@Bharat: 0x notation indicates hexadecimal, so 0x11 = 1*16 + 1 = 17.
Dave
On Sunday, October 14, 2012 12:48:24 AM UTC-5, bharat wrote:
> @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 <verm...@gmail.com<javascript:>
> > 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.bh...@gmail.com<javascript:>
>> > wrote:
>>
>>> #include<stdio.h>
>>> main()
>>> {
>>> int *i,*j;
>>> i=(int*)60;
>>> j=(int*)71;
>>> printf("%d",j-i);
>>> }
>>>
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