std :: atomic的默认值是多少? [英] What's the default value for a std::atomic?

问题描述

我发现在实践中，对于各种C ++ 11/C ++ 14编译器，std::atomic具有未定义的初始值，就像它是原始"类型一样.也就是说，我们希望表达式

I find that in practice, with a variety of C++11/C++14 compilers, a std::atomic has an undefined initial value just as it would if it were a "raw" type. That is, we expect that for the expression

int a;

a可以具有任何值.事实证明，对于表达式

a may have any value. It also turns out to be true that for the expression

std::atomic< int > b;

b也可以具有任何值.换句话说，

b may also have any value. To say it another way,

std::atomic< int > b;         // b is undefined

不等同于

std::atomic< int > b{ 0 };    // b == 0

或

std::atomic< int > b{};       // b == 0

因为在后两种情况下b被初始化为已知值.

because in the latter two cases b is initialized to a known value.

我的问题很简单:这种行为在C ++ 11或C ++ 14规范中记录在哪里?

My question is simple: where in the C++11 or C++14 spec is this behavior documented?

推荐答案

[atomics.types.generic]/5谈到积分专业:

[atomics.types.generic]/5 says this about integral specializations:

原子积分专业化和专业化原子应具有标准布局.它们每个都有一个琐碎的默认构造函数和一个琐碎的析构函数.他们每个人都应该支持汇总 初始化语法.

The atomic integral specializations and the specialization atomic shall have standard layout. They shall each have a trivial default constructor and a trivial destructor. They shall each support aggregate initialization syntax.

此外，同一部分开头的主要模板提要通常将默认构造函数指定为:

Moreover, the primary template synopsis at the beginning of the same section normatively specifies the default constructor as:

atomic() noexcept = default;

效果在[atomic.types.operations]/4中定义为:

The effects are defined in [atomic.types.operations]/4 as:

效果:使原子对象处于未初始化状态.

Effects: leaves the atomic object in an uninitialized state.

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