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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