Pointer to string array in C

Question

Playing with pointers in C is fun (not really).

I have several arrays of strings I want to declare in an easy way, preferably something like:

arrayOfStrings1 = {"word1", "word2", etc. };
arrayOfStrings2 = {"anotherword1", "anotherword2", etc. };
arrayOfStrings3 = etc.
etc.

Something similar to a translation array (but not quite), so I want to be able to swap between these during runtime. For that I want a pointer pointerToArrayOfStrings that I can swap like:

pointerToArrayOfStrings = arrayOfStrings1;
doStuff();
pointerToArrayOfStrings = arrayOfStrings2;
doSomeOtherStuff();

In my naive understanding of arrays of strings and pointers to these, this is what I tried:

// Danish transforms
const unsigned char* da_DK[] = {"b","bb","c","c","cc","d","dd","e","f","ff","g","gg","h","hh","j","j","jj","k","k","kk","l","l","l","l","ll","m","mm","n","n","nn","p","pp","r","r","r","rr","s","s","s","ss","t","t","tt","v","v","vv","æ"};

// British english transforms
const unsigned char* en_GB[] = {"a","a","a","a","a","a","a","a","a","a","a","a","a","age","ai","aj","ay","b","cial","cian","cian","dj","dsj","ea","ee","ege","ei","ei","eigh","eigh","f","f","f","g","g","gs","i","i","i","j","j","k","ks","kw","l","m","n","n","o","r","s","s","sd","sdr","sion","sion","sj","sj","tial","tion","tion","tj","u","u","u","u","w","ye","ye","z"};

    // More languages....

const unsigned char** laguageStrings;

// Assign language
if (streq(language, "da-DK")){
    laguageStrings= da_DK;
}
else if (streq(language, "en-GB")){
    laguageStrings= en_GB;
}
else
         return 0;
}

Language is a char * containing the language "en-GB", "da-DK" etc., streq() is just a home brewed (somewhat faster than strcmp()) string comparison function.

Long story short, depending on compiler this approach may work, report compiler warnings or compile, but give unexpected results.

What would be the correct way to solve this problem?

Answer 1

There are two way of working with array of characters (strings) in C. They are as follows:

char a[ROW][COL];
char *b[ROW];

Pictorial representation is available as an inline comment in the code.

Based on how you want to represent the array of characters (strings), you can define pointer to that as follows

    char (*ptr1)[COL] = a;
    char **ptr2 = b;

They are fundamentally different types (in a subtle way) and so the pointers to them is also slightly different.

The following example demonstrates the different ways of working with strings in C and I hope it helps you in better understanding of array of characters (strings) in C.

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

#define ROW 5
#define COL 10

int main(void) 
{
    int i, j;
    char a[ROW][COL] = {"string1", "string2", "string3", "string4", "string5"};
    char *b[ROW];

    /*

    a[][]

      0   1   2   3   4   5   6     7    8   9
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 1 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 2 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 3 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 4 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 5 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+

    */  

    /* Now, lets work on b */    
    for (i=0 ; i<5; i++) {
        if ((b[i] = malloc(sizeof(char) * COL)) == NULL) {
            printf("unable to allocate memory \n");
            return -1;
        }
    }

    strcpy(b[0], "string1");
    strcpy(b[1], "string2");
    strcpy(b[2], "string3");
    strcpy(b[3], "string4");
    strcpy(b[4], "string5");

    /*

       b[]              0   1   2   3   4   5   6    7     8   9
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 1 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 2 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 3 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 4 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 5 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+

    */

    char (*ptr1)[COL] = a;
    printf("Contents of first array \n");
    for (i=0; i<ROW; i++)
        printf("%s \n", *ptr1++);


    char **ptr2 = b;
    printf("Contents of second array \n");
    for (i=0; i<ROW; i++)
        printf("%s \n", ptr2[i]);

    /* b should be free'd */
    for (i=0 ; i<5; i++)
        free(b[i]);

    return 0;
}

Answer 2

What would be the correct way to solve this problem?

Well, the correct way would be to use a library specifically designed for dealing with multilanguage interfaces - for instance gettext.

Another way, though patchier, would be to use a hash table (also known as "dictionary" or "hash map" or "associative map" in other languages/technologies): Looking for a good hash table implementation in C

It's probably not the answer you were looking for, but you've asked the wrong question to the right problem.