在C ++中，如果两个不同的函数声明相同的静态变量会发生什么？

Question

void foo() {
    static int x;
}

void bar() {
    static int x;
}

int main() {
    foo();
    bar();
}

Answer 1

They each see only their own one. A variable cannot be "seen" from outside the scope that it's declared in.

If, on the other hand, you did this:

static int x;

void foo() {
    static int x;
}

int main() {
    foo();
}

then foo() only sees its local x ; the global x has been "hidden" by it. But changes to one don't affect the value of the other.

Answer 2

The variables are distinct, each function has it's own scope. So although both variables last for the lifetime of the process, they do not interfere with each other.

Answer 3

This is perfectly fine. In practice the actual name of the variable in the output of the compiler can be thought of as something like function_bar_x , ie it is the responsibility of your compiler to ensure that these do not clash.

Answer 4

什么都没发生，两个变量都具有theri范围并保留其值以在call中调用

Answer 5

这两个静态变量不同。

Answer 6

The compilator translates each variable in a unique manner, such as foo_x and bar_x in your example, so they are threated differently.

Don't do this as your code will be hard to read and maintain after some time since you will not able to catch at once of what x are you referring to.