How to read a part of a string

[poison.summer@gmail.com]

if I wanna copy a part of the string to another string.
It is my code, Is it OK? but I have a core dump.
include<stdio.h >
int main()
{
FILE *fp;
fp = fopen("test.txt ","r");
char s1[20],s2[5],s3[2];
fscanf(fp,"%s", s1);
printf("s1 is %s\n",s1);
int i;
for(i=0;i<6;i++ )
{
s2[i] = s1[1+i];
}
s2[6]='\0';
printf("s2 is %s\n",s2);
s3[0]=s1[10];
s3[1]=s1[11];
s3[2]='\0';
printf("s3 is %s\n",s3);
fclose(fp);
}

[Arafangion]

Re: How to read a part of a string

poison.summer@g mail.com wrote:[color=blue]
>
> if I wanna copy a part of the string to another string.
> It is my code, Is it OK? but I have a core dump.
> include<stdio.h >
> int main()
> {
> FILE *fp;
> fp = fopen("test.txt ","r");
> char s1[20],s2[5],s3[2];[/color]
fscanf(fp,"%19s ",s1); /* Correct me if I'm wrong, but this
should limit
the fscanf call to 19 chars */
s1[19] = 0; /* Ensure that the string is null-terminated */[color=blue]
> printf("s1 is %s\n",s1);
> int i;[/color]
for(i=0;i<4;i++ )[color=blue]
> {[/color]
s2[i] = s1[i];[color=blue]
> }[/color]
s2[4] = 0; /* s2[6] does not exist */[color=blue]
> printf("s2 is %s\n",s2);
> s3[0]=s1[10];[/color]
s3[1]= 0; /* 1 is the last char, and s3[2] does not exist */[color=blue]
> printf("s3 is %s\n",s3);
> fclose(fp);
> }
>[/color]

Your code is overly complicated and hard to understand.
Try something like:

void my_strdup(char[] source, char[] dest, int offset, int length)
{
/* Copy <length> chars from source, starting at <offset>
* It is assumed that dest has enough space for length chars */

for(int i=0; i<length && source[offset+i]; i++)
{
dest[i] = source[offset+i];
}
}

Additionally, see my edits to your code, you appear to have alot of
trouble with C-style arrays and null-terminated strings.
A core-dump is almost always a memory access violation (afaik - I don't
usually do C code)

[Jens.Toerring@physik.fu-berlin.de]

Re: How to read a part of a string

poison.summer@g mail.com wrote:

[color=blue]
> if I wanna copy a part of the string to another string.
> It is my code, Is it OK? but I have a core dump.
> include<stdio.h >[/color]

I hope that was

#include <stdio.h>
[color=blue]
> int main()[/color]

Make that

int main( void )
[color=blue]
> {
> FILE *fp;
> fp = fopen("test.txt ","r");
> char s1[20],s2[5],s3[2];[/color]

Do you have a C99 compiler? Otherwise you can't define new variables
after the first executable statement.
[color=blue]
> fscanf(fp,"%s", s1);[/color]

What happens if the string you just read had more than 19 characters?
You better make that

fscanf( fp, "%19s", s1 );

And, of course, you should check the return value of fopen() before
you use 'fp'. Checking the one of fscanf() also won't hurt.
[color=blue]
> printf("s1 is %s\n",s1);
> int i;
> for(i=0;i<6;i++ )
> {
> s2[i] = s1[1+i];
> }
> s2[6]='\0';[/color]

You defined 's2' to be an array of 5 chars, but here you're already
using the 7th element that doesn't exist. That won't do. Moreover,
you don't know how many elements of 's1' are set at all - using un-
itialized values is also nothing you try.
[color=blue]
> printf("s2 is %s\n",s2);
> s3[0]=s1[10];
> s3[1]=s1[11];
> s3[2]='\0';[/color]

And here you do it again. 's3' has two elements but you try to use
three.
[color=blue]
> printf("s3 is %s\n",s3);
> fclose(fp);
> }[/color]

You promised that main() would return an int, but you forgot about
that.
Regards, Jens
--
\ Jens Thoms Toerring ___ Jens.Toerring@p hysik.fu-berlin.de
\______________ ____________ http://www.toerring.de

[Zoran Cutura]

Re: How to read a part of a string

Arafangion <Arafangion@inv alid.email.addr ess.com> wrote:[color=blue]
> poison.summer@g mail.com wrote:[color=green]
>>
>> if I wanna copy a part of the string to another string.
>> It is my code, Is it OK? but I have a core dump.
>> include<stdio.h >
>> int main()
>> {
>> FILE *fp;
>> fp = fopen("test.txt ","r");
>> char s1[20],s2[5],s3[2];[/color]
> fscanf(fp,"%19s ",s1); /* Correct me if I'm wrong, but this
> should limit
> the fscanf call to 19 chars */[/color]

that's fine. I would also suggest testing if there was a conversion at
all, that is check the return value of fscanf(). Further improvement to
the field width would be using some macros to calculate the maximum
field width automagically.
[color=blue]
> s1[19] = 0; /* Ensure that the string is null-terminated */[/color]

No need for this. the scanf-family ensures this allready.

--
Z (zoran.cutura@w eb.de)
"LISP is worth learning for the profound enlightenment experience
you will have when you finally get it; that experience will make you
a better programmer for the rest of your days." -- Eric S. Raymond

[Jaspreet]

Re: How to read a part of a string

poison.summer@g mail.com wrote:[color=blue]
> if I wanna copy a part of the string to another string.
> It is my code, Is it OK? but I have a core dump.
> include<stdio.h >
> int main()
> {
> FILE *fp;
> fp = fopen("test.txt ","r");
> char s1[20],s2[5],s3[2];
> fscanf(fp,"%s", s1);[/color]
Running gdb on the program showed that the above line caused
segmentation fault when I did not have "test.txt" in my directory. It
would help as said above to check for return value of fopen and fscanf.
[color=blue]
> printf("s1 is %s\n",s1);
> int i;
> for(i=0;i<6;i++ )
> {
> s2[i] = s1[1+i];
> }
> s2[6]='\0';
> printf("s2 is %s\n",s2);
> s3[0]=s1[10];
> s3[1]=s1[11];
> s3[2]='\0';
> printf("s3 is %s\n",s3);
> fclose(fp);
> }[/color]
Just ran this program and it does not do what you intended to do. Seems
pretty complex logic.

[Prawit Chaivong]

Re: How to read a part of a string



poison.summer@g mail.com wrote:[color=blue]
> if I wanna copy a part of the string to another string.
> It is my code, Is it OK? but I have a core dump.
> include<stdio.h >
> int main()
> {
> FILE *fp;
> fp = fopen("test.txt ","r");
> char s1[20],s2[5],s3[2];
> fscanf(fp,"%s", s1);[/color]
Is s1 big enough?[color=blue]
> printf("s1 is %s\n",s1);
> int i;
> for(i=0;i<6;i++ )
> {
> s2[i] = s1[1+i];
> }
> s2[6]='\0';[/color]
Declaration s2[5];
there are
s2[0]
s2[1]
s2[2]
s2[3]
s2[4]
5 elements (only) there is no s2[5] or s2[6];
NOTE: If we do research, we would find that
s2[5] = s1[0] and s2[6] = s2[1];
(gcc on x86)
[color=blue]
> printf("s2 is %s\n",s2);
> s3[0]=s1[10];
> s3[1]=s1[11];
> s3[2]='\0';[/color]

There is no s3[2] neither.
[color=blue]
> printf("s3 is %s\n",s3);
> fclose(fp);
> }[/color]

Why don't you do something like this.

snprintf(s2, 2, "%s", &s1[yourindex]);

instead of using assign char by char

Regards,

[Lawrence Kirby]

Re: How to read a part of a string

On Tue, 07 Jun 2005 00:37:42 -0700, Prawit Chaivong wrote:

....
[color=blue]
> Declaration s2[5];
> there are
> s2[0]
> s2[1]
> s2[2]
> s2[3]
> s2[4]
> 5 elements (only) there is no s2[5] or s2[6];
> NOTE: If we do research, we would find that
> s2[5] = s1[0] and s2[6] = s2[1];[/color]

C makes no guarantees about the layout of different objects in memory,
it is pretty much always an error to make use of such information so is
probably a bad idea to even mention it.
[color=blue]
> (gcc on x86)[/color]

Even there it could depend on things like gcc version, optimisation level
etc.
[color=blue][color=green]
>> printf("s2 is %s\n",s2);
>> s3[0]=s1[10];
>> s3[1]=s1[11];
>> s3[2]='\0';[/color]
>
> There is no s3[2] neither.
>[color=green]
>> printf("s3 is %s\n",s3);
>> fclose(fp);
>> }[/color]
>
> Why don't you do something like this.
>
> snprintf(s2, 2, "%s", &s1[yourindex]);
>
> instead of using assign char by char[/color]

Or possibly

sprintf(s2, "%.2s", &s1[yourindex]);

snprintf and sprintf are fairly heavyweight for this, strncat may be
better:

s2[0] = '\0';
strncat(s2, &s1[yourindex], 2);

Unfortunately strncpy() doesn't do what you might expect and isn't as good
for this.

Lawrence

[Prawit Chaivong]

Re: How to read a part of a string



Lawrence Kirby wrote:[color=blue]
> On Tue, 07 Jun 2005 00:37:42 -0700, Prawit Chaivong wrote:
>
> ...
>[color=green]
> > Declaration s2[5];
> > there are
> > s2[0]
> > s2[1]
> > s2[2]
> > s2[3]
> > s2[4]
> > 5 elements (only) there is no s2[5] or s2[6];
> > NOTE: If we do research, we would find that
> > s2[5] = s1[0] and s2[6] = s2[1];[/color]
>
> C makes no guarantees about the layout of different objects in memory,
> it is pretty much always an error to make use of such information so is
> probably a bad idea to even mention it.
>[color=green]
> > (gcc on x86)[/color]
>
> Even there it could depend on things like gcc version, optimisation level
> etc.[/color]
That's a little addition information from my research.
Do not have to use it.

It's always a good idea to know how does your compiler work.
[color=blue]
>[color=green][color=darkred]
> >> printf("s2 is %s\n",s2);
> >> s3[0]=s1[10];
> >> s3[1]=s1[11];
> >> s3[2]='\0';[/color]
> >
> > There is no s3[2] neither.
> >[color=darkred]
> >> printf("s3 is %s\n",s3);
> >> fclose(fp);
> >> }[/color]
> >
> > Why don't you do something like this.
> >
> > snprintf(s2, 2, "%s", &s1[yourindex]);
> >
> > instead of using assign char by char[/color]
>
> Or possibly
>
> sprintf(s2, "%.2s", &s1[yourindex]);
>
> snprintf and sprintf are fairly heavyweight for this, strncat may be
> better:[/color]
Can you explain further why they are heavyweight?
Or recommend me some paper.[color=blue]
>
> s2[0] = '\0';
> strncat(s2, &s1[yourindex], 2);
>
> Unfortunately strncpy() doesn't do what you might expect and isn't as good
> for this.
>
> Lawrence[/color]

[Lawrence Kirby]

Re: How to read a part of a string

On Tue, 07 Jun 2005 03:00:54 -0700, Prawit Chaivong wrote:

....
[color=blue][color=green]
>> Even there it could depend on things like gcc version, optimisation level
>> etc.[/color]
> That's a little addition information from my research.
> Do not have to use it.[/color]

More specifically in the vast majority of situations you SHOULD NOT use
it.
[color=blue]
> It's always a good idea to know how does your compiler work.[/color]

Not always. Knowledge of how your compiler works can lead to false
assumptions and non-portable code. It isn't appropriate to present such
information in a context like this which is about basic correctness for C
code. Code that goes off the end of an array is simply wrong, irrespective
of what happens to be in memory before and after it. Irrelevant details
like that are simply a recipe for confusion.

Lawrence

[D Power]

Re: How to read a part of a string

In article <1118106584.547 406.63610@g47g2 000cwa.googlegr oups.com>,
poison.summer@g mail.com wrote:
[color=blue]
> if I wanna copy a part of the string to another string.
> It is my code, Is it OK? but I have a core dump.
> include<stdio.h >
> int main()
> {
> FILE *fp;
> fp = fopen("test.txt ","r");[/color]

<delurk> In addition to the other replies, it should be noted that
fopen returns a valid file pointer only if test.txt exists, and was
opened for reading, otherwise fp is set to NULL, so here you need to
test fp for NULL, e.g.
if (fp == NULL)
{
/* take appropriate action */
}
</delurk>
[color=blue]
> char s1[20],s2[5],s3[2];
> fscanf(fp,"%s", s1);
> printf("s1 is %s\n",s1);
> int i;
> for(i=0;i<6;i++ )
> {
> s2[i] = s1[1+i];
> }
> s2[6]='\0';
> printf("s2 is %s\n",s2);
> s3[0]=s1[10];
> s3[1]=s1[11];
> s3[2]='\0';
> printf("s3 is %s\n",s3);
> fclose(fp);
> }[/color]