please these questions are compiler dependent and have no standard
answers...........these are rarely asked by companies

On Sun, Aug 7, 2011 at 1:23 PM, Gary Drocella <[email protected]> wrote:

> @puneet The provided faq is garbage, if you want to learn about the
> semantics of the C programming
> language, then refer to this original ISO spec here
> http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1124.pdf
> I also suggest that for all programming languages (OCaml, Ruby, lua
> script, etc)
>
> It is definitely not the OS that determines how to execute a given
> fragment of C code, this is the job of the compiler.
> The compiler converts the code into machine code using grammar parsing
> techniques, and the validity of compiling
> C code such as this
>
> [CODE]
> int i = 1, j = 3;
> printf("%d", i+++++j);
> [/CODE]
>
> will more then likely depend on w/e compiler you're using.  If it's a
> good compiler like gcc you can tell the compiler what standard to use.
> gcc -ansi -o foo foo.c
> gcc -c99 -o foo foo.c
>
> The only time an OS would do any parsing of code is if you built an
> interpreter such as the JVM on the bare metal of a machine. Benefit
> being
> you have a garbage collector cleaning up un-used memory.  The OS is
> more of an abstraction between userspace and the hardware.
>
> On Aug 7, 3:22 pm, Puneet Gautam <[email protected]> wrote:
> > Also guys, this link:http://c-faq.com/~scs/cgi-bin/faqcat.cgi?sec=expr
> >
> > discusses erroneous expns like
> > a[i]=i++;
> >
> > But if u run this code..this gives no error on GNU compiler..
> >
> > So, are we really referring to a reliable document here..?
> http://c-faq.com/~scs/cgi-bin/faqcat.cgi?sec=expr
> >
> > I really doubt that...!
> >
> > Run it on as many different systems as you can...!
> >
> > Lets c what all results we get...!!
> >
> > Pls give ur feedback...
> >
> > On 8/8/11, Puneet Gautam <[email protected]> wrote:
> >
> >
> >
> >
> >
> >
> >
> > > @ Amit: Well, the link you have posted refers that i++*i++ is not an
> > > invalid expression, it just runs differently on different OS's because
> > > every OS has a different implementation on how to solve such
> > > expressions that use the same address space(Address of the integer i).
> >
> > > As far as i know the value of a variable can be increased multiple
> > > times on most OS's ... i have tried it on TUrbo running on WIN 95 as
> > > well as Dev Cpp on WIN 7 OS.
> >
> > > Though it may give different output on Unix OS's like Linux but:\
> >
> > >  Hey, we can only predict the output as we see is apt as far as we
> > > have learnt in books.
> >
> > > And the output in my first post does the processing the bookish way...
> >
> > > BUT THE CODE WILL NOT GIVE ANY ERROR ....!!!
> >
> > > On 8/3/11, Arun <[email protected]> wrote:
> >
> > >> What Amit told is exactly correct. But I would like to know the
> > >> expression evaluation order of this in gcc and turboc
> >
> > >> Arun
> > >> On Aug 3, 6:15 pm, Arun Vishwanathan <[email protected]> wrote:
> > >>> @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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-- 
*UTKARSH SRIVASTAV
CSE-3
B-Tech 3rd Year
@MNNIT ALLAHABAD*

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