大于给定值的最小可表达值

你可以用 std::nextafter .

注意这里我使用 std::numeric_limits<Floating>::max() ,如果希望无穷大有意义的行为,则可能需要修改代码。

#include <iostream>
#include <limits>
#include <cmath>
#include <type_traits>
#include <iomanip>

template <typename Floating,
    std::enable_if_t<std::is_arithmetic_v<Floating> && !std::is_integral_v<Floating>, bool> = false>
Floating generic_next_val(Floating val) {
    return std::nextafter(val, std::numeric_limits<Floating>::max());
}

template <typename Integral,
    std::enable_if_t<std::is_arithmetic_v<Integral> && std::is_integral_v<Integral>, int> = 0>
Integral generic_next_val(Integral val) {
    return static_cast<Integral>(val + static_cast<Integral>(1));
}

int main() {
    int a = 1;
    float f = 0.0f;
    std::cout << std::setprecision(20) << generic_next_val(f) << " " << generic_next_val(a) << std::endl;
    return 0;
}

这个 <cmath> 页眉 std::nexttoward()

所以,给定这样的代码:

T my_val = some_input; // float, double, int, etc.
T next = std::nexttoward(my_val, std::numeric_limits<T>::max());`

my_val 是1.0f, next 将是1之后的下一个最大的浮动;如果 我的价值

(请注意,这里有很多关于数字溢出的细节,上面链接的CppReference页面讨论了这些细节。长短不一是不及格 std::numeric_limits<T>::max() 第一个参数)

#include <type_traits>
#include <limits>
#include <iostream>

/*
NOTE: Untested code for 'proof' of concept only.

This will cover all the is_integral<T> except bool.

This should probably be a set of free functions rather than a class
but just to check out the idea. Class should have constraints but as
the final implementation should probably be free functions they are
omitted.
*/
template<typename T>
class GLB{
public:
    using limits = std::numeric_limits<T>;

    static T get(const T& value){
       return value == limits::max() ? limits::max() : value + static_cast<T>(1);
    }
};

int main(int, char**){

    std::cout << GLB<int>::get(42) << '\n';
    std::cout << GLB<unsigned>::get(42) << '\n';
    std::cout << GLB<wchar_t>::get(42) << '\n';
    return 0;
}

/*
 That was the easy bit now what about the floating point numbers.
*/