Re: operator++ function

From:
Rahul <sam_cit@yahoo.co.in>
Newsgroups:
comp.lang.c++
Date:
Sat, 23 Feb 2008 09:56:44 -0800 (PST)
Message-ID:
<c27b26c5-93fb-42a8-961a-531f2c432d5b@p25g2000hsf.googlegroups.com>
On Feb 23, 6:05 pm, Jeff Schwab <j...@schwabcenter.com> wrote:

Jeff Schwab wrote:

Rahul wrote:

 I have the following code,


[code similar to:]

#include <iostream>

struct A {
    A() { std::cout << "in A\n"; }
    ~A() { std::cout << "in ~A\n"; }

    A operator++(int) {
        std::cout << "in operator function\n";
        return A();
    }
};

int main() {
    A obj;
    std::cout << "1\n";
    obj++;
    std::cout << "2\n";
    return 0;
}

and the output i get is,

in A
1
in A
in operator function
in ~A // Destructor for local object
in operator function
in ~A // Destructor is invoked for
which object???
2
in ~A


I think you may have copied either the code or the output incorrectly.
Your output indicates two invocations of operator++(int), but the code
shows only one.

What you should be seeing is:

in A
1
in operator function
in A
in ~A
2
in ~A

The inner constructor/destructor pair is for the temporary object
returned by the post-increment function.

second, is it possible to have the operator++(int) [postfix operator+
+] to return a reference? As per my program it isn't possible as
returning reference of a local variable doesn't make any sense...


It's possible, but it rarely makes sense. In general, it is better to
define the pre-increment operator first, and return the current object
by reference:

    A& operator++() {
        // ... increment this object ...
        return *this;
    }

If you need post-increment, you can them implement it like this:

    A operator++(int) {


Whoops! That should be:

     A const operator++(int) {

Otherwise, the client of this code might inadvertently call a non-const
method on the copy, but think they are modifying the original object.
For example, the following will by syntactically valid, but probably
will not do what the programmer expects:

     A a;
     ++(a++);

        A const old_value = *this;
        ++*this;
        return old_value;
    }


Isn't the temp object constant by default?

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