C - Pass Array of Strings as Function Parameter

Question

I need to pass a pre-allocated array of strings as a function parameter, and strcpy() to each of the strings within the string array, as in this example:

 static void string_copy(char * pointer[]) {

    strcpy(pointer[0], "Hello ");

    strcpy(pointer[1], "world");

}

int main(int argc, const char * argv[]) {

    char my_array[10][100];

    string_copy(my_array);

    printf("%s%s\n", my_array[0], my_array[1]);

}

And the resulting printed string would be 'Hello world'.

How do I pass a pre-allocated string array and fill out each string within a function as shown above?

As in "static void string_array(char ** pointer[])"? That didn't compile. — James M-Coates
– James M-Coates, Commented Aug 8, 2016 at 6:06

sps · Accepted Answer · 2016-08-08 15:49:07Z

6

When you are doing string_copy(my_array), you are passing a char (*)[100], i.e. pointer to char[100] array to your function. But your function is expecting a char *[], i.e. array of char pointers, because you have defined your function that way.

You can fix this by making changes so that your function (string_copy()) expects a char (*)[100], instead of a char *[].

For this, you can change your function definition as:

/* Your my_array gets converted to pointer to char[100]
   so, you need to change your function parameter
   from `char *pointer[]` to `char (*pointer)[100]` 
*/ 
/* static void string_copy(char *pointer []) */
static void string_copy(char (*pointer) [100])
{
    strcpy(pointer[0], "Hello ");
    strcpy(pointer[1], "world");
}

* Alternative Solution *

A different design/solution would be to change in your main() function so that you are actually passing a char *[], which decays into a char ** - which is fine - to string_copy(). This way you would NOT have to change your string_copy() function.

int main(int argc, const char * argv[]) {

    char my_array[10][100];
    int tot_char_arrs, i;
    char *char_arr_ptr[10];

    /* Get total number of char arrays in my_array */
    tot_char_arrs = sizeof(my_array) / sizeof(my_array[0]);

    // Store all char * 
    for (i = 0; i < tot_char_arrs; i++)
            char_arr_ptr[i] = my_array[i];

    /* Actually passing a char *[].
       it will decay into char **, which is fine
     */
    string_copy(char_arr_ptr);

    printf("%s%s\n", my_array[0], my_array[1]);
}

edited Aug 8, 2016 at 15:49

answered Aug 8, 2016 at 6:08

sps

2,7302 gold badges24 silver badges42 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

2501 Over a year ago

Oh, I see. You have two versions, sorry.

sps Over a year ago

@2501 I edited the post and highlighted that there are two versions, so that it will not confuse future readers. I think the way I had put it before could confuse people.

stzahi · Accepted Answer · 2016-08-08 06:29:01Z

3

you need to use a pointer to the array. here is an example with 1 dimension array:

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

static void string_copy(char **pointer) {

    strcpy(pointer[0], "Hello ");

}

int main(int argc, const char * argv[]) {

    char my_array[10];
    char * p_array = my_array;

    string_copy(&p_array);

    printf("%s\n",  my_array);

}

answered Aug 8, 2016 at 6:29

stzahi

3491 silver badge7 bronze badges

Comments

LPs · Accepted Answer · 2016-08-08 06:35:09Z

Your function can simply accept matrix dimensions and pass a const char * that stores the array of literals (pre-allocated) strings:

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

#define STRINGS_LENGTH 100

static void string_copy(size_t n, size_t m, char pointer[n][m], const char *strings_to_copy[])
{
    for (size_t i=0; i< n; i++)
    {
        strcpy(pointer[i], strings_to_copy[i]);
    }
}

int main( void )
{
    const char *strings[] = { "hello", "World" };
    char my_array[sizeof(strings)/sizeof(strings[0])][STRINGS_LENGTH];

    string_copy(sizeof(strings)/sizeof(strings[0]), STRINGS_LENGTH, my_array, strings);

    printf("%s %s\n", my_array[0], my_array[1]);
}

You can also change the structure of your code using dynamic allocation for your output array like:

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

static bool string_copy(char *pointer[], const char *strings_to_copy[], size_t strings)
{

    for (size_t i=0; i< strings; i++)
    {
        pointer[i] = malloc(strlen(strings_to_copy[i])+1);

        if (pointer[i] != NULL)
            strcpy(pointer[i], strings_to_copy[i]);
        else
            return false;
    }

    return true;
}

int main(void)
{
    const char *strings[] = { "hello", "World" };
    char *my_array[sizeof(strings)/sizeof(strings[0])] = {0};

    if (string_copy(my_array, strings, sizeof(strings)/sizeof(strings[0])) )
    {
        printf("%s %s\n", my_array[0], my_array[1]);
    }

    for (size_t i = 0; i<sizeof(strings)/sizeof(strings[0]); i++)
        free (my_array[i]);
}

Collectives™ on Stack Overflow

C - Pass Array of Strings as Function Parameter

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related