指向成员函数指针调用的问题

[ssdx]

#include <iostream>
#include <cstdio>

class CTest
{
public:
    int fun1(int i)
    {
        std::cout << "fun1" << std::endl;
        return 0;
    }

    int fun2(int i)
    {
        std::cout << "fun2" << std::endl;
        return 0;
    }

    void testPfun(void)
    {
        int (CTest::*pFun)(int) = &CTest::fun1;
        pFun(1);   －－－－－－ error C2064: term does not evaluate to a function taking 1 arguments
        (int(CTest::*)(int))pFun(1); －－－－－ error C2064: term does not evaluate to a function taking 1 arguments
    }
};

int main(int argc, char *argv)
{
    CTest o1;
    o1.testPfun();

    system("pause");
}

两个调用位置都包无法将pFun解释为函数指针调用，想问一下该怎么写通过pFun调用fun1/fun2?

[ssdx]

自己搞定
(this->*pFun)(1);

.*
->*
用这两个操作符

[pengzhixi]

jf

[tan870426]

恩

[traceless]

还是建议你这样：

public:
typedef int (CTest::*pFun)(int );

然后使用：
pFun pfoo1 = &CTest::fun1;
(this->*pfoo1)(1);

[stardust20]

jf

[leewon1988]

你函数指针定义不对啊
定义成 int(CTest::*pf)(int)

[cattycat]

前面加this就行。

[猫已经找不回了]

typedef int (*Fun)(int);
Fun pFun=&CTest::fun1;
即可。

[ssdx]

[Quote=引用 6 楼 leewon1988 的回复:]
你函数指针定义不对啊
定义成 int(CTest::*pf)(int)
[/Quote]

是吗？

[traceless]

LZ 你改了后是对的，相信自己

[ssdx]

[Quote=引用 8 楼 hairetz 的回复:]
typedef int (*Fun)(int);
Fun pFun=&CTest::fun1;
即可。
[/Quote]
我问的怎么调用，不是怎么定义。
而且typedef起个别名而已，采用原帖里面的调用方式也是错的。


1楼我自己回答了，用.* ->*这两个操作符。

c++ primer 4th edtion:
18.3.2. Using a Pointer to Class Member
Analogous to the member access operators, operators. and ->, are two new operators, .* and .->, that let us bind a pointer to member to an actual object. The left-hand operand of these operators must be an object of or pointer to the class type, respectively. The right-hand operand is a pointer to a member of that type:

The pointer-to-member dereference operator (.*) fetches the member from an object or reference.

The pointer-to-member arrow operator (->*) fetches the member through a pointer to an object.

[macrojj]

JF

[linuhuge]

.*
->*

专为成员函数指针准备的