@Ashok : I didn't get this answer ..
i=0x3c --> what is this address .. variables has addresses but not the
values right? we are not storing 60 any where right?
0x11 = 3 in decimal format not 11 base 10.
typecasting to (int*) needs an address right?
I mean
int b=10;
int * a=(int*)&b;
On Sat, Oct 13, 2012 at 9:10 PM, Ashok Varma <verma....@gmail.com> wrote:
> This gives a clear explanation:
>
> #include<stdio.h>
> main(){
> int *i,*j;
> i=(int*)60;
> j=(int*)71;
>
> printf
> <http://www.opengroup.org/onlinepubs/009695399/functions/printf.html>("%p %p
> %d",i,j,j-i);}
>
>
> op: 0x3c 0x47 2
>
> 0x47 - 0x3c = 0x11 and hence j-1 = 2 (11/4 = 2, size of int = 4 bytes)
>
>
> On Sat, Oct 13, 2012 at 3:36 PM, bharat b <bagana.bharatku...@gmail.com>wrote:
>
>> #include<stdio.h>
>> main()
>> {
>> int *i,*j;
>> i=(int*)60;
>> j=(int*)71;
>> printf("%d",j-i);
>> }
>>
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