CODEWITHSUNDEEP
c++templatesc++11
Damon
Damon

Reputation: 70126

Making function template parameter unsigned in C++11

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 (i.e. not make_unsigned<T>) preferred.)

Upvotes: 9

Views: 2125

Answers (1)

bames53
bames53

Reputation: 88155

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;

Upvotes: 10

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