Changing string

[yaru22]

Hi, I'm relatively new to C programming and was quite confused by this
error.

What I tried to do is that if I have a string with double quotes
around it (i.e. char *str = "\"hello\"" ;), I have to remove the
quotes.

So I programmed this as follow:

#include <stdio.h>
#include <string.h>

int change(char *str) {
int len = strlen(str);
if (str[0]=='\"' && str[len-1]=='\"') {
char str_tmp[len+1];
strncpy(str_tmp ,str+1,len-2);
str_tmp[len-2]='\0';
*str = *str_tmp;
}
return 0;
}

int main() {
char *str = "\"hello\"" ;
change(str);
printf("%s\n",s tr);
return 0;
}

However, if I run this code, I get segmentation fault error.

I guess "*str = *str_tmp;" part is wrong, but I don't know how to fix
it.

Doesn't this statement mean, substitute the value of str_tmp into the
memory location that str is pointing to?

I wonder what the reason for the error and how to do it.

Also, the return type of the function "change" should be int.

Thanks in advance.

Brian

[Eric Sosman]

Re: Changing string

yaru22 wrote On 03/06/07 10:27,:

Hi, I'm relatively new to C programming and was quite confused by this
error.
>
What I tried to do is that if I have a string with double quotes
around it (i.e. char *str = "\"hello\"" ;), I have to remove the
quotes.
>
So I programmed this as follow:
>
#include <stdio.h>
#include <string.h>
>
int change(char *str) {
int len = strlen(str);
if (str[0]=='\"' && str[len-1]=='\"') {
char str_tmp[len+1];
strncpy(str_tmp ,str+1,len-2);
str_tmp[len-2]='\0';
*str = *str_tmp;
}
return 0;
}
>
int main() {
char *str = "\"hello\"" ;
change(str);
printf("%s\n",s tr);
return 0;
}
>
However, if I run this code, I get segmentation fault error.

This is Question 1.32 in the comp.lang.c Frequently
Asked Questions (FAQ) list at <http://www.c-faq.com/>.


--
Eric.Sosman@sun .com

[Arun]

Re: Changing string

On Mar 6, 10:31 am, Eric Sosman <Eric.Sos...@su n.comwrote:

yaru22 wrote On 03/06/07 10:27,:
>
>
>
>
>

Hi, I'm relatively new to C programming and was quite confused by this
error.

>

What I tried to do is that if I have a string with double quotes
around it (i.e. char *str = "\"hello\"" ;), I have to remove the
quotes.

>

So I programmed this as follow:

>

#include <stdio.h>
#include <string.h>

>

int change(char *str) {
int len = strlen(str);
if (str[0]=='\"' && str[len-1]=='\"') {
char str_tmp[len+1];
strncpy(str_tmp ,str+1,len-2);
str_tmp[len-2]='\0';
*str = *str_tmp;

=The above statement will assign the first char in str_tmp to first
char of str.
=Use strcpy(str,str_ tmp) instead. This will fix the logical error.

}
return 0;
}

>

int main() {
char *str = "\"hello\"" ;

=Specify to your compiler that this is a writable string by
=char str[] = "\"hello\"" ;

change(str);
printf("%s\n",s tr);
return 0;
}

>

However, if I run this code, I get segmentation fault error.

>
This is Question 1.32 in the comp.lang.c Frequently
Asked Questions (FAQ) list at <http://www.c-faq.com/>.
>
--
Eric.Sos...@sun .com- Hide quoted text -
>
- Show quoted text -

-Arun Joseph

[ComerHides@gmail.com]

Re: Changing string

Replace the code

*str = *str_tmp; by

str = str_tmp;

and check the output

[Flash Gordon]

Re: Changing string

ComerHides@gmai l.com wrote, On 06/03/07 16:28:

Replace the code
>
*str = *str_tmp; by
>
str = str_tmp;
>
and check the output

Firstly, please quote sufficient of what you are replying to so that
people know what you are refering to. There is no guarantee that people
have, or ever will, see the message you are replying to.

Secondly, try your suggestion and you will see that it does not work.
Then search through the comp.lang.c FAQ for the question and answer that
tells you why it does not work.
--
Flash Gordon