c++ - C++ 11 thread_local变量是否自动静态？

Question

这两个代码段有区别吗:

void f() {
    thread_local vector<int> V;
    V.clear();
    ... // use V as a temporary variable
}

和

void f() {
    static thread_local vector<int> V;
    V.clear();
    ... // use V as a temporary variable
}

背景故事:最初我有一个 STATIC vector V(用于保存一些中间值，每次进入函数时它都会被清除)和一个单线程程序。我想把程序变成一个多线程的程序，所以我必须以某种方式摆脱这个静态修饰符。我的想法是把每一个静态都变成thread_local，而不用担心别的？这种方法会适得其反吗？

Answer 1

根据 C++ 标准

When thread_local is applied to a variable of block scope the storage-class-specifier static is implied if it does not appear explicitly

所以意思是这个定义

void f() {
    thread_local vector<int> V;
    V.clear();
    ... // use V as a temporary variable
}

等价于

void f() {
    static thread_local vector<int> V;
    V.clear();
    ... // use V as a temporary variable
}

但是，静态变量不与 thread_local 变量相同。

1 All variables declared with the thread_local keyword have thread storage duration. The storage for these entities shall last for the duration of the thread in which they are created. There is a distinct object or reference per thread, and use of the declared name refers to the entity associated with the current thread

为了区分这些变量，标准引入了一个新术语线程存储持续时间以及静态存储持续时间。