如何定义指向成员函数的指针？

class test
{
void (*p)();
public:
void Print(){cout<<"Test"<<endl;}
test(){p = Print;}
};

我本来想定义指向成员函数的指针*p,可惜不能指向Print函数给出错误
error C2440: '=' : cannot convert from 'void (__thiscall test::*)(void)' to 'void (__cdecl *)(void)'
There is no context in which this conversion is possible

17 个解决方案

#1

这样定义试一下
class test
{
void (*p)(test *);
...
}

#2

class test
{
void (test::*p)();
public:
void Print(){cout<<"Test"<<endl;}
test(){p = &test::Print;}
};

#3

都不行

#4

谢谢 hz129(古雨)　搞定
不过可以不用p=&test::Print；可以直接就用p=Print;

#5

you should declare a point to a member function outside the class, otherwise you would use the point in a ugly express like this:

#include <iostream>

using namespace std;

class test
{
public:
void (test::*p)();
void Print(){cout<<"Test"<<endl;}
test(){p = &test::Print;}
};

void (test::*q)();

main() {
test t;
q = &test::Print;
(t.*q)();
(t.*(t.p))(); //the ugly express
}

#6

I use g++, it must use "p=&test::Print;", and this is more clearly.

#7

But how can I use in the inner of the class?

#8

what's your mean?

#9

I means that How can I used the 'p' in the class funtion?

#10

Oh, I can use it like this:

#include <iostream.h>
class test
{
int idx;
public:
void (test::*p)();
void Print(){cout<<"Test : "<<idx<<endl;}
test(int _idx)
{
idx = _idx;
p = Print;
(this->*(test::p))();  // NOTE:
//NOTE: if I change it to "(*(test::p))();" , the complier gave me a Error:
//       error C2171: '*' : illegal on operands of type 'void (__thiscall test::*)(void)'
//      if I change it to "((test::p))();" ,it still gave me :
//error C2064: term does not evaluate to a function
// I don't know why
        }
};

int main(int argc, char* argv[])
{
test t1(100),t2(200);
void (test::*q)();
    q = &test::Print;
    (t1.*q)();
q = t2.p;
(t2.*q)();
    (t1.*(t1.p))();    //the ugly express
return 0;
}

#11

NOTE行也可改为
（this->*p)();

#12

(this->*(test::p))(); // NOTE:
可写成
(this->*p)(); // NOTE:
用指向成员函数的指针进行调用时，必须指明它的调用者。

"((test::p))();"的错误在于test::p只是限定了调用本类的p函数，但事实上p是指向函数的指针，其本身不是函数，所以出错。

另：
q = t2.p;
这一行可不用，q申明为指向test类的Print函数，对于test类来说，不论生成多少个对象，其Print函数不会因为调用对象的不同而导致行为不同，所以test类只需有一份Print函数的拷贝就行了。这时用q进行调用，只要限定调用它的对象就行了，而不必再次进行赋值，除非你想让它指向另一个不同的函数。

#13

#include <iostream.h>

class test
{
int idx;
public:
void (test::*p)();
void (test::*arrp[10])();
void Print(){cout<<"Test::Print "<<idx<<endl;}
test(int _idx)
{
cout<<endl<<"__________Begin Construct__________"<<endl;
idx = _idx;
cout<<"The value is : "<<idx<<endl;
p = Print;
cout<<"Call (this->*p)() : ======> ";
(this->*p)();  // NOTE: if I change it to "(*(test::p))();"
arrp[0] = p;
cout<<"Call (this->*arrp[0])() =====> ";
(this->*arrp[0])();
cout<<"~~~~~~~~~~End Construct~~~~~~~~~~~~"<<endl;
}
};

int main(int argc, char* argv[])
{
cout<<"Initialize"<<endl;
test t1(100),t2(200);
void (test::*q)();
         q = &test::Print;

// 一般用法：
cout<<endl<<"(t1.*q)() =======> ";          (t1.*q)(); //NOTE
cout<<endl<<"(t2.*q)() =======> ";          (t2.*q)();
// NOTE: q并非是test的成员变量或成员函数, 而*q为test的一个成员函数

// 注意运行结果
cout<<endl<<"(t1.*(t1.p))(); =======> ";    (t1.*(t1.p))();
cout<<endl<<"(t1.*(t2.p))(); =======> ";    (t1.*(t2.p))();
cout<<endl<<"(t2.*(t1.p))(); =======> ";    (t2.*(t1.p))();
cout<<endl<<"(t2.*(t2.p))(); =======> ";    (t2.*(t2.p))();
return 0;
}

上面有关NOTE的一行猪是对吗？

#14

Sorry，猪是——注释

#15

对

#16

consider it as a static function?

#17

可以，没问题

#1