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strTok function (thread safe, supports empty tokens, doesn't change string)

String case reverse function in CGet line from string functionTDD: String Calculator KataC - K&R getint() variationSimple function to generate an HTML-safe stringGeneric Pairing Heap PerformancePattern for writing a generic string transformation functionChange a string into a function/def activatorC++ string tokenizing without streams, with certain conditionsRead consecutive blanks in array

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

2

$begingroup$

I'm new to C language and want to explode a string like we do in PHP explode() function, I searched for a built-in function with the C standard library, and I found strtok , but It doesn't support empty tokens like 1,2,3,,5 . Inspired by the answers I found in this SO question I made this function, it is supposed to be thread safe and support empty tokens and doesn't change the original string

char* strTok(char** newString, char* delimiter)

 char* string = *newString;
 char* delimiterFound = (char*) 0;
 int tokLenght = 0;
 char* tok = (char*) 0;

 if(!string) return (char*) 0;

 delimiterFound = strstr(string, delimiter);

 if(delimiterFound)
 tokLenght = delimiterFound-string;
 else
 tokLenght = strlen(string);
 

 tok = malloc(tokLenght + 1);
 memcpy(tok, string, tokLenght);
 tok[tokLenght] = '';

 *newString = delimiterFound ? delimiterFound + strlen(delimiter) : (char*)0;

 return tok;

I designed it to be used like

char* input = "1,2,3,4,5,6,7,,,10,";
char** inputP = &input;
char* tok;
while( (tok=strTok(inputP, ",")) )
 printf("%sn", tok);

beginner c strings

share|improve this question

asked 4 hours ago

Accountant مAccountant م

1727

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Better user-interface then the original strtok. You may be interested in strsep, too. code.woboq.org/userspace/glibc/string/strsep.c.html
  $endgroup$
  – Neil Edelman
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NeilEdelman thanks I never saw this function before, I will check it.
  $endgroup$
  – Accountant م
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's not in the standard C libraries, but in POSIX, (any type of gcc.) However, like strtok, it obliterates the char to replace it with , so it's not the same.
  $endgroup$
  – Neil Edelman
  2 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

I'm new to C language and want to explode a string like we do in PHP explode() function, I searched for a built-in function with the C standard library, and I found strtok , but It doesn't support empty tokens like 1,2,3,,5 . Inspired by the answers I found in this SO question I made this function, it is supposed to be thread safe and support empty tokens and doesn't change the original string

char* strTok(char** newString, char* delimiter)

 char* string = *newString;
 char* delimiterFound = (char*) 0;
 int tokLenght = 0;
 char* tok = (char*) 0;

 if(!string) return (char*) 0;

 delimiterFound = strstr(string, delimiter);

 if(delimiterFound)
 tokLenght = delimiterFound-string;
 else
 tokLenght = strlen(string);
 

 tok = malloc(tokLenght + 1);
 memcpy(tok, string, tokLenght);
 tok[tokLenght] = '';

 *newString = delimiterFound ? delimiterFound + strlen(delimiter) : (char*)0;

 return tok;

I designed it to be used like

char* input = "1,2,3,4,5,6,7,,,10,";
char** inputP = &input;
char* tok;
while( (tok=strTok(inputP, ",")) )
 printf("%sn", tok);

beginner c strings

share|improve this question

asked 4 hours ago

Accountant مAccountant م

1727

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Better user-interface then the original strtok. You may be interested in strsep, too. code.woboq.org/userspace/glibc/string/strsep.c.html
  $endgroup$
  – Neil Edelman
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NeilEdelman thanks I never saw this function before, I will check it.
  $endgroup$
  – Accountant م
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's not in the standard C libraries, but in POSIX, (any type of gcc.) However, like strtok, it obliterates the char to replace it with , so it's not the same.
  $endgroup$
  – Neil Edelman
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I'm new to C language and want to explode a string like we do in PHP explode() function, I searched for a built-in function with the C standard library, and I found strtok , but It doesn't support empty tokens like 1,2,3,,5 . Inspired by the answers I found in this SO question I made this function, it is supposed to be thread safe and support empty tokens and doesn't change the original string

char* strTok(char** newString, char* delimiter)

 char* string = *newString;
 char* delimiterFound = (char*) 0;
 int tokLenght = 0;
 char* tok = (char*) 0;

 if(!string) return (char*) 0;

 delimiterFound = strstr(string, delimiter);

 if(delimiterFound)
 tokLenght = delimiterFound-string;
 else
 tokLenght = strlen(string);
 

 tok = malloc(tokLenght + 1);
 memcpy(tok, string, tokLenght);
 tok[tokLenght] = '';

 *newString = delimiterFound ? delimiterFound + strlen(delimiter) : (char*)0;

 return tok;

I designed it to be used like

char* input = "1,2,3,4,5,6,7,,,10,";
char** inputP = &input;
char* tok;
while( (tok=strTok(inputP, ",")) )
 printf("%sn", tok);

beginner c strings

share|improve this question

asked 4 hours ago

Accountant مAccountant م

1727

$endgroup$

char* strTok(char** newString, char* delimiter)

 char* string = *newString;
 char* delimiterFound = (char*) 0;
 int tokLenght = 0;
 char* tok = (char*) 0;

 if(!string) return (char*) 0;

 delimiterFound = strstr(string, delimiter);

 if(delimiterFound)
 tokLenght = delimiterFound-string;
 else
 tokLenght = strlen(string);
 

 tok = malloc(tokLenght + 1);
 memcpy(tok, string, tokLenght);
 tok[tokLenght] = '';

 *newString = delimiterFound ? delimiterFound + strlen(delimiter) : (char*)0;

 return tok;

char* input = "1,2,3,4,5,6,7,,,10,";
char** inputP = &input;
char* tok;
while( (tok=strTok(inputP, ",")) )
 printf("%sn", tok);

share|improve this question

asked 4 hours ago

Accountant مAccountant م

1727

share|improve this question

asked 4 hours ago

Accountant مAccountant م

1727

asked 4 hours ago

Accountant مAccountant م

1727

asked 4 hours ago

Accountant مAccountant م

1727

2 Answers
2

active

oldest

votes

1

$begingroup$

• delimiterFound + strlen(delimiter) sounds like a bug. If the delimiter is longer than one character, *newString will point too far into the original, maybe even beyond the end. Correct me if I am wrong, delimiterFound + 1 is what you are actually after.



• 
  

  Modern C allows, and strongly encourages, to declare variables as close to their use a possible. Consider

  
  
  

  char * delimiterFound = strstr(string, delimiter);
  ....
  char * tok = malloc(tokLenght + 1);
  

  
  
  

  etc.

  
  



• Always test that malloc didn't fail.



• 
  

  More spaces - around keywords, braces, etc - definitely improve readability:

  
  
  

   if (....) 
   ....
   else 
   ....
   
  

  
  

share|improve this answer

answered 3 hours ago

vnpvnp

40.6k233103

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thaaank you very much for these precious points, regarding the delimiter length bug, ummmm, I want to support long delimiters more than 1 characters like the boundary string in http requests that has content-type multi-part, and I don't think it's a bug because delimiterFound + strlen(delimiter) can never be after the 0 byte that terminates the original string!, right ? "fooDELIMITER" 3 + 12
  $endgroup$
  – Accountant م
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Accountantم Long delimiters here refer to, say ",;.", in where any character delimits the string on its own right.
  $endgroup$
  – vnp
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't get it, I'm sorry, can you please give me an example input that can break this code, exploiting this bug ?
  $endgroup$
  – Accountant م
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Accountantم Sorry for not being clear. I should realize that your intentions are different (and read man strstr more carefully). Consider it my blinder - since you mentioned strtok, I expected the strtok semantics.
  $endgroup$
  – vnp
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's my fault because I said it's strtok, I wanted strtok that can support delimiters more than 1 characters, because I need this feature a lot. So do you mean it's not a bug ??
  $endgroup$
  – Accountant م
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

From a readability viewpoint, you should use NULL instead of (char*) 0 as it is easier to recognize what you're trying to do. Also, the tokLenght misspells "length", and should probably be tokLength.

You leak memory, as the memory allocated to hold the returned string is never freed.

share|improve this answer

answered 3 hours ago

1201ProgramAlarm1201ProgramAlarm

3,6532925

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much I will use NULL from now on, and I will remember to free() memory , 'I miss PHP garbage collector :(', I didn't get the tokLength spelling note, aren't they the same ?
  $endgroup$
  – Accountant م
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's a matter of style, and including .h if you want to use NULL. However, you don't have to cast (char *)0, just use 0 (or NULL.) It knows from the return type.
  $endgroup$
  – Neil Edelman
  2 hours ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

• delimiterFound + strlen(delimiter) sounds like a bug. If the delimiter is longer than one character, *newString will point too far into the original, maybe even beyond the end. Correct me if I am wrong, delimiterFound + 1 is what you are actually after.



• 
  

  Modern C allows, and strongly encourages, to declare variables as close to their use a possible. Consider

  
  
  

  char * delimiterFound = strstr(string, delimiter);
  ....
  char * tok = malloc(tokLenght + 1);
  

  
  
  

  etc.

  
  



• Always test that malloc didn't fail.



• 
  

  More spaces - around keywords, braces, etc - definitely improve readability:

  
  
  

   if (....) 
   ....
   else 
   ....
   
  

  
  

share|improve this answer

answered 3 hours ago

vnpvnp

40.6k233103

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thaaank you very much for these precious points, regarding the delimiter length bug, ummmm, I want to support long delimiters more than 1 characters like the boundary string in http requests that has content-type multi-part, and I don't think it's a bug because delimiterFound + strlen(delimiter) can never be after the 0 byte that terminates the original string!, right ? "fooDELIMITER" 3 + 12
  $endgroup$
  – Accountant م
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Accountantم Long delimiters here refer to, say ",;.", in where any character delimits the string on its own right.
  $endgroup$
  – vnp
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't get it, I'm sorry, can you please give me an example input that can break this code, exploiting this bug ?
  $endgroup$
  – Accountant م
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Accountantم Sorry for not being clear. I should realize that your intentions are different (and read man strstr more carefully). Consider it my blinder - since you mentioned strtok, I expected the strtok semantics.
  $endgroup$
  – vnp
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's my fault because I said it's strtok, I wanted strtok that can support delimiters more than 1 characters, because I need this feature a lot. So do you mean it's not a bug ??
  $endgroup$
  – Accountant م
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

• delimiterFound + strlen(delimiter) sounds like a bug. If the delimiter is longer than one character, *newString will point too far into the original, maybe even beyond the end. Correct me if I am wrong, delimiterFound + 1 is what you are actually after.



• 
  

  Modern C allows, and strongly encourages, to declare variables as close to their use a possible. Consider

  
  
  

  char * delimiterFound = strstr(string, delimiter);
  ....
  char * tok = malloc(tokLenght + 1);
  

  
  
  

  etc.

  
  



• Always test that malloc didn't fail.



• 
  

  More spaces - around keywords, braces, etc - definitely improve readability:

  
  
  

   if (....) 
   ....
   else 
   ....
   
  

  
  

share|improve this answer

answered 3 hours ago

vnpvnp

40.6k233103

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thaaank you very much for these precious points, regarding the delimiter length bug, ummmm, I want to support long delimiters more than 1 characters like the boundary string in http requests that has content-type multi-part, and I don't think it's a bug because delimiterFound + strlen(delimiter) can never be after the 0 byte that terminates the original string!, right ? "fooDELIMITER" 3 + 12
  $endgroup$
  – Accountant م
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Accountantم Long delimiters here refer to, say ",;.", in where any character delimits the string on its own right.
  $endgroup$
  – vnp
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't get it, I'm sorry, can you please give me an example input that can break this code, exploiting this bug ?
  $endgroup$
  – Accountant م
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Accountantم Sorry for not being clear. I should realize that your intentions are different (and read man strstr more carefully). Consider it my blinder - since you mentioned strtok, I expected the strtok semantics.
  $endgroup$
  – vnp
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's my fault because I said it's strtok, I wanted strtok that can support delimiters more than 1 characters, because I need this feature a lot. So do you mean it's not a bug ??
  $endgroup$
  – Accountant م
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

• delimiterFound + strlen(delimiter) sounds like a bug. If the delimiter is longer than one character, *newString will point too far into the original, maybe even beyond the end. Correct me if I am wrong, delimiterFound + 1 is what you are actually after.



• 
  

  Modern C allows, and strongly encourages, to declare variables as close to their use a possible. Consider

  
  
  

  char * delimiterFound = strstr(string, delimiter);
  ....
  char * tok = malloc(tokLenght + 1);
  

  
  
  

  etc.

  
  



• Always test that malloc didn't fail.



• 
  

  More spaces - around keywords, braces, etc - definitely improve readability:

  
  
  

   if (....) 
   ....
   else 
   ....
   
  

  
  

share|improve this answer

answered 3 hours ago

vnpvnp

40.6k233103

$endgroup$

share|improve this answer

answered 3 hours ago

vnpvnp

40.6k233103

answered 3 hours ago

vnpvnp

40.6k233103

answered 3 hours ago

vnpvnp

40.6k233103

1

$begingroup$

From a readability viewpoint, you should use NULL instead of (char*) 0 as it is easier to recognize what you're trying to do. Also, the tokLenght misspells "length", and should probably be tokLength.

You leak memory, as the memory allocated to hold the returned string is never freed.

share|improve this answer

answered 3 hours ago

1201ProgramAlarm1201ProgramAlarm

3,6532925

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much I will use NULL from now on, and I will remember to free() memory , 'I miss PHP garbage collector :(', I didn't get the tokLength spelling note, aren't they the same ?
  $endgroup$
  – Accountant م
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's a matter of style, and including .h if you want to use NULL. However, you don't have to cast (char *)0, just use 0 (or NULL.) It knows from the return type.
  $endgroup$
  – Neil Edelman
  2 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

From a readability viewpoint, you should use NULL instead of (char*) 0 as it is easier to recognize what you're trying to do. Also, the tokLenght misspells "length", and should probably be tokLength.

You leak memory, as the memory allocated to hold the returned string is never freed.

share|improve this answer

answered 3 hours ago

1201ProgramAlarm1201ProgramAlarm

3,6532925

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much I will use NULL from now on, and I will remember to free() memory , 'I miss PHP garbage collector :(', I didn't get the tokLength spelling note, aren't they the same ?
  $endgroup$
  – Accountant م
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's a matter of style, and including .h if you want to use NULL. However, you don't have to cast (char *)0, just use 0 (or NULL.) It knows from the return type.
  $endgroup$
  – Neil Edelman
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

From a readability viewpoint, you should use NULL instead of (char*) 0 as it is easier to recognize what you're trying to do. Also, the tokLenght misspells "length", and should probably be tokLength.

You leak memory, as the memory allocated to hold the returned string is never freed.

share|improve this answer

answered 3 hours ago

1201ProgramAlarm1201ProgramAlarm

3,6532925

$endgroup$

share|improve this answer

answered 3 hours ago

1201ProgramAlarm1201ProgramAlarm

3,6532925

answered 3 hours ago

1201ProgramAlarm1201ProgramAlarm

3,6532925

answered 3 hours ago

1201ProgramAlarm1201ProgramAlarm

3,6532925