Modifying a string array inside of a function

Question

The following code example, doesn't print the strings test1 - test5 contained in array in the main() function

however it works inside the make() function

I am certain the answer is simple, how would I produce the desired result?

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

#define ELEMENTS 4

void make(char ***array) {

char p2[] = "test1 test2 test3 test4 test5";
char* token = strtok(p2, " ");
int i = 0;
while (token) 
{

    (*array)[i]= token;
    token = strtok(NULL, " ");
    i++;
}

printf("%s\n",(*array)[0]);
printf("%s\n",(*array)[1]);
printf("%s\n",(*array)[2]);
printf("%s\n",(*array)[3]);
printf("%s\n",(*array)[4]);

}

int main(int argc, char **argv) {
char **array;
make(&array);

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

return 0;
}

This code compiles without error or warning and produces the following output:

test1
test2
test3
test4
test5
yf�


���

my expected result was to have test1 - test5 printed twice, once inside the make() function and once in the main() function

as a side note, this is only my second post to stackoverflow, this is modified code from my first question Passing a string array to a function in C

You are still making your arrays too small.

Dark Falcon
– Dark Falcon

2013-04-05 15:26:59 +00:00
Commented Apr 5, 2013 at 15:26 — Dark Falcon
– Dark Falcon, Commented Apr 5, 2013 at 15:26

che · Accepted Answer · 2013-04-12 21:32:38Z

char p2[] = "test1 test2 test3 test4 test5";

defines the string p2 on the stack of make() function. strtok() only returns pointers to the same array, which becomes invalid when make() returns.

It's the same as this:

char * foo()
{
    char array[] = "hello";
    printf("%s\n", array); // works fine
    return array;
}

void main()
{
    char * array = foo(); // just a pointer to invalid data
    printf("%s\n", array); // FAIL
}

How to do it correctly

There are basically two ways to return strings in C.

Either by filling in a buffer:

void fill_buf(char * buf, size_t len)
{
    char string[] = "hello";
    snprintf(buf, len, "%s", string);
}

void main()
{
    char buffer[25];
    fill_buf(buffer, sizeof(buffer));
    printf("%s\n", buffer);
}

Or by returning malloc'd string:

char * get_malloced_or_null()
{
    char my_string[] = "hello";
    char * copied_pointer = strdup(my_string); // might fail and return NULL
    return copied_pointer;
}

void main()
{
    char * string = get_malloced_or_null();
    if (string == NULL) { return; }
    printf("%s\n", string);
    free(string);
}

The advantage of the second approach is that you can use it with arbitrarily long strings. On the other hand, you need to check for allocation failures, and you have to free memory after you're done with it.

Bechir · Accepted Answer · 2013-04-05 15:37:50Z

In the make() function, you are parsing a local variable. As such, the strtok() returns stack address (and you array will reference a content in the stack that could be overwritten any time). Once the function make() is completed, those pointers will refer to anything in your stack and your array content may be any thing.

You may fix this as follow:

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

#define ELEMENTS 4

void make(char ***array)
{
    char p2[] = "test1 test2 test3 test4 test5";
    char* token = strtok(p2, " ");
    int i = 0;
    while (token) 
    {
        (*array)[i]= strdup(token);
        token = strtok(NULL, " ");
        i++;
    }

    printf("%s\n",(*array)[0]);
    printf("%s\n",(*array)[1]);
    printf("%s\n",(*array)[2]);
    printf("%s\n",(*array)[3]);
    printf("%s\n",(*array)[4]);

}

int main(int argc, char **argv)
{
    char **array;
    make(&array);

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

    return 0;
}

MOHAMED · Accepted Answer · 2013-04-05 15:41:09Z

0

In addition of the above answer;

You have to allocate memory for the char ***array at the beginning the make() function

*array = malloc(ELEMENTS * sizeof(char *));

and ELEMENTS should be define as 5 and not as 4

#define ELEMENTS 5

answered Apr 5, 2013 at 15:41

MOHAMED

44k61 gold badges177 silver badges293 bronze badges

Comments

xtof pernod · Accepted Answer · 2013-04-05 15:55:10Z

0

It works with simply defining p2[] static:

static char p2[] = "test1 test2 test3 test4 test5";

also, ELEMENTS should be 5, and the declaration of array in main is a simple "pointer to pointer", it should be "array of pointers":

char *array[ELEMENTS];

then you'll get enough room for your 5 strings.

EDIT: working with an array of pointers simplfies the make() function:

void make(char **array) {
    static char p2[] = "test1 test2 test3 test4 test5";
    char *token = strtok(p2, " ");
    int i = 0;
    while (token) {
        array[i] = token;
        token = strtok(NULL, " ");
        i++;
    }
    for (i = 0; i < ELEMENTS; ++i)
        printf("%d: %s\n", i, array[i]);
}

int main(int argc, char **argv)
{
    char *array[ELEMENTS];
    make(array);
    /* ... */
}

edited Apr 5, 2013 at 15:55

answered Apr 5, 2013 at 15:45

xtof pernod

8835 silver badges7 bronze badges

3 Comments

che Over a year ago

While this works, it's not the best approach to making actual software, since it will break if you use the the function more than once.

xtof pernod Over a year ago

True, but why parse the same string multiple times.. Actual software should not do so :)

che Over a year ago

Yep, but I've seen cases where one function was used to parse different strings.

Ziggy · Accepted Answer · 2013-04-08 04:02:45Z

0

This code compiles cleanly, and produces the desired result, Thanks to all that helped me understand pointers. now i'll work on freeing the memory I allocated to the array

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

#define ELEMENTS 5

void make(char ***array) {

char p2[] = "test1 test2 test3 test4 test5";
char* token = strtok(p2, " ");
int i = 0;
while (token) 
{

    (*array)[i] = malloc(strlen(token) + 1);
    strcpy((*array)[i], token);
    token = strtok(NULL, " ");
    i++;
}

}

int main(int argc, char **argv) {
char **array;
make(&array);

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

return 0;
}

answered Apr 8, 2013 at 4:02

Ziggy

353 bronze badges

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