给定两个数字类型From
和To
。以下代码是否真的确定了 From
类型的任何值是否可以表示为 To
类型的值而不丢失信息?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.
};
最佳答案
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。
您可以根据 To { std::declval<From>() }
编写传统的表达式测试器特征并可能添加额外的检查 std::is_integral
和 std::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 都做不到。
关于c++ - 确定数值类型 A 是否可以转换为数值类型 B,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12567335/