在C ++ 11中使函数模板参数无符号

Question

In a template function that looks like this:

template<typename T> constexpr T foo(T a, T b) { return /*recursive call*/; }

I am getting a warning about comparing signed vs unsigned (due to comparing against sizeof ) which I'd like to eliminate.

Conceptually, one would need something like this:

template<typename T> constexpr T foo(T a, unsigned T b) { ... }
    or
template<typename T> constexpr T foo(T a, std::make_unsigned<T>::type b) { ... }

Unluckily, the first version is not valid C++, and the second version breaks the build because T is not a qualified type when the compiler sees make_unsigned .

Is there is a solution for this that actually works?

(NB: Somehow related to / almost same as Get the signed/unsigned variant of an integer template parameter without explicit traits , though function rather than class (so no typedefs), traits or any feature of C++11 explicitly welcome, and working solution (ie not make_unsigned<T> ) preferred.)

Answer 1

You forgot a 'typename'

template<typename T>
constexpr T foo(T a, typename std::make_unsigned<T>::type b) { ... }

In C++14 you should be able to write

template<typename T>
constexpr T foo(T a, std::make_unsigned_t<T> b) { ... }

Or you can implement this yourself in C++11:

template<typename T>
using make_unsigned_t = typename std::make_unsigned<T>::type;