Flatten array into string

Question

I have an array that is a "NULL-terminated array of NULL-terminated strings". This is given by char **args.

I can access individual elements using args[0], args[1], etc. I wanted to take the entire array and flatten all the elements into a string. If the array contained:

args[0] = "abc"
args[1] = "def"

I want a resulting string to be:

abcdef

I tried to do this by looping through all the elements and concatenating them all together but I do not know how to tell when I have reached the end of the array because sizeof() does not work.

K-ballo · Accepted Answer · 2011-10-08 19:22:03Z

4

I have an array that is a "NULL-terminated array of NULL-terminated strings".

The array ends with NULL, that is as soon as args[i] == NULL you stop your iteration.

answered Oct 8, 2011 at 19:22

K-ballo

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Community · Accepted Answer · 2017-05-23 12:03:35Z

2

As your array is null terminated you know you have reached the end of the array when you get a NULL element.

if (args[i] == NULL){
    //DONE
}

If you wanted to get the length of the array args you could loop through it until you get a null, counting the number of iterations:

int length_of_args = 0;
while (args[length_of_args] != NULL){
    length_of_args++;
}

Someone has posted a similar question Copy argv to string in C(newbie question) with some answers you might find helpful.

edited May 23, 2017 at 12:03

CommunityBot

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answered Oct 8, 2011 at 19:23

Ethan Heilman

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1 Comment

R.. GitHub STOP HELPING ICE Over a year ago

sizeof will determine the size of arrays, but only if you actually give it the array. The problem is the OP only has a pointer to the array, not the array itself.

David Heffernan · Accepted Answer · 2011-10-08 19:39:23Z

1

You are looking for something like this:

char* concat_string_array(char** input)
{
    int i, len;
    char* result;

    len = 1;
    for (i=0; input[i]; i++)
        len += strlen(input[i]);

    result = malloc(len);
    result[0] = '\0';
    for (i=0; input[i]; i++)
        result = strcat(result, input[i]);

    return result;
}

The key part that I believe you are missing is that the array is terminated by a NULL entry. That's what the test in the for loops checks.

edited Oct 8, 2011 at 19:39

answered Oct 8, 2011 at 19:31

David Heffernan

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Comments

RushPL · Accepted Answer · 2011-10-08 19:43:22Z

In C99:

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

int main()
{
    char* args[] = { "hello", "my", "world", NULL};
    size_t str_size = 0;
    char** args_itr = args;
/* calculate resulting size */
    while(*args_itr != NULL) {
        str_size += strlen(*args_itr);
        args_itr++;
    }
    char result[str_size+1];
    result[str_size] = '\0'; // protect against 0 size                                                                                                                                          
    args_itr = args;
    char* result_ptr = result;
    while(*args_itr != NULL) {
        strcpy(result_ptr, *args_itr);
        result_ptr += strlen(*args_itr);
        args_itr++;
    }
    /* if you use it in a lib function you could do */
    // return strdup(result);                                                                                                                                                                   

    printf("%s\n", result);
    return 0;
}

rahmu · Accepted Answer · 2011-10-08 19:51:09Z

0

You could use sprintf to concatenate 2 strings. You would need to control size of the new string though. Something like that:

int main ()
{
    char* s1 = "abc";
    char *s2 = "def";

    char* snew = (char *)malloc (strlen (s1) + strlen (s2) + 1);

    sprintf (snew, "%s%s", s1, s2);
    printf ("%s\n", snew);

    return EXIT_SUCCESS;
}

edited Oct 8, 2011 at 19:51

answered Oct 8, 2011 at 19:30

rahmu

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2 Comments

RushPL Over a year ago

In this case you are under control but as a rule one should never ever use sprintf. Use snprintf. It will profit in less bugs by just using this rule. Also, in your code, snew should alloc 1 byte more to leave space for the ending \0 string termination character.

rahmu Over a year ago

You're right about the second one, I edited my answer. As for the never use sprintf, you're probably right too, it just happens to work. So I thought I'd put it up.

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