Hi all,

int main()
{
char ch[] = "";
cout<<"size is"<<sizeof(ch)<<endl;
return 0;
}

the above code prints size of ch is 1, why is it?? even though ch[] is
empty, what for 1 byte?? Ideally size of ch[] should be zero by looking
at the code, but why 1??


Cheers..

"Radde" <msradder@gmail.com> wrote in message
news:1124196466.727712.321880@z14g2000cwz.googlegr oups.com...[color=blue]
> Hi all,
>
> int main()
> {
> char ch[] = "";
> cout<<"size is"<<sizeof(ch)<<endl;
> return 0;
> }
>
> the above code prints size of ch is 1, why is it?? even though ch[] is
> empty, what for 1 byte?? Ideally size of ch[] should be zero by looking
> at the code, but why 1??
>
>
> Cheers..
>[/color]

"" has a terminating zero ('\0') character so that makes the size 1.

Ben

Radde wrote:
[color=blue]
> Hi all,
>
> int main()
> {
> char ch[] = "";
> cout<<"size is"<<sizeof(ch)<<endl;
> return 0;
> }
>
> the above code prints size of ch is 1, why is it??[/color]

String literals are always terminated with a '\0' character.
Btw, even if this were possible as you want it, the compiler wouldn't accept
it, because objects must always have a size of at least one. So an array
with zero elements is not allowed in C++.
The reason for that is that each object must have a distinct address, and an
object that occupies no storage can't have an address.

In message <ddsrg4$tbf$01$1@news.t-online.com>, Rolf Magnus
<ramagnus@t-online.de> writes[color=blue]
>Radde wrote:
>[color=green]
>> Hi all,
>>
>> int main()
>> {
>> char ch[] = "";
>> cout<<"size is"<<sizeof(ch)<<endl;
>> return 0;
>> }
>>
>> the above code prints size of ch is 1, why is it??[/color]
>
>String literals are always terminated with a '\0' character.
>Btw, even if this were possible as you want it, the compiler wouldn't accept
>it, because objects must always have a size of at least one. So an array
>with zero elements is not allowed in C++.
>The reason for that is that each object must have a distinct address, and an
>object that occupies no storage can't have an address.
>[/color]
( Except when it can: T *p = new T[0]; :-)


--
Richard Herring

Richard Herring wrote:[color=blue]
> In message <ddsrg4$tbf$01$1@news.t-online.com>, Rolf Magnus
> <ramagnus@t-online.de> writes[color=green]
> >Radde wrote:
> >[color=darkred]
> >> Hi all,
> >>
> >> int main()
> >> {
> >> char ch[] = "";
> >> cout<<"size is"<<sizeof(ch)<<endl;
> >> return 0;
> >> }
> >>
> >> the above code prints size of ch is 1, why is it??[/color]
> >
> >String literals are always terminated with a '\0' character.
> >Btw, even if this were possible as you want it, the compiler wouldn't accept
> >it, because objects must always have a size of at least one. So an array
> >with zero elements is not allowed in C++.
> >The reason for that is that each object must have a distinct address, and an
> >object that occupies no storage can't have an address.
> >[/color]
> ( Except when it can: T *p = new T[0]; :-)
>[/color]

Well the object here is a pointer on some type T. The pointer occupies
some storage and have a distinct address. So Rolf`s definition still
hold for this example. It may contain the address on some valid memory
chuck or it may be set to 0 (a memory address that lead to undefined
behaviour if dereferenced).

kwijibo

In message <1124284031.592937.288480@f14g2000cwb.googlegroups .com>,
kwijibo28 <kwijibo28@hotmail.com> writes[color=blue]
>
>Richard Herring wrote:[color=green]
>> In message <ddsrg4$tbf$01$1@news.t-online.com>, Rolf Magnus
>> <ramagnus@t-online.de> writes[color=darkred]
>> >Radde wrote:
>> >
>> >> Hi all,
>> >>
>> >> int main()
>> >> {
>> >> char ch[] = "";
>> >> cout<<"size is"<<sizeof(ch)<<endl;
>> >> return 0;
>> >> }
>> >>
>> >> the above code prints size of ch is 1, why is it??
>> >
>> >String literals are always terminated with a '\0' character.
>> >Btw, even if this were possible as you want it, the compiler wouldn't accept
>> >it, because objects must always have a size of at least one. So an array
>> >with zero elements is not allowed in C++.
>> >The reason for that is that each object must have a distinct address, and an
>> >object that occupies no storage can't have an address.
>> >[/color]
>> ( Except when it can: T *p = new T[0]; :-)
>>[/color]
>
>Well the object here is a pointer on some type T. The pointer occupies
>some storage and have a distinct address. So Rolf`s definition still
>hold for this example. It may contain the address on some valid memory
>chuck or it may be set to 0 (a memory address that lead to undefined
>behaviour if dereferenced).[/color]

What is "it"? My 'p' contains a distinct address which is not 0.


--
Richard Herring

I'm sorry, it seems you are right. I've just took a look at the
standard and even if you allocate a zero length array the new operator
will return a "non-null pointer value". My initial thought was that
"new T[0]" must return a NULL.

This raise another question. Does omitting to delete a zero length
array will lead to a memory leak?
I guess so. Compiler must store the array length somewhere so the
delete[] operator will know how much memory to deallocate.

Kwijibo

In message <1124292013.789533.69760@g44g2000cwa.googlegroups. com>,
kwijibo28 <kwijibo28@hotmail.com> writes[color=blue]
>I'm sorry, it seems you are right. I've just took a look at the
>standard and even if you allocate a zero length array the new operator
>will return a "non-null pointer value". My initial thought was that
>"new T[0]" must return a NULL.
>
>This raise another question. Does omitting to delete a zero length
>array will lead to a memory leak?[/color]

Yes, as Michiel Salters pointed out in
<1124182196.663538.271390@g14g2000cwa.googlegroups .com>
in the thread entitled "the pointer this"...
[color=blue]
>I guess so. Compiler must store the array length somewhere so the
>delete[] operator will know how much memory to deallocate.
>[/color]
Not necessarily: it might maintain separate pools for small objects.

But as MS said, whether that's so or not, each call of new T[0] must
return a *distinct* pointer, so eventually all distinct values will be
used up.

--
Richard Herring