i can see the output coming as f .............

http://ideone.com/pB9ga

but when i ran the same code on ubuntu machine with gcc compiler then it is
also giving segmentation fault i don' t know why.....this is a illustration
taken from K  & R page  no. 132.

i don't think there is any need to allocate memory for the string separately
we have declared the variable of a structure which contains a character
pointer....which points to the given string....so where is the need of
allocating memory ??


--


Amol Sharma
Third Year Student
Computer Science and Engineering
MNNIT Allahabad




On Sun, Aug 7, 2011 at 8:36 PM, sukran dhawan <[email protected]>wrote:

> oh ya i didnt it sorry :)
>
>
> On Sun, Aug 7, 2011 at 8:29 PM, saurabh singh <[email protected]> wrote:
>
>> And also u have not allocated any space for str...so even if u do wat
>> sukran has suggested u wont get an f...
>>
>>
>> On Sun, Aug 7, 2011 at 8:27 PM, sukran dhawan <[email protected]>wrote:
>>
>>> replace ur code with printf("%c",++(p->str));
>>> *ptr is a deferenced value.but mere p is a lvalue
>>>
>>>
>>> On Sun, Aug 7, 2011 at 8:23 PM, RAHUL AGARWAL <[email protected]>wrote:
>>>
>>>> #include<stdio.h>
>>>> struct abc{
>>>>        int len;
>>>>        char *str;
>>>>        }*p;
>>>>        main()
>>>>        {
>>>>        struct abc k;
>>>>        k.len=20;
>>>>        k.str="echo is your command";
>>>>        p=&k;
>>>>        printf("%c",++(*p->str));
>>>>        }
>>>> I expect the output of this code to be "f" but it's displays
>>>> "SEGMENTATION FAULT"..
>>>> PLZ EXPLAIN.WHY?
>>>>
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>>>
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>>
>>
>> --
>> Saurabh Singh
>> B.Tech (Computer Science)
>> MNNIT ALLAHABAD
>>
>>
>>
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