为什么在这个C ++代码中i == 0的值？ [英] Why is the value of i == 0 in this C++ code?

问题描述

我对以下代码感到困惑：

I am confused about the following code:

#include <iostream>

int i = 1;
int main()
{
    int i = i;
    std::cout << "i: " << i << "\n";
    return 0;
}

输出：

i: 0

预期运行上面的代码会打印 1 。

推荐答案

您正在初始化 i 与自身。 i int i = i; 中的 i 都是内层而不是外层。这是未定义的行为，您可能会遇到 0 或任何。

You are initializing i with itself. The both i's in int i = i; are the inner one not the outer one. This is undefined behavior and you may get 0 or anything may happen.

如果您要将外部 i 分配给内部 i ，这是正确的方法。

This is the right way if you want to assign the outer i to the inner i.

#include <iostream>

int i = 1;
int main()
{
    int i = ::i;
    std::cout << "i: " << i << "\n";
    return 0;
}

现场演示

BTW，您应仔细阅读所有编译器警告。如果您自己可以看到问题：

BTW, You should carefully read all the compiler warnings. If you did you could see the problem yourself:

警告'i'在此函数中未初始化

warning 'i' is used uninitialized in this function

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