确保cout具有ctype< char>方面? [英] Is cout Guaranteed to Have the ctype<char> facet?

问题描述

给出:auto foo = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"s我可以通过以下方式将所有字符转换为小写:

Given: auto foo = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"s I can convert all the characters to lowercase by:

use_facet<ctype<char>>(cout.getloc()).tolower(data(foo), next(data(foo), foo.size()));

在线示例

但这取决于包含ctype<char> facet的cout.getloc().

假设我使用的是未修改的cout，我可以假设cout.getloc()将包含facet ctype<char>，还是在使用以下命令之前需要确认这一点:

Presuming that I'm using an unmodified cout can I assume that cout.getloc() will contain the facet ctype<char> or do I need to confirm this before use with:

has_facet<ctype<char>>(cout.getloc())

推荐答案

来自 cppreference :

在C ++程序中构造的每个语言环境至少都具有以下标准方面:[...]

Each locale constructed in a C++ program holds at least the following standard facets [...]:

1. std :: ctype< char>
2. ...

任何语言环境，这意味着即使不是cout对象的语言环境也将支持std::ctype<char>.

Any locale, meaning even locales not of the cout object will support std::ctype<char>.

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