Template use without angle brackets - Overloading?

Question

I'm a little confused about the idea of using template with <> brackets and without them. When I compiled the code I got the output I wasn't expecting, and didn't figure out why.

For example, let's say I have 2 functions and a template with the same prototype:

using namespace std;

template<typename T> void copy(T a, T b)
{
    cout << "template copy" << endl;
}
void copy(int a, int b)
{
    cout << "int copy" << endl;
}
void copy(string a, string b)
{
    cout << "string copy" << endl;
}

And after compiling the main function:

int main()
{
    copy<int>(1, 2);
    copy<string>("ha", "ha");
    copy("ab", "bc");
    copy(1, 2);
    
    return 0;
}

the output looked like this:

template copy
template copy
template copy
int copy

for the record all the code is written on the same CPP file.

Answer 1

You have to remember that literal strings are really (constant) arrays of characters, which decays to (constant) pointers to char, i.e. const char*.

Since your function taking std::string arguments is not a direct match, and the compiler will not do a conversion, the template overload will be used (as copy<const char*>).