[SOLVED] how to truncate a string

atlantis43 · 05-26-2015, 11:14 AM

Help needed to understand how to truncate a string using C.
Specific goal is, eg, to be able to truncate a seven char string down to six char, for which I thought strncmp() might be suitable.
Tried using the following code to no avail:

Code:

#include <stdio.h>
#include <string.h>
#define LIM 5
char *strncpy(char *dest, const char *src, size_t n)
{
    size_t i;

    for (i = 0; i < n && src[i] != '\0'; i++)
        dest[i] = src[i];
    for ( ; i < n; i++)
        dest[i] = '\0';
    return dest;
}

int main(void)
{
    char *path[LIM];
    path[1] = "go to sleep";
    printf("s\n",strncpy(path[2],path[1],1));
}

---as it gives garbage output!
Is this an incorrect application for strncmp()? Any corrections or suggestions as to a proper way to accomplish my goal would be appreciated.(should I simply use a for-loop with strlen - 1?).

jailbait · 05-26-2015, 11:39 AM

Put a null character (string terminator) after the sixth character in the string.

----------------------
Steve Stites

veerain · 05-26-2015, 11:40 AM

A corrected code:

Code:

#include <stdio.h>
#include <string.h>
#define LIM 5
char *strncpy(char *dest, const char *src, size_t n)
{
    size_t i;

    for (i = 0; i < n && src[i] != '\0'; i++)
        dest[i] = src[i];
    if ( (src[i] == '\0') || (i == n) ) dest[i] = '\0';
    return dest;
}

int main(void)
{
    char *path[LIM];
    path[1] = "go to sleep";
    printf("%s\n",strncpy(path[2],path[1],1));
}

Run and post status.

ntubski · 05-26-2015, 11:43 AM

Code:

#define LIM 5
...
int main(void)
{
    char *path[LIM];
    path[1] = "go to sleep";
    printf("%s\n",strncpy(path[2],path[1],1));
}

You didn't make path[2] (or any of path[0], path[3], path[4]) point anywhere, so they are pointing to some random memory location. Writing to some random location isn't going to work very well.

Also, strncpy() is an existing function in libc, so it's a bit confusing to redefine it (you should call yours my_strncpy or something).

Finally, why are you passing 1 as the length to strncpy()?

atlantis43 · 05-26-2015, 11:47 AM

Quote:

Originally Posted by veerain

A corrected code:

Code:

#include <stdio.h>
#include <string.h>
#define LIM 5
char *strncpy(char *dest, const char *src, size_t n)
{
    size_t i;

    for (i = 0; i < n && src[i] != '\0'; i++)
        dest[i] = src[i];
    if ( (src[i] == '\0') || (i == n) ) dest[i] = '\0';
    return dest;
}

int main(void)
{
    char *path[LIM];
    path[1] = "go to sleep";
    printf("%s\n",strncpy(path[2],path[1],1));
}

Run and post status.

Very funny that the code I used was copied & pasted straight from the man pages, yet needed your correction in order to work properly!

NevemTeve · 05-26-2015, 11:50 AM

strncmp and strncpy are different, but you need neither.

Code:

    printf ("First five chars: '%.5s'\n", "long long long text");

atlantis43 · 05-26-2015, 12:00 PM

Quote:

Originally Posted by ntubski

You didn't make path[2] (or any of path[0], path[3], path[4]) point anywhere, so they are pointing to some random memory location. Writing to some random location isn't going to work very well.

Oh! So that's my screw-up!

Quote:

Also, strncpy() is an existing function in libc, so it's a bit confusing to redefine it (you should call yours my_strncpy or something).

Though I thought it should work in that fashion, when it didn't (now obviously because of the above) I went to man pages and found:

Code:

       A simple implementation of strncpy() might be:

           char *
           strncpy(char *dest, const char *src, size_t n)
           {
               size_t i;

               for (i = 0; i < n && src[i] != '\0'; i++)
                   dest[i] = src[i];
               for ( ; i < n; i++)
                   dest[i] = '\0';

               return dest;
           }

----and tried to use that. (From your comment, I now understand why veerain's "code correction" worked)

Quote:

Finally, why are you passing 1 as the length to strncpy()?

Quote:

Just to see it work

atlantis43 · 05-26-2015, 12:04 PM

Quote:

Originally Posted by jailbait

Put a null character (string terminator) after the sixth character in the string.

----------------------
Steve Stites

I guess that would work as well, if I first copied path[1] to path[2]. Path[1] can't be changed once established.

atlantis43 · 05-26-2015, 12:07 PM

Quote:

Originally Posted by NevemTeve

strncmp and strncpy are different, but you need neither.

Code:

    printf ("First five chars: '.5s'\n", "long long long text");

But they are not for viewing, but need transfer to another char*. How would using printf() in that context work?
I'm clearly losing my attentiveness to detail. I didn't notice the "strcmp" confusion until you just mentioned it.

ntubski · 05-26-2015, 12:26 PM

Quote:

Originally Posted by atlantis43

(From your comment, I now understand why veerain's "code correction" worked)

It only works by accident. I guess writing 2 bytes to the random location path[2] is pointing to happens not to break anything important.

The "correction" changes the behaviour of strncpy() such that

Quote:

Originally Posted by man 3 strncpy

If there is no null byte among the first n bytes of src, the string placed in dest will not be null-terminated.

no longer applies. So you have a strncpy() that acts a bit differently from the standard strncpy(). It would be better to give it a different name, or else use the standard strncpy() but handle the ending null byte outside the function.

NevemTeve · 05-26-2015, 02:29 PM

sprintf is the string-wrinting variant. eg

Code:

char p= buff;
p += sprintf (p, "something=d", intval1);
p += sprintf (p, "str='%.5s', strval1);
...
len= p-buff;

schneidz · 05-26-2015, 03:01 PM

Quote:

Originally Posted by atlantis43

Help needed to understand how to truncate a string using C.
Specific goal is, eg, to be able to truncate a seven char string down to six char, ...

Code:

seven_char_string[6] = '\0';

atlantis43 · 05-26-2015, 04:30 PM

Quote:

Originally Posted by schneidz

Code:

seven_char_string[6] = '\0';

forgot to initially emphasize that I need a 'new' string with the truncated chars in it, as I'd attempted with my submitted code.

jpollard · 05-26-2015, 07:42 PM

Or something like:

Code:

char *new_nstr(char *p, size_t n)
{
    char *q;

    q = calloc(n+1,1);         /* make sure default array is zero filled and
                                  space for terminator */
    if (q) {                   /* protect against basic memory allocation failure */
        if (p) strncpy(q,p,n); /* protect against p being NULL */
        q[n]= '\0';            /* force null terminator if q is length n */
    }
    return (q);
}

Granted some may prefer a "if (NULL != q)" construct instead of "if (q)" for testing a null pointer.

atlantis43 · 05-27-2015, 06:59 AM

Quote:

Originally Posted by jpollard

Or something like:

Code:

char *new_nstr(char *p, size_t n)
{
    char *q;

    q = calloc(n+1,1);         /* make sure default array is zero filled and
                                  space for terminator */
    if (q) {                   /* protect against basic memory allocation failure */
        if (p) strncpy(q,p,n); /* protect against p being NULL */
        q[n]= '\0';            /* force null terminator if q is length n */
    }
    return (q);
}

Granted some may prefer a "if (NULL != q)" construct instead of "if (q)" for testing a null pointer.

Style appreciated. Actually, each new array will have additional chars added to array elements, so I don't think that null terminators are needed immediately following strncpy(), but would be added at a later time.