c++ - C++11 constexpr function's argument passed in template argument

Question

This used to work some weeks ago:

template <typename T, T t>
T            tfunc()
{
    return t + 10;
}

template <typename T>
constexpr T       func(T t)
{
    return tfunc<T, t>();
}

int main()
{
    std::cout << func(10) << std::endl;
    return 0;
}

But now g++ -std=c++0x says:

main.cpp: In function ‘constexpr T func(T) [with T = int]’:
main.cpp:29:25:   instantiated from here
main.cpp:24:24: error: no matching function for call to ‘tfunc()’
main.cpp:24:24: note: candidate is:
main.cpp:16:14: note: template<class T, T t> T tfunc()
main.cpp:25:1: warning: control reaches end of non-void function [-Wreturn-type]

clang++ -std=c++11 says that template's parameters of tfunc<T, t>() are ignored because invalid.

Is that a bug, or a fix ?

PS:

g++ --version => g++ (GCC) 4.6.2 20120120 (prerelease)

clang++ --version => clang version 3.0 (tags/RELEASE_30/final) (3.0.1)

Answer 1

The parameter t is not a constant expression. Hence the error. It should be also noted that it cannot be a constant expression.

You can pass the constant expression as argument, but inside the function, the object (the parameter) which holds the value, is not a constant expression.

Since t is not a constant expression, it cannot be used as template argument:

return tfunc<T, t>(); //the second argument must be a constant expression

---

Maybe, you want something like this:

template <typename T, T t>
T  tfunc()
{
    return t + 10;
}

template <typename T, T t>  //<---- t became template argument!
constexpr T  func()
{
    return tfunc<T, t>();
}

#define FUNC(a)  func<decltype(a),a>()

int main()
{
    std::cout << FUNC(10) << std::endl;
}

Now it should work : online demo