std :: cout在具有静态存储持续时间的对象的构造函数中的用法 [英] std::cout usage in constructors of objects with static storage duration

问题描述

在C ++ 98 / C ++ 03中使用statc存储持续时间的对象的构造函数中使用 std :: cout 是否安全？

Is it safe to use std::cout in constructors of objects with statc storage duration in C++98 / C++03?

从这个回答看来，它不是，但不包括任何

It seems from this answer that it isn't but it doesn't include any quotes from the standard.

在C ++ 11和C ++ 14中只能安全吗？

Is it only safe to do that in C++11 and C++14?

推荐答案

从C ++ 14（N3797），§27.4p2：

From C++14 (N3797), §27.4p2:

并且在第一个
时间之前的某个时间建立关联，并且在任何情况下在主体的主体开始exe-
cution之前构建ios_base :: Init类的对象。 295在程序执行期间对象不会被销毁。在翻译单元中包含
的结果应该好像定义了一个带有静态存储
持续时间的ios_base :: Init的实例。类似地，整个程序应当表现为具有静态存储持续时间的至少一个ios_base :: Init
实例。

The objects are constructed and the associations are established at some time prior to or during the first time an object of class ios_base::Init is constructed, and in any case before the body of main begins exe- cution.295 The objects are not destroyed during program execution.296 The results of including in a translation unit shall be as if defined an instance of ios_base::Init with static storage duration. Similarly, the entire program shall behave as if there were at least one instance of ios_base::Init with static storage duration.

C ++ 98使用类似的术语，但没有as if子句。

C++98 uses similar terminology, but without the "as if" clause.

基本上，这 主：

#include <ostream>
extern std::ostream cout;

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