On 8/5/13 10:57 PM, DJ Mills wrote:
> On Sun, Aug 4, 2013 at 4:41 PM, Andreas Schwab wrote:
>
>> Chris Down writes:
>>
>> x+=a is the same as x=x+a. Now replace a by (x=1) and it becomes
>> obvious that 1 is a perfectly valid outcome.
>>
>>
> I still don't see that; for one thing even with:
>
>
Dan Douglas writes:
> Evaluating the expression on the RHS
Now do that strictly from left to right.
Andreas.
--
Andreas Schwab, sch...@linux-m68k.org
GPG Key fingerprint = 58CA 54C7 6D53 942B 1756 01D3 44D5 214B 8276 4ED5
"And now for something completely different."
On Tue, Aug 6, 2013 at 3:20 AM, Andreas Schwab wrote:
> DJ Mills writes:
>
> > I still don't see that;
>
> Go strictly from left to right.
>
>
It doesn't make any kind of logical sense to do so. Yes, then the "(x =
1)" would be evaluated
last, but how do you evaluate the entire statement, for th
On Tuesday, August 06, 2013 09:20:38 AM Andreas Schwab wrote:
> DJ Mills writes:
>
> > I still don't see that;
>
> Go strictly from left to right.
How do you assign something that hasn't been evaluated yet? Evaluating the
expression on the RHS is an absolute prerequisite to evaluating the assi
DJ Mills writes:
> I still don't see that;
Go strictly from left to right.
> gcc agrees with me.
No, it doesn't.
> Especially with the parens there,
Parens have no effect at all.
Andreas.
--
Andreas Schwab, sch...@linux-m68k.org
GPG Key fingerprint = 58CA 54C7 6D53 942B 1756 01D3 44D5 21
On Sun, Aug 4, 2013 at 4:41 PM, Andreas Schwab wrote:
> Chris Down writes:
>
> x+=a is the same as x=x+a. Now replace a by (x=1) and it becomes
> obvious that 1 is a perfectly valid outcome.
>
>
I still don't see that; for one thing even with:
#include
int main(void) {
int x = 0;
printf("
On Sun, Aug 4, 2013 at 9:08 PM, Linda Walsh wrote:
>
>
> Chris Down wrote:
>
>> Yes, I agree, it becomes ambiguous when described in this fashion. I
>> think the
>> aesthetics of x+=y vs x=x+y are important here.
>>
>
> From the bash manpage, it would see that += is higher precedence
> than assig
Dan Douglas writes:
> = and += have equal precedence. Associativity is right to left, as described
> in
> operator(7). x=1 occurs first.
Associativity doesn't define evaluation order, only how the expression
is parsed.
Andreas.
--
Andreas Schwab, sch...@linux-m68k.org
GPG Key fingerprint =
Dan Douglas wrote:
On Sunday, August 04, 2013 06:08:18 PM Linda Walsh wrote:
From the bash manpage, it would see that += is higher precedence
than assignment, so the increment would be done first, followed
by the attempt at an assignment of 1 to 1.
= and += have equal precedence. Associativi
On Sunday, August 04, 2013 06:08:18 PM Linda Walsh wrote:
> From the bash manpage, it would see that += is higher precedence
> than assignment, so the increment would be done first, followed
> by the attempt at an assignment of 1 to 1.
= and += have equal precedence. Associativity is right to left
Chris Down wrote:
Yes, I agree, it becomes ambiguous when described in this fashion. I think the
aesthetics of x+=y vs x=x+y are important here.
From the bash manpage, it would see that += is higher precedence
than assignment, so the increment would be done first, followed
by the attempt at a
On 2013-08-04 22:41, Andreas Schwab wrote:
> x+=a is the same as x=x+a.
In most cases I'd agree, in this case I think it changes the logic when
considering += as an atomic increment (which, of course, += isn't, but
aesthetically it presents itself as such) as opposed to two separate operations.
>
Chris Down writes:
> On 2013-08-03 17:04, Dan Douglas wrote:
>> Is it specified what the value of x should be after this expression?
>>
>> x=0; : $((x+=x=1))
>
> I don't know if it is specified in a standard (I suspect it may be undefined),
> but it looks pretty clear to me that the answer should
On 2013-08-03 17:04, Dan Douglas wrote:
> Is it specified what the value of x should be after this expression?
>
> x=0; : $((x+=x=1))
I don't know if it is specified in a standard (I suspect it may be undefined),
but it looks pretty clear to me that the answer should be 2. Has anyone proposed
logi
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