@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 <puneet.nsi...@gmail.com> 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 <puneet.nsi...@gmail.com> 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 <toarunb...@gmail.com> 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 <aaron.nar...@gmail.com> wrote:
> >>> @amit:+1
>
> >>> On Wed, Aug 3, 2011 at 3:14 PM, amit karmakar
> >>> <amit.codenam...@gmail.com>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 <puneet.nsi...@gmail.com> 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 <ankitsamb...@gmail.com> 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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