left( source, count ) in C

[Dave Kirsch]

> nacer writes:
> 
> Msg-ID: <510007 at hpmcaa.mcm.hp.com>
> Posted: 16 Mar 90 17:55:54 GMT
> 
> Org.  : HP McMinville Division
> Person: Abdenacer Moussaoui
> 
> How do you write a function that returns the left part of a string in C?
> 
> The interface is  left(  source, count ) here are some test cases:
> 
>         left( "123456789", 3 ) returns "123"
>         left( "123456789", 20 ) returns "123456789"
> 
> if    stimef( current_time ) returns "13:45:23:48"
> then  left( stimef( current_time ), 8 ) returns "13:45:23"
> 
> As you can see in the last case, I would't want left(,) to modify the source
> (ie. implementing left something like source[count] = '\0' )
> Thanks for any info.

Try this [note that it is this function uses a static buffer that is
overwritten on each call, so save the results before you call it again]:

char *left(const char *source, int count)
{
static char st[81]; /* Increase this as required */

   strncpy(st, source, count);
   st[count] = 0; /* Shorten it. */
   return st;  /* And return it. */
}

--
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Dave Kirsch           UUCP: {uunet,ubc-cs}!van-bc!rsoft!mindlink!a563
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