parsing string into an array

[John Smith]

I would like to parse a string into an array. I found on the net
the following codes which parse a string and print it. The result
is exactly what I want:

char * pch;
pch = strtok (buffer," ");
while (pch != NULL)
{
printf ("%s\n",pch) ;
pch = strtok (NULL, " ,.");
}

To store the output to a string array, I created a two-dimensional
array and copied the output to the array:

char output[5][55];
char * pch;
int i=0;
pch = strtok (str," ");
while (pch != NULL)
{
strcpy(output[i],pch);
printf ("%s\n",pch) ;
printf("output[%d] is %s.\n";i,output[i++]);
pch = strtok (NULL, " ,.");
}

This doesn't work. What's in the output array is nothing like the
output (on screen). How should it be done?

[Walter Roberson]

In article <lb************ *********@twist er.southeast.rr .com>,
John Smith <js****@company .com> wrote:

printf("output[%d] is %s.\n";i,output[i++]);

If I recall correctly, this uses undefined behaviour.
The post increment may be done at any time between the
sequence points, so it is not defined whether the first
reference to i has its value taken before or after the
increment.

It also is a syntax error to have the semicolon in the middle
of the statement. That line should not compile at all.
--
'ignorandus (Latin): "deserving not to be known"'
-- Journal of Self-Referentialism

[John Smith]

Walter Roberson wrote:

In article <lb************ *********@twist er.southeast.rr .com>,
John Smith <js****@company .com> wrote:

printf("output[%d] is %s.\n";i,output[i++]);

If I recall correctly, this uses undefined behaviour.
The post increment may be done at any time between the
sequence points, so it is not defined whether the first
reference to i has its value taken before or after the
increment.

It also is a syntax error to have the semicolon in the middle
of the statement. That line should not compile at all.

Thanks for the reply. That semicolon was a typo.

Actually, I had also tried the following (a separate i++) and it
did not work either.

char output[5][55];
char * pch;
int i=0;
pch = strtok (str," ");
while (pch != NULL)
{
strcpy(output[i],pch);
printf ("%s\n",pch) ;
printf("output[%d] is %s.\n",i,output[i]);
i++;
pch = strtok (NULL, " ,.");
}

[Al Bowers]

John Smith wrote:

Actually, I had also tried the following (a separate i++) and it
did not work either.

char output[5][55];
char * pch;
int i=0;
pch = strtok (str," ");
while (pch != NULL)
{
strcpy(output[i],pch);
printf ("%s\n",pch) ;
printf("output[%d] is %s.\n",i,output[i]);
i++;
pch = strtok (NULL, " ,.");
}

The code provided here seems to be ok. What do you
mean by "did not work either".
--
Al Bowers
Tampa, Fl USA
mailto: xa******@myrapi dsys.com (remove the x to send email)
http://www.geocities.com/abowers822/

[Joe Estock]

John Smith wrote:

I would like to parse a string into an array. I found on the net
the following codes which parse a string and print it. The result
is exactly what I want:

char * pch;
pch = strtok (buffer," ");
while (pch != NULL)
{
printf ("%s\n",pch) ;
pch = strtok (NULL, " ,.");
}

To store the output to a string array, I created a two-dimensional
array and copied the output to the array:

char output[5][55];
char * pch;
int i=0;
pch = strtok (str," ");
while (pch != NULL)
{
strcpy(output[i],pch);
printf ("%s\n",pch) ;
printf("output[%d] is %s.\n";i,output[i++]);
pch = strtok (NULL, " ,.");
}

This doesn't work. What's in the output array is nothing like the
output (on screen). How should it be done?

Smells like a buffer overflow to me.

[Guillaume]

John Smith wrote:

To store the output to a string array, I created a two-dimensional
array and copied the output to the array:

char output[5][55];
(snip)
This doesn't work. What's in the output array is nothing like the
output (on screen). How should it be done?

Are you aware that this will fail if:
* there are more than 5 "tokens"
* one or more of the tokens is (are) longer than 54 characters

That said, there are a couple more things to know:
* strtok() alters the string it works on. So you can't use
it to parse the same string twice unless you have made a copy
of this string before. This may be what happens to you if
your program tries both of your parsing attempts one after
the other on the same string.
* strtok() is one of those "evilish" C std library functions,
because it maintains an internal structure and thus cannot be
used in nested calls. So you must use it with a lot of care,
and it really is safe only in the simplest programs (unless
you really know what you're doing).
Not worth it in my opinion, since it's so easy to implement
"by hand" in a safe manner.

As a side note, I will add that outside of textbooks, multidimensiona l
arrays in C are not particularly useful. Their syntax is often
misleading (in my opinion) and they can serve only very limited
purpose. I tend to prefer arrays of arrays, which are easier to handle
and more flexible (in my opinion).

[Joe Wright]

John Smith wrote:

Walter Roberson wrote:

In article <lb************ *********@twist er.southeast.rr .com>,
John Smith <js****@company .com> wrote:

printf("output[%d] is %s.\n";i,output[i++]);

If I recall correctly, this uses undefined behaviour.
The post increment may be done at any time between the
sequence points, so it is not defined whether the first
reference to i has its value taken before or after the
increment.

It also is a syntax error to have the semicolon in the middle
of the statement. That line should not compile at all.

Thanks for the reply. That semicolon was a typo.

Actually, I had also tried the following (a separate i++) and it
did not work either.

char output[5][55];
char * pch;
int i=0;
pch = strtok (str," ");
while (pch != NULL)
{
strcpy(output[i],pch);
printf ("%s\n",pch) ;
printf("output[%d] is %s.\n",i,output[i]);
i++;
pch = strtok (NULL, " ,.");
}

Hi John,
(Do us a favor. See if you can turn off the annoying MIME stuff)
Like Al, I can't see what "did not work". Of course, you have not shown
us what the original 'char *str = "...";' looks like, or where you get
it. Enquiring Minds want to know.

--
Joe Wright mailto:jo****** **@comcast.net
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---

[CBFalconer]

Joe Wright wrote:

John Smith wrote:

Walter Roberson wrote:

John Smith <js****@company .com> wrote:

printf("output[%d] is %s.\n";i,output[i++]);

If I recall correctly, this uses undefined behaviour. .... snip ...
It also is a syntax error to have the semicolon in the middle
of the statement. That line should not compile at all.
Thanks for the reply. That semicolon was a typo.

Actually, I had also tried the following (a separate i++) and it
did not work either.

.... snip ...
(Do us a favor. See if you can turn off the annoying MIME stuff)

.... snip ...

I am among the first to bitch and moan about mime and html in
Usenet, but I see none in this thread. Has something stripped it
on the path from there to here?

--
"If you want to post a followup via groups.google.c om, don't use
the broken "Reply" link at the bottom of the article. Click on
"show options" at the top of the article, then click on the
"Reply" at the bottom of the article headers." - Keith Thompson

[John Smith]

Thanks to those who replied to my queston. I appreciate your help.
The code did work. I wasn't aware strtok would change the original
string. That caused some confusion.

(Is MIME turned off this time? Hope it is.)

[Joe Wright]

John Smith wrote:

Thanks to those who replied to my queston. I appreciate your help.
The code did work. I wasn't aware strtok would change the original
string. That caused some confusion.

(Is MIME turned off this time? Hope it is.)

I can't tell about MIME anymore. After I wrote my note to you I've set
Thunderbird to ignore it.

--
Joe Wright mailto:jo****** **@comcast.net
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---