Passing array to function in C++

Question

Why does the following not compile

#include <iostream>

template <typename T, size_t N>
const size_t len(T[N]){
    return N;
}

int main(int argc, char* argv[]) {
    using namespace std;

    int arr[] = {1, 2, 3};
    cout << len(arr);
}

but this does:

#include <iostream>

const size_t foo(int[3]) {
    return 42;
}

int main(int argc, char* argv[]) {
    using namespace std;

    int arr[1] = {123};
    cout << foo(arr);
}

but with obviously incorrect argument and strangely only with the parameter identifier omitted

I am using GCC 4.9.2 with -std=c++1y

edit:
the error message for the first example:

main.cpp:12:24: error: no matching function for call to 'len(int [3])'
         cout << len(arr);
                        ^
main.cpp:12:24: note: candidate is:
main.cpp:4:18: note: template<class T, unsigned int N> const size_t len(T*)
     const size_t len(T[N]){
                  ^
main.cpp:4:18: note:   template argument deduction/substitution failed:
main.cpp:12:24: note:   couldn't deduce template parameter 'N'
         cout << len(arr);
                        ^

Answer 1

A function argument that looks like an array is actually a pointer, for bizarre historical reasons. If you specify an array size, then it's ignored. So the template is equivalent to

template <typename T, size_t N>
const size_t len(T*){
    return N;
}

The argument can be pointer or anything convertible to one, including an array of any size, so N cannot be deduced from the argument. You can fix this by taking the array by reference:

template <typename T, size_t N>
size_t len(T(&)[N]){
    return N;
}

Now the argument can only be an array of known size, and N will be deduced from that size.