调用指向成员函数的指针 C++
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/23686456/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Calling pointer-to-member function C++
提问by wolfPack88
I have a pointer to a member function defined within a class, e.g.:
我有一个指向类中定义的成员函数的指针，例如：
class Example {
void (Example::*foo)();
void foo2();
};
In my main code, I then set foo as:
在我的主代码中，我将 foo 设置为：
Example *a;
a->foo = &Example::foo2;
However, when I try to call foo:
但是，当我尝试调用 foo 时：
a->foo();
I get the following compile time error: "error: expression preceding parentheses of apparent call must have (pointer-to-) function type". I'm assuming I'm getting the syntax wrong somewhere, can someone point it out to me?
我收到以下编译时错误：“错误：明显调用括号前的表达式必须具有（指向）函数类型”。我假设我在某个地方弄错了语法，有人可以向我指出吗？
回答by Bob Fincheimer
to call it you would do: (a->*(a->foo))()
调用它你会这样做： (a->*(a->foo))()
(a->*X)(...)- dereferences a member function pointer - the parens around a->*Xare important for precedence.
(a->*X)(...)- 取消引用成员函数指针 - 周围的括号a->*X对于优先级很重要。
X = a->foo- in your example.
X = a->foo- 在你的例子中。
See ideone herefor working example
有关工作示例，请参见此处的ideone
