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not so obvious pointers

开发者 https://www.devze.com 2022-12-23 15:42 出处:网络
I have a class : class X{ public : void f ( int ) ; int a ; } ; And the task is \"Inside the code provide declarations for :

I have a class :

class X{
 public :
 void f ( int ) ;
 int a ;
} ;

And the task is "Inside the code provide declarations for :

  • pointer to int variable of class X
  • pointer to function void(int) defined inside class X
  • pointer to double variable of class X"

Ok so pointer to int a will be just int *x = &a, right ? If there is no double in X can I already create pointer to double inside this class ? And the biggest pro开发者_如何转开发blem is the second task. How one declares pointer to function ?


These are called pointers to members. They are not regular pointers, i.e. not addresses, but "sort-of" offsets into an instance of the class (it gets a bit tricky with virtual functions.) So:

  1. int X::*ptr_to_int_member;
  2. void (X::*ptr_to_member_func)( int );
  3. double X::*ptr_to_double_member;


You need to declare them as pointer-to-members. Pointers to members are different than usual pointers in that they are the address of a member of a structure or class, not an absolute address like regular pointers.

For more information, read this.


A pointer to any type is declared with an '*' after the type name and the variable name:

Franks_Class * p_franks_class; // Declares a pointer to an instance of Franks_Class.

The more obscure declaration is for the function:

typedef void (*My_Int_Function_Ptr)(int parameter);

which declares a type (synonym), named My_Int_Function_Ptr, a pointer to a function taking an int parameter and returning void.

From this and a good C++ book, you should be able to answer the rest of your questions.

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