size of the pointer
3 //size of the array of 3 chars

Demo : http://www.ideone.com/aXcF5