Question

给定两种数字类型From和To。以下代码是否实际确定类型From的任何值是否可以表示为类型To的值而不会丢失信息？如果是，是否有更短或更易读的确定方法？

template <class From, class To>
struct can_cast
{
    static const bool value = 
        (std::numeric_limits<From>::is_integer || // either From is an integer type OR 
        std::is_floating_point<To>::value) && // ...they're both floating point types AND
        (std::numeric_limits<From>::is_signed == false || // either From is unsigned OR
        std::numeric_limits<To>::is_signed == true) && // ...they're both signed AND
        (std::numeric_limits<From>::digits < std::numeric_limits<To>::digits || // To has more bits for digits than From OR
        std::numeric_limits<From>::digits == std::numeric_limits<To>::digits && // To and From have same number of bits, but
        std::numeric_limits<From>::is_signed == std::numeric_limits<To>::is_signed); // they're either both signed or both unsigned.
};

Answer 1

编译器现在内置了此功能：使用list-initialisation时不允许缩小转换。

您可以根据To { std::declval<From>() }编写传统的表达式测试程序特征，并可能使用std::is_integral和std::is_floating_point添加其他检查。

template <typename T>
struct sfinae_true : std::true_type {};

struct can_cast_tester {
   template <typename From, typename To>
   sfinae_true<decltype(To { std::declval<From>() })> static test(int);
   template <typename...>
   std::false_type static test(...);
}; 

template <typename From, typename To>
struct can_cast // terrible name
: decltype(can_cast_tester::test<From, To>(0)) {};

从理论上讲，这应该有效，但目前似乎GCC和clang都没有做对。

确定数字类型A是否可以转换为数字类型B.

1 个答案: