为什么 c++ 编译器接受这个初始化?静态整数 x = x;
我刚刚发现了这个:
static int x = x;
为什么 C++ 编译器会接受这种初始化?
Why is this initialization accepted by the C++ compiler?
我会称之为编译器异常，但有人可能会对此给出一个很好的解释.
I would call it a compiler anomaly, but someone might come with a good explanation for this.
因此，对于具有静态存储的数据，可以使用自身初始化变量...我已经使用 VS2015 和 VS2017 编译器以及其他一些在线 C++ 编译器进行了尝试.
So for data with static storage, it is possible to initialize a variable with itself... I've tried this with a VS2015 and VS2017 compiler and also some other online C++ compilers.
推荐答案
static和非static变量其实是一样的.
It's actually the same for static and non-static variables.
名称在其声明符之后和初始化之前立即可见(如果有的话).因此在
A name becomes visible immediately after its declarator and before its initialization, if any. Thus in
static int x = x;
名称 x 在其第一次出现后立即变得可见，并且可以在初始化程序中引用.因为它是静态的，所以它的初始值是很好定义的(它是0).
the name x becomes visible immediately after its first occurrence, and can be referred to in the initializer. Since it's static, its initial value is well defined (it's 0).
这也是合法的，即使在块范围内:
This is also legal, even at block scope:
int x = x;
尽管在这里您可能会收到警告，因为 x 正在使用其自己的不确定值进行初始化(在大多数情况下行为未定义).
although here you're likely to get a warning because x is being initialized with its own indeterminate value (the behavior is undefined in most cases).
这是一件很愚蠢的事情，但 C++ 并不会特意阻止你做愚蠢的事情.例如，您可能想要声明一个指向自身的指针:
It's a silly thing to do, but C++ isn't in the business of going out of its way to prevent you from doing silly things. As an example, you might want to declare a pointer that points to itself:
void *p = (void*)&p;
这里初始化器指的是 p 的地址而不是它的值，但名称 p 必须是可见的才能使其工作.添加特殊情况规则被认为不值得.
Here the initializer refers to the address of p rather than its value, but the name p has to be visible to make that work. It wasn't considered worthwhile to add a special-case rule.
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