how can kernel agrees with this ? if we directly access address 60 .. which
is not in our control ... any malicious thing can happen right ?
On Sun, Oct 14, 2012 at 5:39 PM, Dave <dave_and_da...@juno.com> wrote:
> @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> 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>wrote:
>>>
>>>> #include<stdio.h>
>>>> main()
>>>> {
>>>> int *i,*j;
>>>> i=(int*)60;
>>>> j=(int*)71;
>>>> printf("%d",j-i);
>>>> }
>>>>
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