静态初始化顺序fiasco [英] static initialization order fiasco

问题描述

我从一本书中阅读关于SIOF的信息，并给出了一个例子：

I was reading about SIOF from a book and it gave an example :

//file1.cpp
extern int y;
int x=y+1;

//file2.cpp
extern int x;
int y=x+1;  

现在我的问题是：

在上面的代码中， ？

1.在编译file1.cpp时，编译器保留y不变，即不为其分配存储。

2.编译器为x分配存储，但不会初始化它。

3.编译file2.cpp时，编译器将x保留为原样，即不为其分配存储空间。

4.编译器为y分配存储空间，但doesn

5.在链接file1.o和file2.o时，现在先让file2.o初始化，现在：

x的初始值为0 ？

Now My question is :
In above code..will following things happen ?
1. while compiling file1.cpp, compiler leaves y as it is i.e doesn't allocate storage for it.
2. compiler allocates storage for x, but doesn't initialize it.
3. While compiling file2.cpp, compiler leaves x as it is i.e doesn't allocate storage for it.
4. compiler allocates storage for y, but doesn't initialize it.
5. While linking file1.o and file2.o, now let file2.o is initialized first, so now:
Does x gets initial value of 0? or doesn't get initialized?

推荐答案

初始化步骤在3.6.2初始化非本地对象中给出C ++标准：

The initialization steps are given in 3.6.2 "Initialization of non-local objects" of the C++ standard:

步骤1： x 和 y 在任何其他初始化发生之前被初始化。

Step 1: x and y are zero-initialized before any other initialization takes place.

步骤2： x 或 y - 哪一个没有规定的标准。该变量将获得值 1 ，因为其他变量将被初始化为零。

Step 2: x or y is dynamically initialized - which one is unspecified by the standard. That variable will get the value 1 since the other variable will have been zero-initialized.

步骤3：其他变量将被动态初始化，获取值 2 。

Step 3: the other variable will be dynamically initialized, getting the value 2.

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