C - Determining which delimiter used - strtok()
Solution 1
Important: strtok
is not re-entrant, you should use strtok_r
instead of it.
You can do it by saving a copy of the original string, and looking into offsets of the current token into that copy:
char str[] = "Hello there; How are you? / I'm good - End";
char *copy = strdup(str);
char *delim = ";-/";
char *res = strtok( str, delim );
while (res) {
printf("%c\n", copy[res-str+strlen(res)]);
res = strtok( NULL, delim );
}
free(copy);
This prints
;
/
-
EDIT: Handling multiple delimiters
If you need to handle multiple delimiters, determining the length of the current sequence of delimiters becomes slightly harder: now you need to find the next token before deciding how long is the sequence of delimiters. The math is not complicated, as long as you remember that NULL
requires special treatment:
char str[] = "(20*(5+(7*2)))+((2+8)*(3+6*9))";
char *copy = strdup(str);
char *delim = "*+()";
char *res = strtok( str, delim );
while (res) {
int from = res-str+strlen(res);
res = strtok( NULL, delim );
int to = res != NULL ? res-str : strlen(copy);
printf("%.*s\n", to-from, copy+from);
}
free(copy);
Solution 2
You can't. strtok
overwrites the next separator character with a nul character (in order to terminate the token that it's returning this time), and it doesn't store the previous value that it overwrites. The first time you call strtok
on your example string, the ;
is gone forever.
You could do something if you keep an unmodified copy of the string you're modifying with strtok
- given the index of the nul terminator for your current token (relative to the start of the string), you can look at the same index in the copy and see what was there.
That might be worse than just writing your own code to separate the string, of course. You can use strpbrk
or strcspn
, if you can live with the resulting token not being nul-terminated for you.
Solution 3
man 3 strtok
The strtok() and strtok_r() functions return a pointer to the beginning of each subsequent token in the string, after replacing the token itself with a NUL character. When no more tokens remain, a null pointer is returned.
But with a little pointer arithmetic you can do something like:
char* string = "Hello,World!";
char* dup = strdup(string);
char* world = strtok(string, ",");
char delim_used = dup[world - string];
free(dup);
Andrew Backes
Updated on July 18, 2022Comments
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Andrew Backes almost 2 years
Let's say I'm using
strtok()
like this..char *token = strtok(input, ";-/");
Is there a way to figure out which token actually gets used? For instance, if the inputs was something like:
Hello there; How are you? / I'm good - End
Can I figure out which delimiter was used for each token? I need to be able to output a specific message, depending on the delimiter that followed the token.
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Andrew Backes almost 12 yearsThank you, this is the kind of behavior I was hoping to achieve.
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David over 8 years@dasblinkenlight, this wont work if your have multiple separator in succession. e.g consider tokenization of arithmetic expression 20*5+(7*2) where you define your delim as "+-/*()" and you are interested in tokenization of operator and operands. +( will not be tokenized correctly.
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Sergey Kalinichenko over 8 years@David You are right, the code assumes there is always one delimiter. However, there is an easy fix for that - all you need to do is to get the next token to decide how long is the current run of delimiters (see the edit and the demo).
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David over 8 years@dasblinkenlight Thanks . it works but it still misses the leading delimiters. think if you want to evaluate the expression you need all the delimiters (including the leading one). Also you have to split the delimiters again as it is printing multiple delimiter together. I found out writing custom tokenizer by using find_first_of() or find_first_not_of() is much cleaner. Also boost tokens() class achieve this nicely.
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David over 8 yearsSomething like stackoverflow.com/questions/9823263/…