Sizeof and increment operators

[Anjali M]

#include <stdio.h>

int main()
{
int number = 10;
printf("Number is: %d\n", number);

printf("Sizeof of number is: %d\n", sizeof(number++ ));

printf("Number now is: %d\n", number);

return 0;
}


The post-increment operator doesnt seem to work here. The value of the
number remains the same even after the call to the sizeof operator.
Can someone clarify this please?

Thanks,
Anjali.

[Richard Bos]

Re: Sizeof and increment operators

g_managoli@redi ffmail.com (Anjali M) wrote:
[color=blue]
> printf("Sizeof of number is: %d\n", sizeof(number++ ));[/color]
[color=blue]
> The post-increment operator doesnt seem to work here. The value of the
> number remains the same even after the call to the sizeof operator.
> Can someone clarify this please?[/color]

From the last public draft of the C99 Standard, 6.5.3.4#2:

# The sizeof operator... If the type of the operand is a variable length
# array tye, the operand is evaluated; otherwise, the operand is not
# evaluated and the result is an integer constant.

Richard

[smitha]

Re: Sizeof and increment operators

Hi,

The sizeof operator just returns the size of the type or expressions,

so i guess,it is not defined to allow airthmetic operations within it.

Bye,

Smitha

"Anjali M" <g_managoli@red iffmail.com> wrote in message
news:997f953e.0 407052026.38f5e 1f2@posting.goo gle.com...[color=blue]
> #include <stdio.h>
>
> int main()
> {
> int number = 10;
> printf("Number is: %d\n", number);
>
> printf("Sizeof of number is: %d\n", sizeof(number++ ));
>
> printf("Number now is: %d\n", number);
>
> return 0;
> }
>
>
> The post-increment operator doesnt seem to work here. The value of the
> number remains the same even after the call to the sizeof operator.
> Can someone clarify this please?
>
> Thanks,
> Anjali.[/color]

[Sumukh]

Re: Sizeof and increment operators

HI anjali,

Sizeof is an *operator* in C.
The operand expected in sizeof is either an identifier that is a
unary-expression, or a type-cast expression (that is, a type specifier
enclosed in parentheses). The unary-expression cannot represent a
bit-field object, an incomplete type, or a function designator.

A unary expression is an expression whose valuation is computed by
applying its (unary) operator to the valuation of its operand. There are
three kinds of unary operators:
(1) the logical complement unary operator "!",
(2) the plus unary operator "+", and
(3) the minus unary operator "-".

Hope this helps

sumukh


Anjali M wrote:
[color=blue]
> #include <stdio.h>
>
> int main()
> {
> int number = 10;
> printf("Number is: %d\n", number);
>
> printf("Sizeof of number is: %d\n", sizeof(number++ ));
>
> printf("Number now is: %d\n", number);
>
> return 0;
> }
>
>
> The post-increment operator doesnt seem to work here. The value of the
> number remains the same even after the call to the sizeof operator.
> Can someone clarify this please?
>
> Thanks,
> Anjali.[/color]

[Nitin Bhardwaj]

Re: Sizeof and increment operators

g_managoli@redi ffmail.com (Anjali M) wrote in message news:<997f953e. 0407052026.38f5 e1f2@posting.go ogle.com>...[color=blue]
> #include <stdio.h>
>
> int main()
> {
> int number = 10;
> printf("Number is: %d\n", number);
>
> printf("Sizeof of number is: %d\n", sizeof(number++ ));
>
> printf("Number now is: %d\n", number);
>
> return 0;
> }
>
>
> The post-increment operator doesnt seem to work here. The value of the
> number remains the same even after the call to the sizeof operator.
> Can someone clarify this please?
>
> Thanks,
> Anjali.[/color]

Hi Anjali,

The problem is not that of the ++ operator...but the 'sizeof'
operator.
The C Standard states that the operand of sizeof can be either a type
name or an expression.

e.g. sizeof (int) OR sizeof (a+b)

If it is an expression,then it will not be evaluated; See section
A.7.4.8 of K&R2 - [The C Programming Language 2/ed - Kernighan &
Ritchie ], Appendix A.

For the sake of convenience I'm reproducing the contents verbatim
here :

A.7.4.8 Sizeof Operator
The sizeof operator yields the number of bytes required to store an
object of the type of its operand. The operand is either an
expression,***_ which is not evaluated_**, or a parenthesized type
name. When sizeof is applied to a char, the result is 1; when applied
to an array, the result is the total number of bytes in the array.
When applied to a structure or union, the result is the number of
bytes in the object, including any padding required to make the object
tile an array: the size of an array of n elements is n times the size
of one element. The operator may not be applied to an operand of
function type, or of incomplete type, or to a bit-field. The result is
an unsigned integral constant; the particular type is
implementation-defined.

Hence your expresion number++ never gets evaluated.So the value of
number doesn't increase.

HTH
Nitin

[Nitin Bhardwaj]

Re: Sizeof and increment operators

"smitha" <smithas@cisco. com> wrote in message news:<108909700 7.531247@sj-nntpcache-3>...[color=blue]
> Hi,
>
> The sizeof operator just returns the size of the type or expressions,
>
> so i guess,it is not defined to allow airthmetic operations within it.[/color]

Wrong, the operand of sizeof operator is either type-names OR
expressions - which also includes arithmetic expressions.If it were
not defined for arithmetic expressions then the compiler must have
complained !!

You can even include a function call in the argument to sizeof !!![color=blue]
>
> Bye,
>
> Smitha
>
> "Anjali M" <g_managoli@red iffmail.com> wrote in message
> news:997f953e.0 407052026.38f5e 1f2@posting.goo gle.com...[color=green]
> > #include <stdio.h>
> >
> > int main()
> > {
> > int number = 10;
> > printf("Number is: %d\n", number);
> >
> > printf("Sizeof of number is: %d\n", sizeof(number++ ));
> >
> > printf("Number now is: %d\n", number);
> >
> > return 0;
> > }
> >
> >
> > The post-increment operator doesnt seem to work here. The value of the
> > number remains the same even after the call to the sizeof operator.
> > Can someone clarify this please?
> >
> > Thanks,
> > Anjali.[/color][/color]

[smitha]

Re: Sizeof and increment operators

yes... sizeof is an operator and not a function. In a function call each
argument will be evaluated but this is not the case in sizeof.

"Nitin Bhardwaj" <nitinbhardwaj8 0@hotmail.com> wrote in message
news:17fa8780.0 407060427.4a1f3 a99@posting.goo gle.com...[color=blue]
> "smitha" <smithas@cisco. com> wrote in message[/color]
news:<108909700 7.531247@sj-nntpcache-3>...[color=blue][color=green]
> > Hi,
> >
> > The sizeof operator just returns the size of the type or expressions,
> >
> > so i guess,it is not defined to allow airthmetic operations within it.[/color]
>
> Wrong, the operand of sizeof operator is either type-names OR
> expressions - which also includes arithmetic expressions.If it were
> not defined for arithmetic expressions then the compiler must have
> complained !!
>
> You can even include a function call in the argument to sizeof !!![color=green]
> >
> > Bye,
> >
> > Smitha
> >
> > "Anjali M" <g_managoli@red iffmail.com> wrote in message
> > news:997f953e.0 407052026.38f5e 1f2@posting.goo gle.com...[color=darkred]
> > > #include <stdio.h>
> > >
> > > int main()
> > > {
> > > int number = 10;
> > > printf("Number is: %d\n", number);
> > >
> > > printf("Sizeof of number is: %d\n", sizeof(number++ ));
> > >
> > > printf("Number now is: %d\n", number);
> > >
> > > return 0;
> > > }
> > >
> > >
> > > The post-increment operator doesnt seem to work here. The value of the
> > > number remains the same even after the call to the sizeof operator.
> > > Can someone clarify this please?
> > >
> > > Thanks,
> > > Anjali.[/color][/color][/color]

[Dan Pop]

Re: Sizeof and increment operators

In <1089097007.531 247@sj-nntpcache-3> "smitha" <smithas@cisco. com> writes:
[color=blue]
>The sizeof operator just returns the size of the type or expressions,[/color]

Without evaluating them (except for VLAs in C99).
[color=blue]
>so i guess,it is not defined to allow airthmetic operations within it.[/color]

It is perfectly defined and allowed, they just won't be performed: the
compiler will merely figure out the type of the result.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Dan.Pop@ifh.de

[Charles Richmond]

Re: Sizeof and increment operators

Anjali M wrote:[color=blue]
>
> #include <stdio.h>
>
> int main()
> {
> int number = 10;
> printf("Number is: %d\n", number);
>
> printf("Sizeof of number is: %d\n", sizeof(number++ ));
>
> printf("Number now is: %d\n", number);
>
> return 0;
> }
>
> The post-increment operator doesnt seem to work here. The value of the
> number remains the same even after the call to the sizeof operator.
> Can someone clarify this please?
>[/color]
What's the matter??? Can't you *do* the job after you
suck it up from the U. S.???

--
+-------------------------------
| Charles and Francis Richmond
| richmond at plano dot net
| Re-Defeat Bush!!!
+-------------------------------

[Richard Bos]

Re: Sizeof and increment operators

Charles Richmond <richmond@plano .net> wrote:
[color=blue]
> What's the matter??? Can't you *do* the job after you
> suck it up from the U. S.???[/color]

Got fired, didya?

Richard

[Joona I Palaste]

Re: Sizeof and increment operators

Charles Richmond <richmond@plano .net> scribbled the following:[color=blue]
> Anjali M wrote:[color=green]
>> #include <stdio.h>
>>
>> int main()
>> {
>> int number = 10;
>> printf("Number is: %d\n", number);
>>
>> printf("Sizeof of number is: %d\n", sizeof(number++ ));
>>
>> printf("Number now is: %d\n", number);
>>
>> return 0;
>> }
>>
>> The post-increment operator doesnt seem to work here. The value of the
>> number remains the same even after the call to the sizeof operator.
>> Can someone clarify this please?
>>[/color]
> What's the matter??? Can't you *do* the job after you
> suck it up from the U. S.???[/color]

You are incredibly rude. What makes you so sure the OP sucked a job from
the US? What makes you so sure the OP has a job as a programmer at all?
This might just as well be a question from a genuine C newbie. And
besides, even if he did suck up the job from the US, what business is
that of yours? This is a technical newsgroup. If you need to vent your
patriotism, go to a politics newsgroup.

--
/-- Joona Palaste (palaste@cc.hel sinki.fi) ------------- Finland --------\
\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
"You could take his life and..."
- Mirja Tolsa

[Nitin Bhardwaj]

Re: Sizeof and increment operators

>Charles Richmond <richmond@plano .net> wrote in message >news:<40ECF393 .79AED7B5@plano .net>...
[color=blue][color=green]
> >[/color]
> What's the matter??? Can't you *do* the job after you
> suck it up from the U. S.???[/color]


It seems frm yr language that u hate Indians !!

Did u lost ur job to an Indian, buddy?

Cough up yr technical skills to match-up to ours & take on the
competition instead of making political transgressions on a technical
group like c.l.c!!

[Richard Bos]

Re: Sizeof and increment operators

nitinbhardwaj80 @hotmail.com (Nitin Bhardwaj) wrote:
[color=blue]
>Charles Richmond <richmond@plano .net> wrote in message >news:<40ECF393 .79AED7B5@plano .net>...
>[color=green]
> > suck it up from the U. S.???[/color]
>
> It seems frm yr language that u hate Indians !!
>
> Did u lost ur job to an Indian, buddy?
>
> Cough up yr technical skills to match-up to ours & take on the
> competition instead of making political transgressions on a technical
> group like c.l.c!![/color]

Ow! I know the English did some pretty awful things in India, but do you
_have_ to take revenge by murdering their language? Leave that to the
USAnians, like Mr. Richmond ;->

Richard

[Dan Pop]

Re: Sizeof and increment operators

In <40ECF393.79AED 7B5@plano.net> Charles Richmond <richmond@plano .net> writes:
[color=blue]
>Anjali M wrote:[color=green]
>>
>> #include <stdio.h>
>>
>> int main()
>> {
>> int number = 10;
>> printf("Number is: %d\n", number);
>>
>> printf("Sizeof of number is: %d\n", sizeof(number++ ));
>>
>> printf("Number now is: %d\n", number);
>>
>> return 0;
>> }
>>
>> The post-increment operator doesnt seem to work here. The value of the
>> number remains the same even after the call to the sizeof operator.
>> Can someone clarify this please?
>>[/color]
>What's the matter??? Can't you *do* the job after you
>suck it up from the U. S.???[/color]

How many job assignments like this have you seen? This is not only
a newbie question, it is a perfectly legitimate newbie question.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Dan.Pop@ifh.de

[Anjali M]

Re: Sizeof and increment operators

Charles Richmond <richmond@plano .net> wrote in message news:<40ECF393. 79AED7B5@plano. net>...[color=blue]
> Anjali M wrote:[color=green]
> >
> > #include <stdio.h>
> >
> > int main()
> > {
> > int number = 10;
> > printf("Number is: %d\n", number);
> >
> > printf("Sizeof of number is: %d\n", sizeof(number++ ));
> >
> > printf("Number now is: %d\n", number);
> >
> > return 0;
> > }
> >
> > The post-increment operator doesnt seem to work here. The value of the
> > number remains the same even after the call to the sizeof operator.
> > Can someone clarify this please?
> >[/color]
> What's the matter??? Can't you *do* the job after you
> suck it up from the U. S.???[/color]

I don't know what you have got against me. I asked a very valid
question on this newsgroup and I _dont_ have to take this crap from
you. If you can be rude, so can _I_. So, better be careful before you
say something like this again.