非静态与静态函数和变量 [英] non-static vs. static function and variable

问题描述

我有一个关于静态和非静态函数和变量的问题。

I have one question about static and non-static function and variable.

1）非静态函数访问静态变量。

1) non-static function access static variable.

没关系。

class Bar
{
public:

     static int i;

     void nonStaticFunction() {

         Bar::i = 10;

     }

};

int Bar::i=0;

2）非静态函数访问非静态变量

2) non-static function access non-static variable

当然OK！

3）静态函数访问静态变量& funciton

3) static function access static variable&funciton

4）静态函数访问非静态函数

4) static function access non-static function

没关系

class Bar
{
public:
     static void staticFunction( const Bar & bar)
     {
         bar.memberFunction();

     }

     void memberFunction() const
     {

     }

}

5）静态函数访问非静态变量

5) static function access non-static variable

还是不行？我对此感到困惑。

It's OK or not OK? I am puzzled about this!

此示例如何

class Bar
{
public:
     static void staticFunction( Bar & bar)
     {
         bar.memberFunction();

     }

     void memberFunction()
     {

         i = 0;
     }

     int i;

};

推荐答案

-static
变量

static function access non-static variable

没关系吗？我对这个
感到困惑！

It's OK or not OK? I am puzzled about this!

调用时，静态函数不绑定到类的实例。类实例（对象）将是保存非静态变量的实体。因此，从静态函数，你将无法访问它们，而不实际传递或存储在其他地方的特定实例进行操作。

When called, a static function isn't bound to an instance of the class. Class instances (objects) are going to be the entities that hold the "non-static" variables. Therefore, from the static function, you won't be able to access them without actually being passed or storing elsewhere a specific instance to operate on.

是的，你最后一个例子中的代码是有效的，因为你是在一个实例中传递的。但是，你不能这样做：

So yes, the code in your last example is valid, because you are passed in an instance. However, you could not do:

static void staticFunction()
{
    // error, this function is static, and is therefore
    // not bound to a specific instance when called
    i = 5;


}

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