Why does this C program not function properly?

Question

This C program splits the string "1 2 3 4 5 6 7 8 9 10" into tokens, stores these in buf, and prints the contents of buf.

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

/* Fills an array with pointers to the tokens of the given string.
 * string: A null-terminated char*
 * buf: A buffer to be filled with pointers to each token. 
 */
void get_tokens(char * string, char ** buf) {
    char tmp[100];
    strncpy(tmp, string, 100);
    char * tok = strtok(tmp, " \n");
    int i = 0;
    while (tok != NULL) {
        buf[i] = tok;
        tok = strtok(NULL, " \n");
        i++;
    }
}

int main() {
    char ** buf = malloc(10 * sizeof(char*));
    char * string = "1 2 3 4 5 6 7 8 9 10";
    get_tokens(string, buf);

    int i;
    for (i = 0; i < 10; i++) {
        printf("  %s\n", buf[i]);
    }
}

The output:

Why is my output being mangled?

The root cause is tmp being out of scope after returning from get_tokens. — Ctx
– Ctx, Commented Feb 22, 2016 at 23:04
char tmp[100]; is local variable. Address of that part invalid at out of scope. — BLUEPIXY
– BLUEPIXY, Commented Feb 22, 2016 at 23:05
But the pointers returned by strtok are not pointers to tmp, they are independent strings made with malloc — Jamie
– Jamie, Commented Feb 22, 2016 at 23:06
"they are independent strings made with malloc". Really? Which part of the code does that? char * tok = strtok(tmp, " \n"); buf[i] = tok;. tok points to a position within tmp. — kaylum
– kaylum, Commented Feb 22, 2016 at 23:07
I think I get it now, thank you. The reason the input string to strtok is mangled is that strtok just replaces the delimiter with \0 and returns a pointer to the start of each token within tmp. — Jamie
– Jamie, Commented Feb 22, 2016 at 23:09

Vlad from Moscow · Accepted Answer · 2016-02-22 23:35:14Z

The array tmp is a local array of the function with the automatic storage duration. It is destroyed after the function exit. So all pointers that point to elements of the array become invalid.

I can suggest the following solution

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

size_t get_tokens( char ** buf, size_t n, const char *string ) 
{
    char tmp[100];
    strncpy( tmp, string, 100 );
    tmp[99] = '\0';

    size_t i = 0;
    char * tok = strtok( tmp, " " );

    while ( i < n && tok != NULL ) 
    {
        buf[i] = malloc( strlen( tok ) + 1 );
        strcpy( buf[i++], tok );
        tok = strtok( NULL, " " );
    }

    return i;
}

int main( void ) 
{
    const size_t N = 10;

    char ** buf = malloc( N * sizeof( char * ) );
    char * string = "1 2 3 4 5 6 7 8 9 10";

    size_t n = get_tokens( buf, N, string );

    for ( size_t i = 0; i < n; i++ ) 
    {
        puts( buf[i] );
    }

    for ( size_t i = 0; i < n; i++ ) free( buf[i] );
    free( buf );

    return 0;
}

The program output is

As for your program then at least you should declare the array with storage specifier static.

For example

static char tmp[100];
^^^^^^
strncpy( tmp, string, 100 );
tmp[99] = '\0';
^^^^^^^^^^^^^^^

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