@amit:+1

On Wed, Aug 3, 2011 at 3:14 PM, amit karmakar <[email protected]>wrote:

> You are wrong.
> The above program invokes undefined behavior. Read the standard
> language draft to know about sequence points, side effects and
> undefined behavior.
>
> Between a previous and next sequence point a variable's value cannot
> be modified twice.
> c-faq should be quite useful
> http://c-faq.com/~scs/cgi-bin/faqcat.cgi?sec=expr
>  On Aug 3, 5:20 pm, Puneet Gautam <[email protected]> wrote:
> > As we know:
> >                      In an expression, if pre n post occur
> > simultaneously, pre inc the value then n there only n post executes it
> > after that expression...and expression evaluates right to left...
> >
> >  Also, the value of a variable in  an expression can be modified
> > multifold times...there is no restriction on dat...
> >
> > Here in this code:
> > Print statement No.:
> >
> > 1.  i++*i++ is equivalent to:
> >          output i*i(7*7)
> >        followed by
> > i=i+1;
> > i=i+1;
> > prior to 2nd printf statement..that makes i=9
> >
> > 2. i++*++i
> >     expn. evaluates right to left: i inc. by one due to pre..
> > i is now 10 .
> > output i*i(10*10)
> > i=i+1 (due to post inc., it inc. the value after the output)
> >  i is now 11
> >
> > 3. ++i*i++
> >      right to left evaluation, but post inc. increases value only after
> output..
> > coming to ++i in the expn., i inc. to 12
> > output: 12*12
> > i=i+1(due to postinc.)
> >  i is now 13
> >
> > 4. ++i*++i
> > both pre inc operators, order of evaluation doesnt ,matter:
> > i=i+1
> > i=i+1
> > output: 15*15
> >
> > i finishes at 15
> >
> > Hence the output:
> > 49
> > 100
> > 144
> > 225
> >
> > I think i made it clear..
> > Feel free to point any loopholes..
> >
> > Thanks.
> >
>  > On 8/3/11, ankit sambyal <[email protected]> wrote:
> >
> >
> >
> >
> >
> >
> >
> > > Its compiler dependent. Acc. to the C standard an object's stored value
> can
> > > be modified only once in an expression.
> >
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