cli类方法参数 [英] cli class method parameters

问题描述

如果我有类似的东西:

If I have something that resembles:

public ref class ManagedFoo sealed
{
  void method(ManagedFoo^& mfoo,ManagedGoo^& mgoo)
  {
    //...//
  } 
}

public ref class ManagedGoo sealed
{
  //...//
}

它给了我错误

it gives me error

Error   4   error C2061: syntax error : identifier 'ManagedGoo'  path  55  1  ...

谁知道为什么?

anyone knows why is that ?

推荐答案

C ++要求您在使用它们之前知道类型.类型声明也是一条语句，需要用分号终止.
将您的代码更改为此:

C++ requires you to know the type before you use them. Also a type declaration is a statement and needs to be terminated with a semi-colon.
Change your code to this:

public ref class ManagedGoo sealed
{
  //...//
};

public ref class ManagedFoo sealed
{
  void method(ManagedFoo^& mfoo,ManagedGoo^& mgoo)
  {
    //...//
  } 
};

希望这会有所帮助，
弗雷德里克·博纳德(Fredrik Bornander)

Hope this helps,
Fredrik Bornander

这篇关于cli类方法参数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！