undefined behaviour

[Dan Cernat]

Someone posted earlier an example that looked like taken from the
standard. It was: i = 7, i++, i++; and the result should be 9 (i = 9
in the end)
I thought that the compiler isn't required to do the assignement until
the right hand side expression is fully evaluated. Apparently, the
above expression is handled as:
i = 7;
i++;
i++;
which I don't think is right. Could someone explain, please?

/dan

[Josh Sebastian]

Re: undefined behaviour

On Tue, 11 Nov 2003 08:24:28 -0800, Dan Cernat wrote:
[color=blue]
> Someone posted earlier an example that looked like taken from the
> standard. It was: i = 7, i++, i++; and the result should be 9 (i = 9
> in the end)
> I thought that the compiler isn't required to do the assignement until
> the right hand side expression is fully evaluated. Apparently, the
> above expression is handled as:
> i = 7;
> i++;
> i++;
> which I don't think is right. Could someone explain, please?[/color]

Yes, it's handled like that. You're almost right about assignment:
operator= has higher precedence than operator, so it's the same as if you
wrote

(i=7),i++,i++

The assignment happens first, followed by a post-increment, followed by a
post-increment. The comma operator inserts a sequence point, so the
behavior isn't undefined.

Josh