Postfix and Prefix Override

Question

0.00/5 (No votes)

See more:

Hi,

I have a little problem that I can not figure out the solution, if exist (I really hope so :) ). I have one base class, Base, and inside of it, I have two nested classes, A and B (class B is derived from A). In class A, I have two operatores: increment posfix and addition assignment. In class B, I only have the increment prefix (code below).

My problem is: Why it is call the increment prefix operator instead of the increment posfix operator when I do b++?

Best regards,
Filipe Marques

Code:

C++

#include <stdio.h>

class Base
{
public:
	class A
	{
	protected:
		A() {}
	public:
		A& operator++(int)
		{
			printf("1\n");
			return A();
		}

		A& operator+=(unsigned int __num)
		{
			printf("2\n");
			return A();
		}
	};

	class B : public A
	{
	public:
		B() : A() {}

		B& operator++()
		{
			printf("3\n");
			return B();
		}
	};

	B get_b()
	{
		return B();
	}
};

void main()
{
	Base base;
	Base::B b = base.get_b();

	// No problem
	b += 1;

	// warning C4620:	no postfix form of 'operator ++' found
	// for type 'Base::B', using prefix form
	b++;
}

The output id:

2
3

Posted 17-Oct-13 12:26pm

Filipe Marques

Add a Solution

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Andreas Gieriet · Accepted Answer · 2013-10-17T13:07:00

What happens here is "hiding".

First an example for normal functions:

C++

class A
{
public:
  void f(); // 1
};
class B: public A
{
public:
  void f(int); // 2
};

The function at 2 hides the one at 1. You can circumvent this by a using A::f; in B.

The operator++ has two signatures: one without an argument and the other with argument. When calling one or the other, you have two options, e.g. for operators as class members:

implicit call	explicit call
C++ ++a;	C++ a.operator++();
C++ a++;	C++ a.operator++(0);

The definition in the class is:

C++

class C
{
public:
  C& operator++();    // prefix overload:  ++a
  C  operator++(int); // postfix overload: a++
};

So, finally: since the two operators are overloads of each other, you hide in your B class the one from the A class.
Cheers
Andi

PS: The warning says it all ;-). That the compiler decides to take the "wrong" operator is debatable, but the warning tells that it does so.

User 59241 · Accepted Answer · 2013-10-17T16:54:00

Solution 2

Andreas Gieriet has given the explanation very clearly.

If you wish to reintroduce operator++ from A into the scope of B you may employ

C++

using

.

http://publib.boulder.ibm.com/infocenter/comphelp/v101v121/index.jsp?topic=/com.ibm.xlcpp101.aix.doc/language_ref/using_declaration_class_members.html[^]

So then:

C++

	class B : public A
	{
	public:
		using A::operator++;
		B() : A() {}
 
		B& operator++()
		{
			printf("3\n");
			return B();
		}
	};
 
	B get_b()
	{
		return B();
	}
};

Posted 17-Oct-13 16:54pm

User 59241

Updated 17-Oct-13 17:09pm

v2

Comments

nv3 18-Oct-13 5:45am

A 5 to both of you guys for the interesting explanation.

[no name] 18-Oct-13 8:59am

Thankyou

Filipe Marques 19-Oct-13 10:25am

Andreas Gieriet and Pwasser, thanks for yours answers. I did not know the hide concept. I thought that a method is declared as public in base class, keeps accessible from derived class (for same methods with differents signatures). One more time, thanks for your answers :) . Best regards, Filipe Marques

[no name] 19-Oct-13 19:07pm

Thanks & Cheers