无效使用非静态数据成员 [英] invalid use of non-static data member

问题描述

对于像这样的代码：

class foo {
  protected:
    int a;
  public:
    class bar {
      public:
        int getA() {return a;}   // ERROR
    };
    foo()
      : a (p->param)
};

我收到此错误：

 invalid use of non-static data member 'foo::a'

目前，在 foo 的构造函数中初始化变量 a 。

currently the variable a is initialized in the constructor of foo.

如果我使它静态，则说：

if I make it static, then it says:

 error: 'int foo::a' is a static data member; it can only be initialized at its definition

但是我想传递一个值到 a 。
那么是什么解决方案？

However I want to pass a value to a in the constructor. What is the solution then?

推荐答案

在C ++中，本质上不属于包含类的任何实例。因此 bar :: getA 没有 foo 的任何特定实例，其 code>它可以返回。我猜你想要的是什么样的：

In C++, unlike (say) Java, an instance of a nested class doesn't intrinsically belong to any instance of the enclosing class. So bar::getA doesn't have any specific instance of foo whose a it can be returning. I'm guessing that what you want is something like:

    class bar {
      private:
        foo * const owner;
      public:
        bar(foo & owner) : owner(&owner) { }
        int getA() {return owner->a;}
    };

但是即使这样，你可能需要做一些修改，因为在C ++ 11，不像（再次说）Java，嵌套类没有对它的封闭类的特殊访问，所以它不能看到 protected member a 。这将取决于您的编译器版本。 （

But even for this you may have to make some changes, because in versions of C++ before C++11, unlike (again, say) Java, a nested class has no special access to its enclosing class, so it can't see the protected member a. This will depend on your compiler version. (Hat-tip to Ken Wayne VanderLinde for pointing out that C++11 has changed this.)

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