如何包括不是参数的c ++模板类型

Question

I want to have a type in a template like T_signed in the code below

template <typename T_unsigned, typename T_signed> bool foo(T_unsigned input)
{
    T_signed temp= ((T_signed) input)-100;    
    //use temp for calculations to figure out myBool
    return myBool;
}

While the above is a simplification of the actual code I am writing and greatly believe it is what is preventing the code from compiling. How do I get the the compiler to figure out the type of T_signed implicitly based on what the type input is? Any help appreciated.

Answer 1

Something like this , using std::make_signed :

#include <iostream>
#include <type_traits>

template <typename Tu>
bool foo(Tu input) {
    std::cout << std::is_signed<Tu>::value << std::endl;

    typedef typename std::make_signed<Tu>::type Ts;
    Ts temp = input - 100;

    return (temp < 0);
}

int main() {
    std::cout << foo(32u) << std::endl;
}

You can also add std::enable_if or static_assert to ensure that the passed in type really is unsigned.