Seg fault + string manipulation

[kumarvis@gmail.com]

int main ()
{
char *str = *Aamit ;
*str='R' ;
printf("%s \n " ,str);
}

it's giving segfault
why ?????

[Richard Heathfield]

Re: Seg fault + string manipulation

kumarvis@gmail. com said:

int main ()
{
char *str = *Aamit ;
*str='R' ;
printf("%s \n " ,str);
}
>
it's giving segfault
why ?????

What you have written doesn't compile, so it's hard to see how it could
segfault.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[pete]

Re: Seg fault + string manipulation

kumarvis@gmail. com wrote:

int main ()
{
char *str = *Aamit ;
*str='R' ;
printf("%s \n " ,str);
}
>
it's giving segfault
why ?????

int main(void)
{
char *str = "Aamit" ;

*str = 'R' ;
puts(str);
return 0;
}

--
pete

[Joachim Schmitz]

Re: Seg fault + string manipulation

kumarvis@gmail. com wrote:

int main ()
{
char *str = *Aamit ;
*str='R' ;
printf("%s \n " ,str);
}
>
it's giving segfault
why ?????

If you meant to write
char *str = "Aamit";

Then the
*str ="R";
possibly writes into read-only memory, but surely invokes undefined
behavoir.
segvault is one possible outcome of that.

Another fault (and causing undefined behavoir) is the lack of #include
<stdio.hresp. the resulting lack of a prototype for the varadic function
printf().

Bye, Jojo

[pete]

Re: Seg fault + string manipulation

pete wrote:

kumarvis@gmail. com wrote:

>int main ()
>{
> char *str = *Aamit ;
> *str='R' ;
> printf("%s \n " ,str);
>}
>>
>it's giving segfault
>why ?????

>
int main(void)
{
char *str = "Aamit" ;

Should be:

char str[] = "Aamit" ;

instead.

*str = 'R' ;
puts(str);
return 0;

--
pete

[Keith Thompson]

Re: Seg fault + string manipulation

pete <pfiland@mindsp ring.comwrites:

kumarvis@gmail. com wrote:

>int main ()
>{
> char *str = *Aamit ;
> *str='R' ;
> printf("%s \n " ,str);
>}
>it's giving segfault
>why ?????

>
int main(void)
{
char *str = "Aamit" ;
>
*str = 'R' ;
puts(str);
return 0;
}

pete, what exactly was the point of your followup?

You corrected the OP's compilation error and presumably reproduced
something close to his actual code (which the OP should have posted
himself). Meanwhile, you haven't answered his actual question, or
said anything at all about it. You also forgot the required
"#include <stdio.h>" and replaced the OP's printf call with a puts
call, which subtly changed the behavior of the program.

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Keith Thompson]

Re: Seg fault + string manipulation

kumarvis@gmail. com writes:

int main ()
{
char *str = *Aamit ;
*str='R' ;
printf("%s \n " ,str);
}
>
it's giving segfault

No, it isn't. I'm sure the program you're actually running gives you
a seg fault, but the code you posted won't even compile.

Never re-type code when you post it here. Always copy-and-paste the
*exact* code that you're actually compiling.

You also have a few problems that your compiler might not warn you about:

If you're going to use printf, you *must* have "#include <stdio.h>".

"int main()" is ok, but "int main(void)" is better.

You should add a "return 0;" at the end of main. It's not required in
all circumstances, but it can't hurt.

The space after the "%s" in your printf format is fairly harmless, but
almost certainly useless. It causes an extra blank to be printed
after your string.

The way you format your code is odd and makes it more difficult to
read. The compiler doesn't care about whitespace between tokens, but
for the sake of your readers it's generally best to have a space after
each comma and surrounding binary operators like "=", and *not* to
have a blank preceding a semicolon or comma.

And see question 1.32 of the comp.lang.c FAQ, <http://www.c-faq.com/>.

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[bert]

Re: Seg fault + string manipulation

On 29 May, 13:54, kumar...@gmail. com wrote:

int main ()
{
   char *str = *Aamit ;
   *str='R' ;
    printf("%s \n " ,str);
>
}
>
it's giving segfault
why ?????

Apart from the problems other posters have
already pointed out, there is a major one
which they haven't :-

printf() expects its (str) argument to
point to a zero-terminated string. Your
code modifies the first character which
str is pointing at, but does nothing about
the following ones. If they are all non-zero
junk, and you have fixed the other problems,
it is quite possible for printf() to run off
the end of available memory looking for the
zero terminator.
--

[Chad]

Re: Seg fault + string manipulation

On May 29, 9:18 am, bert <bert.hutchi... @btinternet.com wrote:

On 29 May, 13:54, kumar...@gmail. com wrote:
>

int main ()
{
   char *str = *Aamit ;
   *str='R' ;
    printf("%s \n " ,str);

>

}

>

it's giving segfault
why ?????

>
Apart from the problems other posters have
already pointed out, there is a major one
which they haven't :-
>
printf() expects its (str) argument to
point to a zero-terminated string.  Your
code modifies the first character which
str is pointing at, but does nothing about
the following ones.  If they are all non-zero
junk, and you have fixed the other problems,
it is quite possible for printf() to run off
the end of available memory looking for the
zero terminator.
--

Maybe I'm not thinking this through enough, but I REALLY don't see the
difference between
m-net% more mod.c
#include <stdio.h>

int main (void)
{
char str[] = "Aamit";
*str='R';
printf("%s \n " ,str);
return 0;
}

m-net% gcc -g -Wall mod.c -o mod
m-net% ./mod
Ramit

%
m-net%


Versus something like
m-net% more mod.c
#include <stdio.h>

int main (void)
{
char str[] = "Aamit";
*str='R';
puts(str);
return 0;
}

m-net% gcc -g -Wall mod.c -o mod
m-net% ./mod
Ramit
m-net%

[pete]

Re: Seg fault + string manipulation

Chad wrote:

Maybe I'm not thinking this through enough, but I REALLY don't see the
difference between
m-net% more mod.c
#include <stdio.h>
>
int main (void)
{
char str[] = "Aamit";
*str='R';
printf("%s \n " ,str);
return 0;
}
>
m-net% gcc -g -Wall mod.c -o mod
m-net% ./mod
Ramit
>
%
m-net%
>
>
Versus something like
m-net% more mod.c
#include <stdio.h>
>
int main (void)
{
char str[] = "Aamit";
*str='R';
puts(str);
return 0;
}
>
m-net% gcc -g -Wall mod.c -o mod
m-net% ./mod
Ramit
m-net%

The printf version, outputs a space character (' ') before the newline.

--
pete

[Keith Thompson]

Re: Seg fault + string manipulation

bert <bert.hutchings @btinternet.com writes:

On 29 May, 13:54, kumar...@gmail. com wrote:

>int main ()
>{
>   char *str = *Aamit ;
>   *str='R' ;
>    printf("%s \n " ,str);
>>
>}
>>
>it's giving segfault
>why ?????

>
Apart from the problems other posters have
already pointed out, there is a major one
which they haven't :-
>
printf() expects its (str) argument to
point to a zero-terminated string. Your
code modifies the first character which
str is pointing at, but does nothing about
the following ones. If they are all non-zero
junk, and you have fixed the other problems,
it is quite possible for printf() to run off
the end of available memory looking for the
zero terminator.

The code as posted will not compile, so it clearly isn't the same code
that the OP actually compiled and ran. I think you're making a
different assumption than the rest of us about just how the posted
code differs from the real code.

Are you assuming that there's a declared object of type char** named
``Aamit''?

I think the rest of us assumed that ``*Aamit'' should have been the
string literal "Aamit". Given that assumption, *if* we ignore the
fact that modifying the contents of a string literal invokes undefined
behavior, then the array is already properly terminated with a '\0'.

Unfortunately, the original poster has not seen fit to clear this up
by posting his actual code. Until and unless he does so, I suggest
there's no point in pursuing this further.

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[CBFalconer]

Re: Seg fault + string manipulation

Chad wrote:

>

.... snip ...

>
Maybe I'm not thinking this through enough, but I REALLY don't
see the difference between
>

.... snip ...

>
Versus something like
>
#include <stdio.h>
int main (void) {
char str[] = "Aamit";
*str='R';
puts(str);
return 0;
}

The difference is between:

char *str = "Aamit"; /* 1 */
and
char str[] = "Aamit"; /* 2 */

in /* 1 */ str is a char pointer to a non-modifiable string. In /*
2 */ str is a fully modifiable 6 char array, holding "Aamit\0".

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.


** Posted from http://www.teranews.com **

[Joachim Schmitz]

Re: Seg fault + string manipulation

CBFalconer wrote:

Chad wrote:

>>

... snip ...

>>
>Maybe I'm not thinking this through enough, but I REALLY don't
>see the difference between
>>

... snip ...

>>
>Versus something like
>>
>#include <stdio.h>
>int main (void) {
> char str[] = "Aamit";
> *str='R';
> puts(str);
> return 0;
>}

>
The difference is between:
>
char *str = "Aamit"; /* 1 */
and
char str[] = "Aamit"; /* 2 */
>
in /* 1 */ str is a char pointer to a non-modifiable string. In /*
2 */ str is a fully modifiable 6 char array, holding "Aamit\0".

Read again. Chad was using char str[] in both cases, his only difference was
printf vs. puts

Bye, Jojo

[Joachim Schmitz]

Re: Seg fault + string manipulation

Chad wrote:

On May 29, 9:18 am, bert <bert.hutchi... @btinternet.com wrote:

>On 29 May, 13:54, kumar...@gmail. com wrote:
>>

>>int main ()
>>{
>>char *str = *Aamit ;
>>*str='R' ;
>>printf("%s \n " ,str);

>>

>>}

>>

>>it's giving segfault
>>why ?????

>>
>Apart from the problems other posters have
>already pointed out, there is a major one
>which they haven't :-
>>
>printf() expects its (str) argument to
>point to a zero-terminated string. Your
>code modifies the first character which
>str is pointing at, but does nothing about
>the following ones. If they are all non-zero
>junk, and you have fixed the other problems,
>it is quite possible for printf() to run off
>the end of available memory looking for the
>zero terminator.
>--

>
>
Maybe I'm not thinking this through enough, but I REALLY don't see the
difference between

Then look again.

m-net% more mod.c
#include <stdio.h>
>
int main (void)
{
char str[] = "Aamit";
*str='R';
printf("%s \n " ,str);
return 0;
}
>
m-net% gcc -g -Wall mod.c -o mod
m-net% ./mod
Ramit
>

See this linefeed?

%
m-net%
>
>
Versus something like
m-net% more mod.c
#include <stdio.h>
>
int main (void)
{
char str[] = "Aamit";
*str='R';
puts(str);
return 0;
}
>
m-net% gcc -g -Wall mod.c -o mod
m-net% ./mod
Ramit

Here it is not.

m-net%

On top you got a space before and anotzher one after the newline, but these
are 'invisible'.

Bye, Jojo

[CBFalconer]

Re: Seg fault + string manipulation

Joachim Schmitz wrote:

CBFalconer wrote:

>Chad wrote:

>>>

>... snip ...

>>>
>>Maybe I'm not thinking this through enough, but I REALLY don't
>>see the difference between
>>>

>... snip ...

>>>
>>Versus something like
>>>
>>#include <stdio.h>
>>int main (void) {
>> char str[] = "Aamit";
>> *str='R';
>> puts(str);
>> return 0;
>>}

>>
>The difference is between:
>>
> char *str = "Aamit"; /* 1 */
>and
> char str[] = "Aamit"; /* 2 */
>>
>in /* 1 */ str is a char pointer to a non-modifiable string. In /*
>2 */ str is a fully modifiable 6 char array, holding "Aamit\0".

>
Read again. Chad was using char str[] in both cases, his only
difference was printf vs. puts

That was in his last manual copy of the original problem. The

char* = "constant";

was the actual problem.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.


** Posted from http://www.teranews.com **