think in terms of pointers...
they are same :-
p[-1] = *(p - 1)
On Thu, Jan 26, 2012 at 11:15 PM, rahul sharma <rahul23111...@gmail.com>wrote:
> [-1] in end is same as -1 ??????
>
>
> On Thu, Jan 26, 2012 at 11:11 PM, atul anand <atul.87fri...@gmail.com>wrote:
>
>> btw your compiler has sizeof(int)=4;
>> thats why o/p = fg
>>
>> On Thu, Jan 26, 2012 at 11:09 PM, atul anand <atul.87fri...@gmail.com>wrote:
>>
>>> output depends on sizeof(int)....so it may be different if you run on
>>> different compilers.
>>>
>>> considering *sizeof(int) = 2;*
>>>
>>> argv[] is array of pointers.
>>> (p+=sizeof(int))[-1];
>>> p=p+2 // 2=sizeof(int);
>>>
>>> now p will be pointing at index *argv[2];
>>> then you are doing....
>>>
>>> p=p-1;
>>>
>>> i.e p will point to *argv[1]
>>>
>>> hence output will be
>>> o/p = cd
>>>
>>> On Thu, Jan 26, 2012 at 10:53 PM, rahul sharma
>>> <rahul23111...@gmail.com>wrote:
>>>
>>>> #include<stdio.h>
>>>> #include<conio.h>
>>>> void fun(char **);
>>>>
>>>> int main()
>>>> {
>>>> char *argv[]={"ab","cd","de","fg"};
>>>> fun(argv);
>>>> getch();
>>>> return 0;
>>>> }
>>>>
>>>> void fun(char **p)
>>>> {
>>>> char *t;
>>>> t=(p+=sizeof(int))[-1];
>>>> printf("%s\n",t);
>>>> }
>>>>
>>>> o/p: fg
>>>>
>>>> can nyone xplain
>>>>
>>>> the 2nd statement in fun?????
>>>>
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>>>
>>>
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