这个自初始化有效吗？
我有这个问题，我之前考虑过，但我认为回答这个问题并不简单。
int x = x + 1;
int main() {
return x;
}
我的问题是程序的行为是否已定义或未定义（如果它完全有效）。如果已定义，x 的值在 main 中是否已知？
I have this question, which i thought about earlier, but figured it's not trivial to answer
int x = x + 1;
int main() {
return x;
}
My question is whether the behavior of the program is defined or undefined if it's valid at all. If it's defined, is the value of x known in main?
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我很确定它已定义，并且 x 的值应该为 1。§3.6.2/1 说：“具有静态存储持续时间 (3.7.1) 的对象应在任何其他初始化发生之前进行零初始化 (8.5)。 ”
之后，我认为一切都非常简单。
I'm pretty sure it's defined, and x should have the value 1. §3.6.2/1 says: "Objects with static storage duration (3.7.1) shall be zero-initialized (8.5) before any other initialization takes place."
After that, I think it's all pretty straightforward.
这段代码绝对不干净，但对我来说它应该可以预见地工作。
int x将变量放入数据段中，该数据段在程序开始时定义为零。在main()之前，调用静态初始化器。对于x，即代码x = x + 1。x = 0 + 1 = 1。因此 main() 将返回 1。如果 x 是在堆栈上分配的局部变量，则代码肯定会以不可预测的方式工作。与数据段不同，堆栈状态几乎可以保证包含未定义的垃圾。
This code is definitely not clean, but to me it should work predictably.
int xputs the variable into the data segment which is defined to be zero at the program start. Beforemain(), static initializers are called. Forxthat is the codex = x + 1.x = 0 + 1 = 1. Thus the main() would return 1.The code would definitely work in unpredictable fashion if
xis a local variable, allocated on stack. State of stack, unlike the data segment, is pretty much guaranteed to contain undefined garbage.'x' 变量存储在 .bss 中，加载程序时会用 0 填充。因此，当程序加载到内存中时，“x”的值为 0。
然后在调用 main 之前，执行“x = x + 1”。
我不知道它是否有效，但行为并非未定义。
The 'x' variable in stored in the .bss, which is filled with 0s when you load the program. Consequently, the value of 'x' is 0 when the program gets loaded in memory.
Then before main is called, "x = x + 1" is executed.
I don't know if it's valid or not, but the behavior is not undefined.
在 main 调用之前 x 必须初始化为 0，因此它的值必须为 1，当您输入 main 时，您将返回 1。这是一个已定义的行为。
Before the main call x must be initialized to 0 therefore it's value must be 1 one you enter main, and you will return 1. It's a defined behavior.