26

I have been trying to tokenize a string using SPACE as delimiter but it doesn't work. Does any one have suggestion on why it doesn't work?

Edit: tokenizing using:

strtok(string, " ");

The code is like the following

pch = strtok (str," ");
while (pch != NULL)
{
  printf ("%s\n",pch);
  pch = strtok (NULL, " ");
}
6
  • Are you using strtok or something you grew yourself? cplusplus.com/reference/clibrary/cstring/strtok.html If you are using strtok are you trying to do it on a constant string? Commented Nov 5, 2008 at 19:48
  • your example will get the first token, look to either gbjbaanb's or my answers for proper usage. Commented Nov 5, 2008 at 20:02
  • OK. Now we're getting somewhere. What behavior do you expect that you are not getting? Commented Nov 5, 2008 at 20:08
  • 5
    BTW, kombo. Many people who work help desks or teach see the phrase "it doesn't work" as marking a user who hasn't read the furnished manual, or doesn't know what they actually want, or is deeply confused. The form you want is "I'm doing X, and I expected Y, but I got Z. What's wrong?" Commented Nov 5, 2008 at 20:17
  • 2
    @dmckee: good point. Canonical x-ref: catb.org/~esr/faqs/smart-questions.html Commented Nov 5, 2008 at 22:46

8 Answers 8

48

Do it like this:

char s[256];
strcpy(s, "one two three");
char* token = strtok(s, " ");
while (token) {
    printf("token: %s\n", token);
    token = strtok(NULL, " ");
}

Note: strtok modifies the string its tokenising, so it cannot be a const char*.

Sign up to request clarification or add additional context in comments.

Comments

39

Here's an example of strtok usage, keep in mind that strtok is destructive of its input string (and therefore can't ever be used on a string constant

char *p = strtok(str, " ");
while(p != NULL) {
    printf("%s\n", p);
    p = strtok(NULL, " ");
}

Basically the thing to note is that passing a NULL as the first parameter to strtok tells it to get the next token from the string it was previously tokenizing.

3 Comments

strtok has an internal state variable tracking the string being tokenized. When you pass NULL to it, strtok will continue to use this state variable. When you pass a non-null value, the state variable is reset. So in other words: passing NULL means "continue tokenizing the same string".
you're right, that's why many implementations offer strtok_r which atr the very least offers a way to use it in a thread safe way.
@Gnuey, p will point to characters in the string being tokenized. Additionally, strtok replaces the delimiter found with a '\0' character so that p will effectively be a valid NUL terminated string. So if you were to run it on char[] s = "hello world"; The first call would return a pointer to the h character and the buffer would then contain "hello\0world".
7

strtok can be very dangerous. It is not thread safe. Its intended use is to be called over and over in a loop, passing in the output from the previous call. The strtok function has an internal variable that stores the state of the strtok call. This state is not unique to each thread - it is global. If any other code uses strtok in another thread, you get problems. Not the kind of problems you want to track down either!

I'd recommend looking for a regex implementation, or using sscanf to pull apart the string.

Try this:

char strprint[256];
char text[256];
strcpy(text, "My string to test");
while ( sscanf( text, "%s %s", strprint, text) > 0 ) {
   printf("token: %s\n", strprint);
}

Note: The 'text' string is destroyed as it's separated. This may not be the preferred behaviour =)

4 Comments

In fact, if you look at modern strtok implementations, they tend to use thread-local storage (MSVC has certainly done this for years and years), so they are thread-safe. It's still an archaic function which I would avoid, though...
strtok_r is a thread-safe version of strtok pubs.opengroup.org/onlinepubs/009695399/functions/strtok.html
I agree with the first paragraph but the sentence after that is terrible. scanf is hard to use properly as shown in your example; you forget to pass a size (%255s).
strtok() is fine for non-threaded legacy systems, though. Archaic code for retro systems.
2

You can simplify the code by introducing an extra variable.

#include <string.h>
#include <stdio.h>

int main()
{
    char str[100], *s = str, *t = NULL;

    strcpy(str, "a space delimited string");
    while ((t = strtok(s, " ")) != NULL) {
        s = NULL;
        printf(":%s:\n", t);
    }
    return 0;
}

Comments

2

I've made some string functions in order to split values, by using less pointers as I could because this code is intended to run on PIC18F processors. Those processors does not handle really good with pointers when you have few free RAM available:

#include <stdio.h>
#include <string.h>

char POSTREQ[255] = "pwd=123456&apply=Apply&d1=88&d2=100&pwr=1&mpx=Internal&stmo=Stereo&proc=Processor&cmp=Compressor&ip1=192&ip2=168&ip3=10&ip4=131&gw1=192&gw2=168&gw3=10&gw4=192&pt=80&lic=&A=A";

int findchar(char *string, int Start, char C) {
    while((string[Start] != 0)) { Start++; if(string[Start] == C) return Start; }
    return -1;
}

int findcharn(char *string, int Times, char C) {
   int i = 0, pos = 0, fnd = 0;

    while(i < Times) {
       fnd = findchar(string, pos, C);
        if(fnd < 0) return -1;
        if(fnd > 0) pos = fnd;
       i++;
   }
   return fnd;
}

void mid(char *in, char *out, int start, int end) {
    int i = 0;
    int size = end - start;

    for(i = 0; i < size; i++){
        out[i] = in[start + i + 1];
    }
    out[size] = 0;
}

void getvalue(char *out, int index) {
    mid(POSTREQ, out, findcharn(POSTREQ, index, '='), (findcharn(POSTREQ, index, '&') - 1));
}

void main() {
   char n_pwd[7];
   char n_d1[7];

   getvalue(n_d1, 1);

   printf("Value: %s\n", n_d1);
} 

Comments

0

When reading the strtok documentation, I see you need to pass in a NULL pointer after the first "initializing" call. Maybe you didn't do that. Just a guess of course.

Comments

0

Here is another strtok() implementation, which has the ability to recognize consecutive delimiters (standard library's strtok() does not have this)

The function is a part of BSD licensed string library, called zString. You are more than welcome to contribute :)

https://github.com/fnoyanisi/zString

char *zstring_strtok(char *str, const char *delim) {
    static char *static_str=0;      /* var to store last address */
    int index=0, strlength=0;       /* integers for indexes */
    int found = 0;                  /* check if delim is found */

    /* delimiter cannot be NULL
    * if no more char left, return NULL as well
    */
    if (delim==0 || (str == 0 && static_str == 0))
        return 0;

    if (str == 0)
        str = static_str;

    /* get length of string */
    while(str[strlength])
        strlength++;

    /* find the first occurance of delim */
    for (index=0;index<strlength;index++)
        if (str[index]==delim[0]) {
            found=1;
            break;
        }

    /* if delim is not contained in str, return str */
    if (!found) {
        static_str = 0;
        return str;
    }

    /* check for consecutive delimiters
    *if first char is delim, return delim
    */
    if (str[0]==delim[0]) {
        static_str = (str + 1);
        return (char *)delim;
    }

    /* terminate the string
    * this assignmetn requires char[], so str has to
    * be char[] rather than *char
    */
    str[index] = '\0';

    /* save the rest of the string */
    if ((str + index + 1)!=0)
        static_str = (str + index + 1);
    else
        static_str = 0;

        return str;
}

As mentioned in previous posts, since strtok(), or the one I implmented above, relies on a static *char variable to preserve the location of last delimiter between consecutive calls, extra care should be taken while dealing with multi-threaded aplications.

Comments

-1
int not_in_delimiter(char c, char *delim){

    while(*delim != '\0'){
            if(c == *delim) return 0;
            delim++;
    }
    return 1;
}

char *token_separater(char *source, char *delimiter, char **last){

char *begin, *next_token;
char *sbegin;

/*Get the start of the token */
if(source)
  begin = source;
else
  begin = *last;

sbegin = begin;

/*Scan through the string till we find character in delimiter. */
while(*begin != '\0' && not_in_delimiter(*begin, delimiter)){
       begin++;
}

/* Check if we have reached at of the string */
if(*begin == '\0') {
/* We dont need to come further, hence return NULL*/
   *last = NULL;
    return sbegin;
}
/* Scan the string till we find a character which is not in delimiter */
 next_token  = begin;
 while(next_token != '\0' && !not_in_delimiter(*next_token, delimiter))    {
    next_token++;
 }
 /* If we have not reached at the end of the string */
 if(*next_token != '\0'){
  *last = next_token--;
  *next_token = '\0';
   return sbegin;
}
}

 void main(){

    char string[10] = "abcb_dccc";
    char delim[10] = "_";
    char *token = NULL;
    char *last = "" ;
    token  = token_separater(string, delim, &last);
    printf("%s\n", token);
    while(last){
            token  = token_separater(NULL, delim, &last);
            printf("%s\n", token);
    }

}

You can read detail analysis at blog mentioned in my profile :)

2 Comments

Nice, @jitsceait, but what happens if I have two delimiters together on input? I'll change a little your code.
I think i have added a test case for consecutive delimiters and it was working. Could you please highlight the code you have changed?

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.