How can I force template parameter type to be signed?

Question

I'll use the following example to illustrate my question:

template<typename T> T diff(T a, T b) {   return a-b; } 

I expect this template function works only when the type T is signed. The only solution I can figure out is to use delete keyword for all the unsigned types:

template<> unsigned char diff(unsigned char,unsigned char) == delete; template<> unsigned char diff(unsigned char,unsigned char) == delete; 

Are there other solutions?

Answer 1

You can use std::is_signed together with std::enable_if:

template<typename T> T diff(T a, T b);  template<typename T> std::enable_if_t<std::is_signed<T>::value, T> diff(T a, T b) {     return a - b; } 

Here std::is_signed<T>::value is true if and only if T is signed (BTW, it is also true for floating-point types, if you don't need it, consider combining with std::is_integral).

std::enable_if_t<Test, Type> is the same as std::enable_if<Test, Type>::type. std::enable_if<Test, Type> is defined as an empty struct in case Test is false and as a struct with an only typedef type equal to template parameter Type otherwise.

So, for signed types, std::enable_if_t<std::is_signed<T>::value, T> is equal to T, while for unsigned it's not defined and compiler uses SFINAE rule, so, if you need to specify an implementation for a particular non-signed type, you can easily do that:

template<> unsigned diff(unsigned, unsigned) {     return 0u; } 

Some relevant links: enable_if, is_signed.