在C ++ 11中sqrt是否定义为constexpr？ [英] In C++11 is sqrt defined as constexpr?

问题描述

在C ++ 11中， std :: sqrt 定义为 constexpr ，即可以合法地从其他 constexpr 函数还是在编译时上下文（如数组大小或模板参数）中？ g ++似乎允许（使用 -std = c ++ 0x ），但是鉴于c ++ 0x / c +，我不确定我是否可以将其视为权威+11支持仍然不完整。我似乎无法在Internet上找到任何东西的事实使我不确定。

In C++11 is std::sqrt defined as constexpr, i.e. can it legally be used from other constexpr functions or in compile-time contexts like array sizes or template arguments? g++ seems to allow it (using -std=c++0x), but I'm not sure I can take that as authoritative given that c++0x/c++11 support is still incomplete. The fact that I can't seem to find anything on the Internet makes me unsure.

似乎应该可以轻松地使用Google找出这一点，但是我已经尝试了（现在已经40分钟了……），却找不到任何东西。我可以找到一些将constexpr添加到标准库各个部分的建议（例如此），但与 sqrt 或其他数学函数无关。

It seems like this should be something one could easily find out using Google, but I've tried (for 40 minutes now...) and couldn't find anything. I could find several proposals for adding constexpr to various parts of the standard library (like for example this one), but nothing about sqrt or other math functions.

推荐答案

std :: sqrt 未定义为 constexpr ，根据N3291的第26.8节：C ++ 11 FDIS（我怀疑他们之后将其添加到最终标准中）。可能可以编写这样的版本，但是标准库版本不是 constexpr 。

std::sqrt is not defined as constexpr, according to section 26.8 of N3291: the C++11 FDIS (and I doubt they added it to the final standard after that). One could possibly write such a version, but the standard library version is not constexpr.

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