问题描述
根据C ++ Primer一书,作者提到我们可以将一个类成员函数指定为另一个类的朋友,而不是整个类(第634页)。
According to the C++ Primer book, the author mentioned that We can specify a class member function as a friend of another class, instead of the entire class (page 634).
然后,我测试了这段代码:
Then, I tested this code:
class A
{
public:
friend void B::fB(A& a);
void fA(){}
};
class B
{
public:
void fB(A& a){};
void fB2(A& a){};
};
我只是想让fB()成为A类的朋友,而不是整个B类的朋友。但是关于代码产生错误:'B':不是类或名称空间名称
。
(我正在使用Visual C ++ 2005)
I just wanted the fB() to be friend of class A, not the entire class B. But the about code produced an error: 'B' : is not a class or namespace name
.(I am using Visual C++ 2005)
推荐答案
尝试将B定义放在A之前:
Try putting the B definition before A's:
class A; // forward declaration of A needed by B
class B
{
public:
// if these require full definition of A, then put body in implementation file
void fB(A& a); // Note: no body, unlike original.
void fB2(A& a); // no body.
};
class A
{
public:
friend void B::fB(A& a);
void fA(){}
};
A
需要完整的 B
。但是, B
需要了解 A
,但不需要完整的定义,因此需要前向声明 A
。
A
needs the full definition of B
. However, B
needs to know about A
, but does not need the full definition, so you need the forward declaration of A
.
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