Question

The following code compiles fine:

template<typename T>
void f(const T &item) { return; }

int main() 
{
  f("const string literal");
}

Compilation succeeded at ideone : http://ideone.com/dR6iZ

But when I mention the return type, it doesn’t compile:

template<typename T>
T f(const T &item) { return item; }

int main() 
{
  f("const string literal");
}

Now it gives error:

prog.cpp:6: error: no matching function for call to ‘f(const char [21])’

Code at ideone : http://ideone.com/b9aSb

Even if I make the return type const T, it doesn’t compile.

My question is :

• Why does it not compile?
• What does the return type has to do with the error and the function template instantiation?

Answer 1

You cannot return an array from a function, so template instantiation fails, and there’s no matching function.

You get this particular error because of SFINAE – It’s not really an error that the compiler cannot instantiate your function, it is an error that there’s no matching function.

You can return a reference to an array – returning T const & will work.

EDIT: In response to comments:

First, this is actually a decent example of SFINAE.

template<typename T> T f(const T &item) { return item; }
char const * f(void const * item) { return 0; }
int main() {
  f("abc");
}

When the compiler compiles this, it’ll first try to instantiate the templated f, to create an exact match for the type const char [3]. This fails, because of the mentioned reasons. It’ll then select the inexact match, the plain function, and in the call decay the const char [3] to a const char *.