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给定两个数字类型From和To. 下面的代码是否真的确定了任何类型的值是否From可以表示为类型的值而To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.
};
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1 回答 1

2

编译器现在内置了这个功能：使用列表初始化时不允许缩小转换。

您可以编写基于 的传统表达式测试器特征，并可能使用和To { std::declval<From>() }添加额外的检查。std::is_integralstd::is_floating_point

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，这应该可行，但目前看来 GCC 和 clang 都没有做对。

于 2012-09-24T14:53:30.727 回答