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ryt

C++ Base class variables in inherited

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Consider the following classes and pInt declaration:

class A { ... };
class B : public A
{
  void function()
  {
    int A::*pInt;
    
    // use pInt
  }
};

Where does pInt belongs, is it local to B::function() or it's a member of A?

Edited by ryt

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pInt belongs to B::function() . That's to say, its scope is limited in B::function() .

Though it's a pointer to member of A.

 

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7 hours ago, Alberth said:

I don't think that compiles at all.

 

It compiles on VC++ 2017 but it's anyone's guess what it actually does.

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3 hours ago, Gnollrunner said:

It compiles on VC++ 2017 but it's anyone's guess what it actually does.

It's a pointer to class A int field.

class A {
public:
	int foo;
};

class B : public A
{
public:
	void function()
	{
		// set pointer to class field
		int A::*pInt = &A::foo;

		// use pointer to modify some field
		A a;
		a.*pInt = 5;
	}
};

 

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4 minutes ago, Zaoshi Kaba said:

It's a pointer to class A int field.

 

Cool, I messed with it a bit and got as far as doing the same assignment you did but I I didn't figure out putting the * after the a. and not before. In any case I'll almost for sure never need it. It's pretty esoteric, but at least I know what it is now.

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@Zaoshi Kaba thanks for the code, but why would I have to do this. Can't I just use "int *pInt2 = &A::foo;" instead of giving it "A::"?

Oh, I see, than I could not use "a.*pInt = 5;". Now I'm confused even more. ".*pInt" is not defined in A but we still use a dot "." for it??
So I guess, only B::function() thinks A has a member pInt, if I'm right.

So what's the use of all this?

Edited by ryt

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34 minutes ago, ryt said:

@Zaoshi Kaba thanks for the code, but why would I have to do this. Can't I just use "int *pInt2 = &A::foo;" instead of giving it "A::"?

Oh, I see, than I could not use "a.*pInt = 5;". Now I'm confused even more. ".*pInt" is not defined in A but we still use a dot "." for it??
So I guess, only B::function() thinks A has a member pInt, if I'm right.

So what's the use of all this?

Errr  sorry. We kind of hijacked your thread.  You probably didn't mean to to code what you coded. I was just so surprised it actually compiled that I wanted to know why.  The way you declared pInt,  it's simply  a local variable of function. It's of a very odd type, one that most programmers would never use, but a local variable none the less.  Putting the type int in front of it in the function makes it local to that function.  Perhaps you meant to put it in class A which would make it a member of A which you could access from function since B inherits from A as follows:

class A {
protected:
  int *pInt;
};

class B : public A
{
public:
  void function()
  {
    pInt = new int();
  }
};

I'm not really sure what you are tying to do, but you stumbled across a seldom used feature of the language.

Edited by Gnollrunner

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7 hours ago, Gnollrunner said:

Errr  sorry. We kind of hijacked your thread.  You probably didn't mean to to code what you coded. I was just so surprised it actually compiled that I wanted to know why.  The way you declared pInt,  it's simply  a local variable of function. It's of a very odd type, one that most programmers would never use, but a local variable none the less.  Putting the type int in front of it in the function makes it local to that function.  Perhaps you meant to put it in class A which would make it a member of A which you could access from function since B inherits from A as follows:

I'm not really sure what you are tying to do, but you stumbled across a seldom used feature of the language.

No worries, the thread is not hijacked, it's actually what I wanted. This is not my code, I'm reading a book and the author used this to explain something else though I did not understand this part and that's why I'm asking.
Declaration of "int A::*pInt;" in B::function() was intentional.

So, does putting "A::" in front of it makes it only "bound" to objects of A, so the pointer can access only A members? I'm still confused about it and using the pointer with a "." as it's not declared in original A class.

Edited by ryt

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38 minutes ago, ryt said:

So, does putting "A::" in front of it makes it only "bound" to objects of A, so the pointer can access only A members?

Apparently so. I'm not sure why anyone would need that level of protection but there you have it. You can use a plain old int * to do the same thing, the  only difference being it can point to any int, and not just one which is a member of class A.

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