pass reference to array in C++

Question

Can any one help me understand the following code

#include <iostream>

void foo(const char * c)
{
   std::cout << "const char *" << std::endl;
}

template <size_t N>
void foo(const char (&t) [N])
{
   std::cout << "array ref" << std::endl;
   std::cout << sizeof(t) << std::endl;
}

int main()
{
    const char t[34] = {'1'};
    foo(t);

    char d[34] = {'1'};
    foo(d);
}

The output is

const char *
array ref
34

Why does the first foo calls the const char * version ? How can I make it call the reference version ?

Answer 1

Conversion of const char[N] to const char* is considered an "exact match" (to make literals easier, mainly), and between two exact matches a non-template function takes precedence.

You can use enable_if and is_array to force it to do what you want.

---

A messy way to force it might be:

#include <iostream>

template <typename T>
void foo(const T* c)
{
   std::cout << "const T*" << std::endl;
}

template <typename T, size_t N>
void foo(const T (&t) [N])
{
   std::cout << "array ref" << std::endl;
}

int main()
{
    const char t[34] = {'1'};
    foo(t);

    char d[34] = {'1'};
    foo(d);
}

/*
array ref
array ref
*/

I realise that the OP had char not some generic T, but nonetheless this demonstrates that the problem lay in one overload being a template and not the other.