c++ - protected 成员的默认值 C++

Question

代码:

#include <iostream>

using namespace std;

class Example
{
public:
    int _public;

    friend ostream& operator<< (ostream& stream, Example& o);

protected:
    int _protected;

private:
    int _private;
};

ostream& operator<< (ostream& stream, Example& o) {
    stream <<
    "_public=" << o._public << endl <<
    "_protected=" << o._protected << endl <<
    "_private=" << o._private << endl;
    return stream;
}


int main(int argc, char const *argv[])
{
    Example e;
    cout << e << endl;
    return 0;
}

输出

_public=4196960
_protected=0
_private=4196368

问题:

所有三个成员都未初始化。但是只有 public 和 private 成员中有垃圾值。为什么 protected 成员初始化为零？这是有原因的吗？

---

g++ 版本和标志

g++ (Ubuntu 5.4.0-6ubuntu1~16.04.11) 5.4.0 20160609

-std=c++11

Answer 1

首先，注意阅读 uninitialized variables是未定义的行为。您需要为您的类定义一个构造函数。这与访问说明符无关。它恰好(在您的情况下)protected 成员位于之前包含值 0 的地址。

最好不要使用以下划线开头的变量名。他们是保留的。允许在类范围内使用它们，但切记不要在全局范围内使用它们。

5.10 Identifiers [lex.name]

3 In addition, some identifiers are reserved for use by C++ implementations and shall not be used otherwise; no diagnostic is required.
...

(3.2) Each identifier that begins with an underscore is reserved to the implementation for use as a name in the global namespace.