为什么std :: chrono :: system_clock :: to_time_t（）不是constexpr？
[英]Why is std::chrono::system_clock::to_time_t() not constexpr?
问题描述
The C++ standard (github.com/cplusplus/draft) has the time_t conversion functions ( std::chrono::system_clock::to_time_t and std::chrono::system_clock::from_time_t ) for listed as static and noexcept but not constexpr . 
C ++标准（github.com/cplusplus/draft）具有time_t转换函数（ std::chrono::system_clock::to_time_t和std::chrono::system_clock::from_time_t ），列为static和noexcept但不是constexpr 。
Given that essentially all of the operations on time_point and duration are constexpr (including duration_cast and time_point_cast ), I can't think of any reason to exclude them. 鉴于time_point和duration基本上所有操作都是constexpr （包括duration_cast和time_point_cast ），我想不出有任何理由要排除它们。 A quick inspection of the libstdc++ sources on my local machine confirms that these functions are implemented as simple duration/time_point casts. 快速检查本地计算机上的libstdc ++源代码确认这些函数是作为简单的duration / time_point强制转换实现的。
Is there any reason that these two functions should not be constexpr ? 是否有任何理由认为这两个功能不应该是constexpr ？ Is this just a case of "because no one proposed they should be"? 这只是“因为没有人提出他们应该”的情况吗？
1 个解决方案
解决方案1
3 已采纳 2014-01-23 15:52:15
Is this just a case of "because no one proposed they should be"?
Yes, I think that is exactly right. 是的，我认为这是完全正确的。
I'm curious: Once you get a constexpr time_t , what are you going to do with it? 我很好奇：一旦你得到一个constexpr time_t ，你打算用它做什么？ None of the C functions taking time_t are constexpr . 考虑time_t的C函数都不是constexpr 。
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