No occurrence founded

[nick048]

Hi,

This is a function that count the number of occurrences of word in a
string:
int wordCounter(char *a, char *b)
{
int cnt = 0;
char *sptr = a;
/* ONLY FOR TEST
int c = strlen(a);
int d = strlen(b);
printf("a = %d e b = %d\n",c,d); END TEST*/
while ( sptr != '\0' && ( sptr = strstr( sptr, b ) ) != NULL )
{
cnt++;
sptr += strlen( b );
}
return(cnt);
}

If I pass, for example:
for a "test1 test2 test3 test4"
for b "test"
the function return cnt = 0.
I have verified that the lenght of a is the number of char+1 and for b
the number of char + 1.
How can modify my function, in order to count correctly the word
number ?

Thank You
Gaetano

[Eric Sosman]

nick048 wrote On 04/11/07 13:32,:

Hi,

This is a function that count the number of occurrences of word in a
string:
int wordCounter(char *a, char *b)
{
int cnt = 0;
char *sptr = a;
/* ONLY FOR TEST
int c = strlen(a);
int d = strlen(b);
printf("a = %d e b = %d\n",c,d); END TEST*/
while ( sptr != '\0' && ( sptr = strstr( sptr, b ) ) != NULL )
{
cnt++;
sptr += strlen( b );
}
return(cnt);
}

If I pass, for example:
for a "test1 test2 test3 test4"
for b "test"
the function return cnt = 0.
I have verified that the lenght of a is the number of char+1 and for b
the number of char + 1.
How can modify my function, in order to count correctly the word
number ?

I am not sure what is wrong, but here are three observations:

- Did you remember to #include <string.h>?

- The first part of the `while' test is probably wrong.

- The function will loop indefinitely if `b' is "".

--
Er*********@sun.com

[Chris Dollin]

nick048 wrote:

Hi,

This is a function that count the number of occurrences of word in a
string:
int wordCounter(char *a, char *b)
{
int cnt = 0;
char *sptr = a;
/* ONLY FOR TEST
int c = strlen(a);
int d = strlen(b);
printf("a = %d e b = %d\n",c,d); END TEST*/
while ( sptr != '\0' && ( sptr = strstr( sptr, b ) ) != NULL )
{
cnt++;
sptr += strlen( b );
}
return(cnt);
}

If I pass, for example:
for a "test1 test2 test3 test4"
for b "test"
the function return cnt = 0.

I got 4.

I'll lay odds you got the parameters in the wrong order,
or otherwise mis-specified them. Since you didn't /show
code/ for the call, we can't see what you did wrong.

Why, for heaven's sake, call one parameter `a` and the
other `b`? Why not something more evocative, like
`lookFor` and `lookIn`? Why call the counter `cnt`
and not `count` or `result` -- I can see that your `o`
and `u` keys work? Why compare `sptr` (another dubious
name, with neither the virtue of brevity nor that of
comprehensibility) with '\0' (a character literal)
rather than `NULL` or `0` or my preferred choice,
nothing at all, just using `sptr`?

(I'd also compute `strlen(lookFor)` once and give it a
nice name.)

--
Prickly Hedgehog
The "good old days" used to be much better.

[Fred Kleinschmidt]

"nick048" <ni*************@moonsoft.itwrote in message
news:11**********************@q75g2000hsh.googlegr oups.com...

Hi,

This is a function that count the number of occurrences of word in a
string:
int wordCounter(char *a, char *b)
{
int cnt = 0;
char *sptr = a;
/* ONLY FOR TEST
int c = strlen(a);
int d = strlen(b);
printf("a = %d e b = %d\n",c,d); END TEST*/
while ( sptr != '\0' && ( sptr = strstr( sptr, b ) ) != NULL )

sptr is a pointer, you cannot compare it to a char.
You want (*sptr && ( sptr = strstr( sptr, b ) ) != NULL )

{
cnt++;
sptr += strlen( b );
}
return(cnt);
}

If I pass, for example:
for a "test1 test2 test3 test4"
for b "test"
the function return cnt = 0.
I have verified that the lenght of a is the number of char+1 and for b
the number of char + 1.
How can modify my function, in order to count correctly the word
number ?

--
Fred L. Kleinschmidt
Boeing Associate Technical Fellow
Aero Stability and Controls Computing

[B. Augestad]

Fred Kleinschmidt wrote:

"nick048" <ni*************@moonsoft.itwrote in message

[snip]

> while ( sptr != '\0' && ( sptr = strstr( sptr, b ) ) != NULL )

sptr is a pointer, you cannot compare it to a char.

'\0' is an integer, not a char. '\0' == 0 == NULL (in many/most cases).
So you're absolutely right, the OP wants to compare *sptr to '\0', but
comparing sptr to '\0' is legal on most platforms.
Bjørn
[snip]
--
Looking for an embeddable web server?
http://www.metasystems.no/products/h...der/index.html

[Keith Thompson]

"B. Augestad" <bo*@metasystems.nowrites:

Fred Kleinschmidt wrote:

>"nick048" <ni*************@moonsoft.itwrote in message

[snip]

>> while ( sptr != '\0' && ( sptr = strstr( sptr, b ) ) != NULL )

sptr is a pointer, you cannot compare it to a char.

'\0' is an integer, not a char. '\0' == 0 == NULL (in many/most
cases). So you're absolutely right, the OP wants to compare *sptr to
'\0', but comparing sptr to '\0' is legal on most platforms.

'\0' is an integer constant expression with value 0, which means it is
by definition a null pointer constant. Assuming sptr is a pointer,
comparing it to '\0' is legal (but horribly bad style) on *all*
platforms.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Flash Gordon]

B. Augestad wrote, On 11/04/07 20:56:

Fred Kleinschmidt wrote:

>"nick048" <ni*************@moonsoft.itwrote in message

[snip]

>> while ( sptr != '\0' && ( sptr = strstr( sptr, b ) ) != NULL )

sptr is a pointer, you cannot compare it to a char.

'\0' is an integer, not a char. '\0' == 0 == NULL (in many/most cases).

NULL could be ((void*)0) which is a rather different type. In fact, on
some common implementations it is.

So you're absolutely right, the OP wants to compare *sptr to '\0', but
comparing sptr to '\0' is legal on most platforms.

It is legal on all implementations since '\0' is an integer constant
with a value of 0 and all such constants are null pointer constants.
Note that other integer values are not compatible with pointers.
--
Flash Gordon

[Old Wolf]

On Apr 12, 7:56 am, "B. Augestad" <[e-mail]:

Fred Kleinschmidt wrote:

"nick048" <[e-mail] in message

while ( sptr != '\0' && ( sptr = strstr( sptr, b ) ) != NULL )

sptr is a pointer, you cannot compare it to a char.

'\0' is an integer, not a char. '\0' == 0 == NULL (in many/most cases).
So you're absolutely right, the OP wants to compare *sptr to '\0', but
comparing sptr to '\0' is legal on most platforms.

Are you sure he wants to compare *sptr to '\0' ? It would
be a redundant test, since the strstr() function would
return NULL in that case anyway.

As written, the code will prevent calling strstr with a null
pointer (in case the code prior to this loop is later
modified, for example) which would cause undefined behaviour.

[B. Augestad]

Keith Thompson wrote:

"B. Augestad" <bo*@metasystems.nowrites:

>>Fred Kleinschmidt wrote:

>>>"nick048" <ni*************@moonsoft.itwrote in message

[snip]

>>> while ( sptr != '\0' && ( sptr = strstr( sptr, b ) ) != NULL )

sptr is a pointer, you cannot compare it to a char.

'\0' is an integer, not a char. '\0' == 0 == NULL (in many/most
cases). So you're absolutely right, the OP wants to compare *sptr to
'\0', but comparing sptr to '\0' is legal on most platforms.

'\0' is an integer constant expression with value 0, which means it is
by definition a null pointer constant. Assuming sptr is a pointer,
comparing it to '\0' is legal (but horribly bad style) on *all*
platforms.

A dash of c.l.c paranoia got the better of me, as I wanted to stress
that NULL not necessarily is equal to 0. I guess I overdid it.

Thanks for clarifying.

Bjørn
--
Looking for an embeddable web server?
http://www.metasystems.no/products/h...der/index.html

[Chris Dollin]

Eric Sosman wrote:

nick048 wrote On 04/11/07 13:32,:

>Hi,

This is a function that count the number of occurrences of word in a
string:

(fx:snip)

I am not sure what is wrong, but here are three observations:

(fx:snip)

- The function will loop indefinitely if `b' is "".

That's plausibly correct behaviour: the empty string occurs infinitely
often in the test string, C can't represent infinite values, the
answer is the "undefined" of denotational semantics, ie bottom,
representing a non-terminating computation ...
--
Chris "not entirely serious" Dollin
The second Jena user conference! http://hpl.hp.com/conferences/juc2007/
"A facility for quotation covers the absence of original thought."/Gaudy Night/

Hewlett-Packard Limited registered office: Cain Road, Bracknell,
registered no: 690597 England Berks RG12 1HN

[Army1987]

"nick048" <ni*************@moonsoft.itha scritto nel messaggio
news:11**********************@q75g2000hsh.googlegr oups.com...

Hi,

This is a function that count the number of occurrences of word in a
string:
int wordCounter(char *a, char *b)
{
int cnt = 0;
char *sptr = a;
/* ONLY FOR TEST
int c = strlen(a);
int d = strlen(b);
printf("a = %d e b = %d\n",c,d); END TEST*/
while ( sptr != '\0' && ( sptr = strstr( sptr, b ) ) != NULL )
{
cnt++;
sptr += strlen( b );
}
return(cnt);
}

To me, this function seems valid, even if it has horrible style. I would
write:
int word_counter(char *str, char *word)
{
int cnt = 0;
size_t length;
if (word == NULL || *word == '\0')
return -1;
length = strlen(word);
if (str != NULL) {
while ((str = strstr(str, word)) != NULL) {
cnt++;
word += length;
}
}
return cnt;
}

which is perfectly equivalent (except for style, and for checking that word
isn't a null pointer or an empty string.) I think the error is in the
main...

[Army1987]

"Army1987" <pl********@for.itha scritto nel messaggio
news:ev**********@tdi.cu.mi.it...

"nick048" <ni*************@moonsoft.itha scritto nel messaggio
news:11**********************@q75g2000hsh.googlegr oups.com...

>Hi,

This is a function that count the number of occurrences of word in a
string:
int wordCounter(char *a, char *b)
{
int cnt = 0;
char *sptr = a;
/* ONLY FOR TEST
int c = strlen(a);
int d = strlen(b);
printf("a = %d e b = %d\n",c,d); END TEST*/
while ( sptr != '\0' && ( sptr = strstr( sptr, b ) ) != NULL )
{
cnt++;
sptr += strlen( b );
}
return(cnt);
}

To me, this function seems valid [snip] I think the error is in the
main...

I wrote:
#include <string.h>
#include <stdio.h>
int wordCounter(char *a, char *b)
{
/* copied and pasted from yours */
}
int main(void)
{
printf("%d\n", wordCounter("test1 test2 test3 test4", "test"));
return 0;
}
and it writes 4. Please show us the rest of the program.

[Barry]

"Army1987" <pl********@for.itwrote in message
news:ev**********@tdi.cu.mi.it...

"nick048" <ni*************@moonsoft.itha scritto nel messaggio
news:11**********************@q75g2000hsh.googlegr oups.com...

>Hi,

This is a function that count the number of occurrences of word in a
string:
int wordCounter(char *a, char *b)
{
int cnt = 0;
char *sptr = a;
/* ONLY FOR TEST
int c = strlen(a);
int d = strlen(b);
printf("a = %d e b = %d\n",c,d); END TEST*/
while ( sptr != '\0' && ( sptr = strstr( sptr, b ) ) != NULL )
{
cnt++;
sptr += strlen( b );
}
return(cnt);
}

To me, this function seems valid, even if it has horrible style. I would
write:
int word_counter(char *str, char *word)
{
int cnt = 0;
size_t length;
if (word == NULL || *word == '\0')
return -1;
length = strlen(word);
if (str != NULL) {
while ((str = strstr(str, word)) != NULL) {
cnt++;
word += length;
}
}
return cnt;
}

which is perfectly equivalent (except for style, and for checking that
word isn't a null pointer or an empty string.) I think the error is in the
main...

Your reply isn't clear. Did you just make a mistake in your code
or are you implying the OP had the parameters reversed?

Either way this only returns some of the matches and the OPs
question needs to be refined. I am sure someone pointed out,
if haystack is "abababababab" and needle is "abab" what is the
expected return value?

[Army1987]

"Barry" <ba****@nullhighstream.netha scritto nel messaggio
news:13*************@corp.supernews.com...

>
"Army1987" <pl********@for.itwrote in message
news:ev**********@tdi.cu.mi.it...

>"nick048" <ni*************@moonsoft.itha scritto nel messaggio
news:11**********************@q75g2000hsh.googleg roups.com...

>>Hi,

This is a function that count the number of occurrences of word in a
string:
int wordCounter(char *a, char *b)
{
int cnt = 0;
char *sptr = a;
/* ONLY FOR TEST
int c = strlen(a);
int d = strlen(b);
printf("a = %d e b = %d\n",c,d); END TEST*/
while ( sptr != '\0' && ( sptr = strstr( sptr, b ) ) != NULL )
{
cnt++;
sptr += strlen( b );
}
return(cnt);
}

To me, this function seems valid, even if it has horrible style. I would
write:
int word_counter(char *str, char *word)
{
int cnt = 0;
size_t length;
if (word == NULL || *word == '\0')
return -1;
length = strlen(word);
if (str != NULL) {
while ((str = strstr(str, word)) != NULL) {
cnt++;
word += length;
}
}
return cnt;
}

which is perfectly equivalent (except for style, and for checking that
word isn't a null pointer or an empty string.) I think the error is in
the main...

Your reply isn't clear. Did you just make a mistake in your code
or are you implying the OP had the parameters reversed?

Which mistake? I'm implying that the OP's error is not in the function
wordCounter, as it is correct (except when b is a null pointer or an empty
string, which is not the case in the OP's example).

Either way this only returns some of the matches and the OPs
question needs to be refined. I am sure someone pointed out,
if haystack is "abababababab" and needle is "abab" what is the
expected return value?

The OP's statement

>> sptr += strlen( b );

makes it clear that (s)he wants to return 3 in that case.

[Army1987]

"Army1987" <pl********@for.itha scritto nel messaggio
news:ev**********@tdi.cu.mi.it...

>
"Barry" <ba****@nullhighstream.netha scritto nel messaggio
news:13*************@corp.supernews.com...

>>
"Army1987" <pl********@for.itwrote in message
news:ev**********@tdi.cu.mi.it...

>>int word_counter(char *str, char *word)
{
int cnt = 0;
size_t length;
if (word == NULL || *word == '\0')
return -1;
length = strlen(word);
if (str != NULL) {
while ((str = strstr(str, word)) != NULL) {
cnt++;
word += length;

I meant str += length;

>> }
}
return cnt;
}

>Your reply isn't clear. Did you just make a mistake in your code
or are you implying the OP had the parameters reversed?

Which mistake?

Sorry. Replied too soon.