c# Post-incrementing problem

[Pat]

Re: c# Post-incrementing problem

Yes, I agree that for the x++ portion, x will take on the
value of x before the statement (ie. x=0), but my
understanding was that x++ has an implicit assignment. In
other words, I thought x++ is equivalent to x=x+1

Regardless, it seems wrong to me, but may be right
according to the c# spec.

[color=blue]
>-----Original Message-----
>According to the C# language specification, in 14.5.9[/color]
Paragraph 5,[color=blue]
>line 2, states:
>
>"The result of x++ or x--is the value of x before the[/color]
operation,[color=blue]
>whereas the result of ++x or --x is the value of x after[/color]
the[color=blue]
>operation."
>
>You can check this out yourself:
>
>int x=0;
>MessageBox.Sho w(x++.ToString( )); //shows "0"
>
>And the reason "x=x++;" doesn't do anything is because[/color]
assignment[color=blue]
>occurs before increment (hence post-*), and the *value* of[/color]
x is[color=blue]
>increased, but is not stored (as assignment has already[/color]
occured).[color=blue]
>
>Austin Ehlers
>
>
>On Wed, 10 Sep 2003 12:48:25 -0700, "Patrick Wood"
><pwood-nospam@boeing.c om> wrote:
>[color=green]
>>I found a problem with C# and post increments. I was[/color][/color]
going[color=blue][color=green]
>>through some source code in c++ and found someone did a
>>post increment:
>>
>>int x=0;
>>for(int i=0; i<10; i++)
>>{
>>x = x++;
>>}
>>
>>In c++, you'd get 0,1,...,9 (as expected). This doesn't
>>work in C#. In C#, you get 0,...,0. It seems the post
>>increment is ignored in this case. However, the lines:
>>x=4;
>>y=x++;
>>produces y=4, x=5 (as expected)
>>
>>Should c# be behaving this way or is there a compiler
>>error? After looking at the assembly code in c++ and c#,[/color][/color]
it[color=blue][color=green]
>>looks like a bug in c#.[/color]
>
>.
>[/color]

[Jon Skeet]

Re: c# Post-incrementing problem

Pat <pwood-nospam@boeing.c om> wrote:[color=blue]
> Yes, I agree that for the x++ portion, x will take on the
> value of x before the statement (ie. x=0), but my
> understanding was that x++ has an implicit assignment.[/color]

x++ has an implicit assignment to x, yes. However, that's effectively
overridden by the later assignment to x.
[color=blue]
> In other words, I thought x++ is equivalent to x=x+1[/color]

Yes, it is - but the value of the expression is the *original* value,
and that's what then gets assigned by the "x=" part of x=x++;
[color=blue]
> Regardless, it seems wrong to me, but may be right
> according to the c# spec.[/color]

Basically remember that the increment happens immediately the
expression is evaluated, *not* after the whole statement is evaluated.

--
Jon Skeet - <skeet@pobox.co m>

http://www.pobox.com/~skeet

If replying to the group, please do not mail me too

[Jack Hanebach]

Re: c# Post-incrementing problem

"Pat" <pwood-nospam@boeing.c om> wrote in message
news:072e01c378 57$eab600b0$a50 1280a@phx.gbl.. .[color=blue]
> Yes, I agree that for the x++ portion, x will take on the
> value of x before the statement (ie. x=0), but my
> understanding was that x++ has an implicit assignment. In
> other words, I thought x++ is equivalent to x=x+1
>
> Regardless, it seems wrong to me, but may be right
> according to the c# spec.[/color]

Look at it this way:

x = x++; is equivalent to x = x.operator++();

where int.operator++( ) translates to something like

{
int temp = this;
this = this + 1;
return temp;
}

Now the behaviour you see makes sense, doesn't it.

[Stu Smith]

Re: c# Post-incrementing problem

This is the crucial point, and one I was just about to make.

People assume if their C++ compiler does something, then every other C++
compiler should do the same.

I'm not sure why you'd want to do this in C# anyway, it's absolutely
horrible code.


"David Olsen" <qg4h9ykc5m@yah oo.com> wrote in message
news:uMyvY4%23d DHA.3096@TK2MSF TNGP11.phx.gbl. ..[color=blue]
> Patrick Wood wrote:[color=green]
> > I found a problem with C# and post increments. I was going
> > through some source code in c++ and found someone did a
> > post increment:
> >
> > int x=0;
> > for(int i=0; i<10; i++)
> > {
> > x = x++;
> > }
> >
> > In c++, you'd get 0,1,...,9 (as expected).[/color]
>
> In C++, the behavior of this code is undefined. 'x' is modified twice
> between sequence points (once by ++, once by =), so the C++ standard
> makes no guarantees about the code's behavior. You can't expect the
> code to do anything in particular.
>
> --
> David Olsen
> qg4h9ykc5m@yaho o.com
>[/color]