Segmentation Fault when using strtok_r
Asked Answered
D

6

18

Can anyone explain why I am getting segmentation fault in the following example?

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

int main(void) {
  char *hello = "Hello World, Let me live.";
  char *tokens[50];
  strtok_r(hello, " ,", tokens);
  int i = 0;
  while(i < 5) {
    printf("%s\n", tokens[i++]);
  }
}
Daveta answered 9/2, 2010 at 6:24 Comment(0)
H
25

Try this:

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

int main(void) {
        char hello[] = "Hello World, Let me live."; // make this a char array not a pointer to literal.
        char *rest; // to point to the rest of the string after token extraction.
        char *token; // to point to the actual token returned.
        char *ptr = hello; // make q point to start of hello.

        // loop till strtok_r returns NULL.
        while(token = strtok_r(ptr, " ,", &rest)) {

                printf("%s\n", token); // print the token returned.
                ptr = rest; // rest contains the left over part..assign it to ptr...and start tokenizing again.    
        }
}
/*
Output:
Hello
World
Let
me
live.
*/
Hutson answered 9/2, 2010 at 6:54 Comment(3)
This example gives me a segfault on the printf line. gdb print token shows 0xffffffffffffdad0 <Address 0xffffffffffffdad0 out of bounds> Also get these two warning when I compile proj1.c:33:2: warning: implicit declaration of function ‘strtok_r’ [-Wimplicit-function-declaration] proj1.c:33:14: warning: assignment makes pointer from integer without a cast [enabled by default]French
Sorry I'm a newbie, why shouldn't it be char *ptr = Hello; with a capital H? Also, Alok's answer says the first time the first parameter needs to "tokenized" and then subsequent calls it needs to be NULL, but it appears your example just calls it the one way in the while loop? Thanks for the code btwSawmill
*ptr = hello; because hello refers to char hello[]....Sines
D
17
  • You need to call strtok_r in a loop. The first time you give it the string to be tokenized, then you give it NULL as the first parameter.
  • strtok_r takes a char ** as the third parameter. tokens is an array of 50 char * values. When you pass tokens to strtok_r(), what gets passed is a char ** value that points to the first element of that array. This is okay, but you are wasting 49 of the values that are not used at all. You should have char *last; and use &last as the third parameter to strtok_r().
  • strtok_r() modifies its first argument, so you can't pass it something that can't be modified. String literals in C are read-only, so you need something that can be modified: char hello[] = "Hello World, Let me live."; for example.
Deborahdeborath answered 9/2, 2010 at 6:55 Comment(3)
Thanks for the answer. I wish SO allowed marking multiple answers as correct. :)Daveta
@Scrub: glad to be of help. Make sure you understand my second point above (about char *tokens[50]; being equivalent to char ** when passed to a function).Deborahdeborath
SAVED me; +1 for reminding me, that param 1 cannot be const char *. For others: Don't feed in data from the .RODATA section. Or: try to see if it works with strcpy'ing to a temp buffer beforehand.Xe
V
5

A bunch of things wrong:

  1. hello points to a string literal, which must be treated as immutable. (It could live in read-only memory.) Since strtok_r mutates its argument string, you can't use hello with it.

  2. You call strtok_r only once and don't initialize your tokens array to point to anything.

Try this:

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

int main(void) {
  char hello[] = "Hello World, Let me live.";
  char *p = hello;
  char *tokens[50];
  int i = 0;

  while (i < 50) {
     tokens[i] = strtok_r(p, " ,", &p);
     if (tokens[i] == NULL) {
        break;
     }
     i++;
  }

  i = 0;
  while (i < 5) {
    printf("%s\n", tokens[i++]);
  }

  return 0;
}
Vegetarian answered 9/2, 2010 at 6:54 Comment(1)
Thanks for the answer. I wish SO allowed marking multiple answers as correct. :)Daveta
L
3

strtok_r tries to write null characters into hello (which is illegal because it is a const string)

Lashley answered 9/2, 2010 at 6:31 Comment(1)
I tried char hello[50] . The segmentation fault is vanished but now the problem is printf just prints sad blank lines. :(Daveta
M
2

You have understood the usage of strtok_r incorrectly. Please check this example and documentation

And try & see this:

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

int main(void)
{
    char hello[] = "Hello World, let me live.";

    char *tmp;
    char *token = NULL;
    for(token = strtok_r(hello, ", ", &tmp);
        token != NULL;
        token = strtok_r(NULL, ", ", &tmp))
    {
        printf("%s\n", token);
    }

    return 0;
}
Mraz answered 9/2, 2010 at 6:46 Comment(0)
S
0

I think it might be the char *tokens[50]; because you are declaring it a pointer when it is already a pointer. An array is already a pointer upon declaration. You mean to say char tokens[50];. That should do the trick.

Sophister answered 9/2, 2010 at 6:26 Comment(0)

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