Re: operator++ function

From:
Rahul <sam_cit@yahoo.co.in>
Newsgroups:
comp.lang.c++
Date:
Sat, 23 Feb 2008 05:04:04 -0800 (PST)
Message-ID:
<9cd03519-336e-4645-b84f-c70b07eaba92@o10g2000hsf.googlegroups.com>
On Feb 23, 5:59 pm, Jeff Schwab <j...@schwabcenter.com> 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) {
         A const old_value = *this;
         ++*this;
         return old_value;
     }


Sorry, that was a editing mistake

in A
1
in A
in operator function
in ~A
in ~A /* this doesn't make sense */
2
in ~A

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