is there an easier way?
Tim Daneliuk
tundra at tundraware.com
Mon Feb 18 23:41:46 UTC 2008
Tim Daneliuk wrote:
> Tim Daneliuk wrote:
>> Gary Kline wrote:
>>> To my fellow C nerds,
>>>
>>> It's been a great manny years since I wrote this appended
>>> snippet. Now I can't remember why (of if ) I need all the
>>> strcpy() calls. Is there a simpler, more logical way of
>>> printing a bunch of string by snipping off the left-most?
>>>
>>> In short,, can anyone 'splain why strtok needs all this?
>>>
>>> tia,
>>>
>>> gary
>>>
>>>
>>
>> I don't think you need the copies. This works just as well:
>>
>> #include <stdio.h>
>> #include <string.h>
>>
>> main()
>> {
>> char *bp, *tok;
>> char *delim=" ", s1[256]="abc def ghi jkl mno.";
>>
>> bp = s1; /* Now both point to the literal string to be tokenized */
>>
>> while ((tok = strtok(bp, delim)) != NULL)
>> {
>> bp = NULL;
>> printf("tok = [%s]\n", tok);
>> }
>> }
>>
>
>
> Ooops ... wasn't paying attention. While the printed output is the
> same, doing
> it this way is destructive to the original s1 string - which may matter
> (or not)...
So, to protect the original string, you do have to copy the original
string to a new buffer:
#include <stdio.h>
#include <string.h>
main()
{
char *bp, *tok, buf[256];
const char *delim=" ", s1[256]="abc def ghi jkl mno.";
bp = strcpy(buf, s1);
while ((tok = strtok(bp, delim)) != NULL)
{
bp = NULL;
printf("tok = [%s]\n", tok);
}
}
So, at least we eliminated the first strcpy in your code:
char *bp, buf[512], *tok, tstr[512];
static char *delim=" ", s1[256]="abc def ghi jkl mno.";
bp = strcpy(buf, tstr);
This accomplishes nothing and may not even be safe - strcpy() copies
until it hits \0. I do not recall if the declaration of tstr[512]
intializes it to \0. If not, then the strcpy() is going to keep
running until it finds one, overwriting the end of buf[] and probably
causing core dumps or other evil...
--
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Tim Daneliuk tundra at tundraware.com
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