Do I need to free the strtok resulting string?

Question

Or rather, how does strtok produce the string to which it's return value points? Does it allocate memory dynamically? I am asking because I am not sure if I need to free the token in the following code:

The STANDARD_INPUT variables is for exit procedure in case I run out of memory for allocation and the string is the tested subject.

int ValidTotal(STANDARD_INPUT, char *str)
{
    char *cutout = NULL, *temp, delim = '#';
    int i = 0; //Checks the number of ladders in a string, 3 is the required number
    temp = (char*)calloc(strlen(str),sizeof(char));
    if(NULL == temp)
        Pexit(STANDARD_C); //Exit function, frees the memory given in STANDARD_INPUT(STANDARD_C is defined as the names given in STANDARD_INPUT)
    strcpy(temp,str);//Do not want to touch the actual string, so copying it
    cutout = strtok(temp,&delim);//Here is the lynchpin - 
    while(NULL != cutout)
    {
        if(cutout[strlen(cutout) - 1] == '_')
            cutout[strlen(cutout) - 1] = '\0'; \\cutout the _ at the end of a token
        if(Valid(cutout,i++) == INVALID) //Checks validity for substring, INVALID is -1
            return INVALID;
        cutout = strtok(NULL,&delim);
        strcpy(cutout,cutout + 1); //cutout the _ at the beginning of a token
    }
    free(temp);
return VALID; // VALID is 1
}

Answer 1

strtok manipulates the string you pass in and returns a pointer to it, so no memory is allocated.

Please consider using strsep or at least strtok_r to save you some headaches later.

Answer 2

According to the docs:

Return Value

A pointer to the last token found in string.

Since the return pointer just points to one of the bytes in your input string where the token starts, whether you need to free depends on whether you allocated the input string or not.

Answer 3

The first parameter to the strtok(...) function is YOUR string:

str
C string to truncate. Notice that this string is modified by being broken into smaller strings (tokens). Alternativelly, a null pointer may be specified, in which case the function continues scanning where a previous successful call to the function ended.

It puts '\0' characters into YOUR string and returns them as terminated strings. Yes, it mangles your original string. If you need it later, make a copy.

Further, it should not be a constant string (e.g. char* myStr = "constant string";). See here.

It could be allocated locally or by malloc/calloc.

If you allocated it locally on the stack (e.g. char myStr[100];), you don't have to free it.

If you allocated it by malloc (e.g. char* myStr = malloc(100*sizeof(char));), you need to free it.

Some example code:

#include <string.h>
#include <stdio.h>
int main()
{
   const char str[80] = "This is an example string.";
   const char s[2] = " ";
   char *token;

   /* get the first token */
   token = strtok(str, s);

   /* walk through other tokens */
   while( token != NULL ) 
   {
      printf( " %s\n", token );

      token = strtok(NULL, s);
   }

   return(0);
}

NOTE: This example shows how you iterate through the string...since your original string was mangled, strtok(...) remembers where you were last time and keeps working through the string.