C++ 函数指针（类成员）到非静态成员函数
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原文地址: http://stackoverflow.com/questions/990625/
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C++ function pointer (class member) to non-static member function
提问by Girish
class Foo {
public:
Foo() { do_something = &Foo::func_x; }
int (Foo::*do_something)(int); // function pointer to class member function
void setFunc(bool e) { do_something = e ? &Foo::func_x : &Foo::func_y; }
private:
int func_x(int m) { return m *= 5; }
int func_y(int n) { return n *= 6; }
};
int
main()
{
Foo f;
f.setFunc(false);
return (f.*do_something)(5); // <- Not ok. Compile error.
}
How can I get this to work?
我怎样才能让它发挥作用？
采纳答案by James Curran
The line you want is
你想要的线是
return (f.*f.do_something)(5);
(That compiles -- I've tried it)
（编译 - 我已经试过了）
"*f.do_something" refers to the pointer itself --- "f" tells us where to get the do_something value from. But we still need to give an object that will be the this pointer when we call the function. That's why we need the "f." prefix.
“ *f.do_something”指的是指针本身---“F”告诉我们哪里得到do_something值从。但是我们仍然需要在调用函数时给出一个对象作为 this 指针。这就是为什么我们需要“ f.”前缀。
回答by Nick Dandoulakis
class A{
public:
typedef int (A::*method)();
method p;
A(){
p = &A::foo;
(this->*p)(); // <- trick 1, inner call
}
int foo(){
printf("foo\n");
return 0;
}
};
void main()
{
A a;
(a.*a.p)(); // <- trick 2, outer call
}
回答by Volkan Ozyilmaz
class A {
int var;
int var2;
public:
void setVar(int v);
int getVar();
void setVar2(int v);
int getVar2();
typedef int (A::*_fVar)();
_fVar fvar;
void setFvar(_fVar afvar) { fvar = afvar; }
void insideCall() { (this->*fvar)(); }
};
void A::setVar(int v)
{
var = v;
}
int A::getVar()
{
std::cout << "A::getVar() is called. var = " << var << std::endl;
return var;
}
void A::setVar2(int v2)
{
var2 = v2;
}
int A::getVar2()
{
std::cout << "A::getVar2() is called. var2 = " << var2 << std::endl;
return var2;
}
int main()
{
A a;
a.setVar(3);
a.setVar2(5);
// a.fvar = &A::getVar;
a.setFvar(&A::getVar);
(a.*a.fvar)();
a.setFvar(&A::getVar2);
(a.*a.fvar)();
a.setFvar(&A::getVar);
a.insideCall();
a.setFvar(&A::getVar2);
a.insideCall();
return 0;
}
I extended Nick Dandoulakis's answer. Thank you.
我扩展了尼克丹杜拉基斯的回答。谢谢你。
I added a function which set the member function pointer from outside of the class. I added another function which can be called from outside to show inner call of member function pointer.
我添加了一个从类外部设置成员函数指针的函数。我添加了另一个可以从外部调用的函数来显示成员函数指针的内部调用。
回答by SaeidMo7
#include<iostream>
using namespace std;
class A {
public:
void hello()
{
cout << "hello" << endl;
};
int x = 0;
};
void main(void)
{
//pointer
A * a = new A;
void(A::*pfun)() = &A::hello;
int A::*v1 = &A::x;
(a->*pfun)();
a->*v1 = 100;
cout << a->*v1 << endl << endl;
//-----------------------------
A b;
void(A::*fun)() = &A::hello;
int A::*v2 = &A::x;
(b.*fun)();
b.*v2 = 200;
cout << b.*v2 << endl;
}
回答by Suvesh Pratapa
Try (f.*do_something)(5);
试试 (f.*do_something)(5);
回答by Pradip Bhojania
I think calling a non static member of the class could also be done using a static member function.
我认为调用类的非静态成员也可以使用静态成员函数来完成。
