I think it allocates on each process's stack, so it is not an issue,
as each process has got its own stack.
On 10/15/12, bharat b <bagana.bharatku...@gmail.com> wrote:
> 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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--
With love and regards,
Sairam Ravu
I M.Tech(CS)
Sri Sathya Sai Institute of Higher Learning
"To live life, you must think it, measure it, experiment with it, dance it,
paint it, draw it, and calculate it"
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