函数指针生成“无效使用非静态成员函数”错误 [英] function pointers generate 'invalid use of non-static member function' error
问题描述
我试图以更好的方式掌握指针函数概念。所以我有一个非常简单和工作的例子:
#include< iostream&
using namespace std;
int add(int first,int second)
{
return first + second;
}
int subtract(int first,int second)
{
return first-second;
}
int操作(int first,int second,int(* functocall)(int,int))
{
return(* functocall)第二);
}
int main()
{
int a,b;
int(* plus)(int,int);
int(* minus)(int,int);
plus =& add;
minus =& subtract;
a = operation(7,5,add);
b = operation(20,a,minus);
cout<< a =<< a<< 和b =< b<< endl;
return 0;
}
到目前为止很好,
现在我需要将函数在类中,并根据我使用的函数指针选择加或减。所以我只是做一个小修改为:
#include< iostream>
using namespace std;
class A
{
public:
int add(int first,int second)
{
return first + second;
}
int subtract(int first,int second)
{
return first-second;
}
int操作(int first,int second,int(* functocall)(int,int))
{
return(* functocall)第二);
}
};
int main()
{
int a,b;
A a_plus,a_minus;
int(* plus)(int,int)= A :: add;
int(* minus)(int,int)= A :: subtract;
a = a_plus.operation(7,5,plus);
b = a_minus.operation(20,a,minus);
cout<< a =<< a<< 和b =< b<< endl;
return 0;
}
,显而易见的错误是:
ptrFunc.cpp:在函数'int main()':
ptrFunc.cpp:87:29:error:invalid使用非静态成员函数' int A :: add(int,int)'
ptrFunc.cpp:88:30:error:invalid使用非静态成员函数'int A :: subtract(int,int)'
coz我没有指定要调用的对象(我现在不想使用静态方法) p>
编辑:
几个意见和答案建议不可能是非静态版本(如我所写) all)
所以,
以下面的方式修改类也不起作用:
#include< ; iostream>
using namespace std;
class A
{
int res;
public:
A(int choice)
{
int(* plus)(int,int)= A :: add;
int(* minus)(int,int)= A :: subtract;
if(choice == 1)
res = operation(7,5,plus);
if(choice == 2)
res = operation(20,2,minus);
cout<< operation of operation =< res;
}
int add(int first,int second)
{
return first + second;
}
int subtract(int first,int second)
{
return first-second;
}
int操作(int first,int second,int(* functocall)(int,int))
{
return(* functocall)第二);
}
};
int main()
{
int a,b;
A a_plus(1);
A a_minus(2);
return 0;
}
产生此错误:
ptrFunc.cpp:在构造函数'A :: A(int)':
ptrFunc.cpp:11:30:从类型'int(A ::)(int,int)'到类型'int(*)(int,int)'
ptrFunc.cpp:12:31:从类型'int(A ::)(int,int)'到类型'int(*)(int,int)'
我可以知道如何解决这个问题吗?
感谢
声明一个指向成员方法的函数指针的语法是:
int加)(int,int)=& A :: add;
int(A :: * minus)(int,int)=& A :: subtract;
要调用成员方法,请使用*或 - > *运算符:
(a_plus。* plus)(7,5);
另请参阅 http://msdn.microsoft.com/en-us/library/b0x1aatf(v = vs.80).aspx < a>
希望这有助。
完成代码:
$ b b
#include< iostream>
using namespace std;
class A
{
public:
int add(int first,int second)
{
return first + second;
}
int subtract(int first,int second)
{
return first-second;
}
int操作(int first,int second,int(A :: * functocall)(int,int))
{
return(this-> ; * functocall)(first,second);
}
};
int main()
{
int a,b;
A a_plus,a_minus;
int(A :: * plus)(int,int)=& A :: add;
int(A :: * minus)(int,int)=& A :: subtract;
a = a_plus.operation(7,5,plus);
b = a_minus.operation(20,a,minus);
cout<< a =<< a<< 和b =< b<< endl;
return 0;
}
I am trying to grasp pointer function concept in a better way. So I have a very simple and working example as:
#include <iostream>
using namespace std;
int add(int first, int second)
{
return first + second;
}
int subtract(int first, int second)
{
return first - second;
}
int operation(int first, int second, int (*functocall)(int, int))
{
return (*functocall)(first, second);
}
int main()
{
int a, b;
int (*plus)(int, int);
int (*minus)(int, int);
plus = &add;
minus = &subtract;
a = operation(7, 5, add);
b = operation(20, a, minus);
cout << "a = " << a << " and b = " << b << endl;
return 0;
}
So far so good, Now I need to group the functions in a class, and select add or subtract based on the function pointer that i use. So I just make a small modification as:
#include <iostream>
using namespace std;
class A
{
public:
int add(int first, int second)
{
return first + second;
}
int subtract(int first, int second)
{
return first - second;
}
int operation(int first, int second, int (*functocall)(int, int))
{
return (*functocall)(first, second);
}
};
int main()
{
int a, b;
A a_plus, a_minus;
int (*plus)(int, int) = A::add;
int (*minus)(int, int) = A::subtract;
a = a_plus.operation(7, 5, plus);
b = a_minus.operation(20, a, minus);
cout << "a = " << a << " and b = " << b << endl;
return 0;
}
and the obvious error is:
ptrFunc.cpp: In function ‘int main()’:
ptrFunc.cpp:87:29: error: invalid use of non-static member function ‘int A::add(int, int)’
ptrFunc.cpp:88:30: error: invalid use of non-static member function ‘int A::subtract(int, int)’
coz I haven't specified which object to invoke(and I don't want to use static methods for now)
EDIT: several comments and answers suggested that the non-static version(as I have written) is not possible.(thanks to all) So, Modifying the class in the following manner also wont work:
#include <iostream>
using namespace std;
class A
{
int res;
public:
A(int choice)
{
int (*plus)(int, int) = A::add;
int (*minus)(int, int) = A::subtract;
if(choice == 1)
res = operation(7, 5, plus);
if(choice == 2)
res = operation(20, 2, minus);
cout << "result of operation = " << res;
}
int add(int first, int second)
{
return first + second;
}
int subtract(int first, int second)
{
return first - second;
}
int operation(int first, int second, int (*functocall)(int, int))
{
return (*functocall)(first, second);
}
};
int main()
{
int a, b;
A a_plus(1);
A a_minus(2);
return 0;
}
generated this error:
ptrFunc.cpp: In constructor ‘A::A(int)’:
ptrFunc.cpp:11:30: error: cannot convert ‘A::add’ from type ‘int (A::)(int, int)’ to type ‘int (*)(int, int)’
ptrFunc.cpp:12:31: error: cannot convert ‘A::subtract’ from type ‘int (A::)(int, int)’ to type ‘int (*)(int, int)’
may I know how to solve this issue please?
thanks
The syntax to declare a function pointer to member methods is:
int (A::*plus)(int, int) = &A::add;
int (A::*minus)(int, int) = &A::subtract;
To invoke member methods use .* or ->* operator:
(a_plus.*plus)(7, 5);
Also have a look at http://msdn.microsoft.com/en-us/library/b0x1aatf(v=vs.80).aspx
Hope this helps.
Complete code:
#include <iostream>
using namespace std;
class A
{
public:
int add(int first, int second)
{
return first + second;
}
int subtract(int first, int second)
{
return first - second;
}
int operation(int first, int second, int (A::*functocall)(int, int))
{
return (this->*functocall)(first, second);
}
};
int main()
{
int a, b;
A a_plus, a_minus;
int (A::*plus)(int, int) = &A::add;
int (A::*minus)(int, int) = &A::subtract;
a = a_plus.operation(7, 5, plus);
b = a_minus.operation(20, a, minus);
cout << "a = " << a << " and b = " << b << endl;
return 0;
}
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