2D character array initialization in C

Question

I am trying to build a list of strings that I need to pass to a function expecting char **

How do I build this array? I want to pass in two options, each wi

Answer 1

char **options[2][100];

declares a size-2 array of size-100 arrays of pointers to pointers to char. You'll want to remove one *. You'll also want to put your string literals in double quotes.

Answer 2

How to create an array size 5 containing pointers to characters:

char *array_of_pointers[ 5 ];        //array size 5 containing pointers to char
char m = 'm';                        //character value holding the value 'm'
array_of_pointers[0] = &m;           //assign m ptr into the array position 0.
printf("%c", *array_of_pointers[0]); //get the value of the pointer to m

How to create a pointer to an array of characters:

char (*pointer_to_array)[ 5 ];        //A pointer to an array containing 5 chars
char m = 'm';                         //character value holding the value 'm'
*pointer_to_array[0] = m;             //dereference array and put m in position 0
printf("%c", (*pointer_to_array)[0]); //dereference array and get position 0

How to create an 2D array containing pointers to characters:

char *array_of_pointers[5][2];          
//An array size 5 containing arrays size 2 containing pointers to char

char m = 'm';                           
//character value holding the value 'm'

array_of_pointers[4][1] = &m;           
//Get position 4 of array, then get position 1, then put m ptr in there.

printf("%c", *array_of_pointers[4][1]); 
//Get position 4 of array, then get position 1 and dereference it.

How to create a pointer to an 2D array of characters:

char (*pointer_to_array)[5][2];
//A pointer to an array size 5 each containing arrays size 2 which hold chars

char m = 'm';                            
//character value holding the value 'm'

(*pointer_to_array)[4][1] = m;           
//dereference array, Get position 4, get position 1, put m there.

printf("%c", (*pointer_to_array)[4][1]); 
//dereference array, Get position 4, get position 1

To help you out with understanding how humans should read complex C/C++ declarations read this: http://www.programmerinterview.com/index.php/c-cplusplus/c-declarations/

Answer 3

I think what you originally meant to do was to make an array only of characters, not of pointers:

char options[2][100];

options[0][0]='t';
options[0][1]='e';
options[0][2]='s';
options[0][3]='t';
options[0][4]='1';
options[0][5]='\0';  /* NUL termination of C string */

/* A standard C library function which copies strings. */
strcpy(options[1], "test2");

The code above shows two distinct methods of setting the character values in memory you have set aside to contain characters.

Answer 4

C strings are enclosed in double quotes:

const char *options[2][100];

options[0][0] = "test1";
options[1][0] = "test2";

Re-reading your question and comments though I'm guessing that what you really want to do is this:

const char *options[2] = { "test1", "test2" };