for compiler giving 8 bit for pointers, shouldn't int also be of 8 bits??
On Thu, Jul 28, 2011 at 12:26 PM, Jnana Sagar <supremeofki...@gmail.com>wrote:
> @aditya..the answer may vary, because c is machine dependent
> language..in few machines int is of 2 bytes and char is of 1 byte..u
> can't say..it varies with diff machines
>
> On 7/26/11, aditya kumar <aditya.kumar130...@gmail.com> wrote:
> > char *s[5] is a array of pointers of type char . but the thing is size of
> > pointers is 4byte irrespective of its type . coz address is always an
> > unsigned int which is of 4byte.
> > On Tue, Jul 26, 2011 at 7:11 PM, Puneet Gautam
> > <puneet.nsi...@gmail.com>wrote:
> >
> >> @everyone:
> >> I have this mind strangling doubt..!!!
> >>
> >> Why is "char *s[5]" of 20 bytes...?
> >>
> >> yes the output is 28...
> >>
> >> On 7/26/11, Don <dondod...@gmail.com> wrote:
> >> > A reasonable guess would be 28 bytes. But the size of a structure is
> >> > implementation dependent, and therefore, some other result could be
> >> > correct as well.
> >> > Don
> >> >
> >> > On Jul 26, 7:40 am, Puneet Gautam <puneet.nsi...@gmail.com> wrote:
> >> >> #include<stdio.h>
> >> >> #include<stddef.h>
> >> >> struct node{
> >> >> int a;
> >> >> char *b[5];
> >> >> struct node *link;
> >> >> };
> >> >> main()
> >> >> {
> >> >> int a;
> >> >> a=sizeof(struct node);
> >> >> printf("%d",a);
> >> >> getchar();
> >> >> return 0;
> >> >> }
> >> >>
> >> >> Whats the output..?
> >> >
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Tushar Bindal
Computer Engineering
Delhi College of Engineering
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E-Mail : tushicom...@gmail.com
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