--- Additional Comments From d_picco at hotmail dot com 2005-02-23 23:27
---
I won't press the issue further because I have other things more pressing ;)
But I think the decision to not change the behaviour here is wrong. I cannot
create an Integer class that acts as an int due t
--- Additional Comments From d_picco at hotmail dot com 2005-02-23 20:46
---
Here is a better clarification:
Case 1
==
int a = 0;
int b = a++ + a++;
printf("b = %d\n", b); // output is 0
Case 2
==
class A
{
int a_;
public:
A() : a_(0) {}
int operator++() {
--- Additional Comments From d_picco at hotmail dot com 2005-02-23 20:38
---
The point I was making with my example is that the native types (int, long,
char, etc...) have different behaviour than a user-defined class with the
operator++. If it is compiler dependent which way the
--- Additional Comments From d_picco at hotmail dot com 2005-02-23 20:24
---
(In reply to comment #0)
> Consider these situations:
>
> int a = 0;
> int b = a++ + a++;
> int c = (a++) + (a++);
> int d = a++ + (a++);
> int e = (a++) + a++;
>
> b == c == d ==
Version: 3.4.3
Status: UNCONFIRMED
Severity: normal
Priority: P2
Component: c
AssignedTo: unassigned at gcc dot gnu dot org
ReportedBy: d_picco at hotmail dot com
CC: gcc-bugs at gcc dot gnu dot org
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=20181
