我何时会在constexpr上使用std :: integral_constant？

Question

#include <iostream>
#include <type_traits>

int main(){

    //creating an integral constant with constexpr
    constexpr unsigned int speed_of_light{299792458};

    //creating an integral constant with std::integral_constant
    typedef std::integral_constant<unsigned int, 299792458> speed_of_light_2;

    //using them
    std::cout << speed_of_light/2 << '\n';
    std::cout << speed_of_light_2::value/2 << '\n';

}

有关std :: integral_constant的特别之处,我会选择在constexpr上使用它吗？
他们的行为和用例看起来与我相同.我正在尝试考虑某种模板场景,其中constexpr可能不够.

Answer 1

模板integral_constant定义一个类型,关键字constexpr定义一个常量.例如std::true_type是std::integral_constant<bool, true>.

其中一个用法示例是tag-dispatching.

template<typename T>
void use_impl(const T&, std::false_type)
{
}

template<typename T>
void use_impl(const T&, std::true_type)
{
}

template<typename T>
void use(const T& v)
{
   use_impl(v, typename std::is_integral<T>::type());
}

实例

Answer 2

它可以与三元运算符一起使用,例如

void gotoN_impl(std::integral_constant<int,0>::type)
{
    std::cout << "GoTo 0" << '\n';
}

void gotoN_impl(std::integral_constant<int,1>::type)
{
    std::cout << "GoTo 1" << '\n';
}

void gotoN_impl(std::integral_constant<int,2>::type)
{
    std::cout << "GoTo 2" << '\n';
}

void gotoN_impl(std::integral_constant<int,3>::type)
{
    std::cout << "GoTo 3" << '\n';
} 

template<int N>
void gotoN()
{
    gotoN_impl(typename std::integral_constant<int, N>::type());
}


int main()
{
    gotoN<0>();
    gotoN<1>();
    gotoN<2>();
    gotoN<3>();

    constexpr auto x = 99;

    gotoN<x<4?x:3>(); // with a ternary operator
}