C Program: newbie trying to pass 2D string array into a function

Question

Programming in C ( with -std=C89), and running into errors trying to pass a character string array into a function.

In main() , I've declared the array as follows:

#define ROWS 501
#define COLS 101
void my_function( char **);
...
char my_array[ROWS][COLS];
...
my_function(my_array);

In my_function , I've declared the array as:

void my_function( char **my_array )
{
...
}

I'm getting this error:

warning: passing argument 1 of 'my_function' from incompatible pointer type, note: expected 'char **' but argument is of type 'char (*)[101]

Answer 1

A two-dimensional array of characters is still a character array and would have a prototype of char *my_array . So just change your function definition to this:

void my_function(char *my_array)

Note that this will flatten the array. There are different techniques to keep the two-dimensional-ness of the array, an easy way is to use this alternative prototype:

void my_function(char my_array[][COLS])

Which will preserve your array's dimensions when passed.

char **my_array means something completely different (pointer to an array, for example).

Answer 2

You can pass a char[] variable as a char* , but you can't pass a char[][] as a char** . When you use the argument char** my_array , you are saying that *my_array has type 'pointer-to-char'. In reality, it has type 'array-of-char'. You would use an argument of type char** if you were using an array declared like char* x[]; and each element was a pointer to a dynamically-allocated buffer.

When working with multidimensional arrays, you have to remember that you can only replace the "innermost" dimension of array with * . If you try to abstract away more than one dimension, the compiler won't know how to do the array arithmetic. If you need a function that takes a multidimensional array with arbitrary sizes in all dimensions, then pass the array as a void* , pass the array dimensions as additional arguments, and then do all of the array arithmetic manually.

Answer 3

You can have a function signature with multi dimensional arrays, ie:

my_fun(char my_array[][COLS]);

You might get some out of this:

A Tutorial on Pointers and Arrays in C , see Ie chapter 7.

---

Edit: I suspect you are trying to do something you do not need.

#include <stdio.h>

#define ROWS 501
#define COLS 101

char my_arr[ROWS][COLS];

void foo(char arr[][COLS])
{
    arr[44][23] = 'a';
    printf("foo_1:  %p\n", (void*) arr);
    printf("foo_2:  %p\n", (void*) &arr);
    printf("foo_3:  %p\n", (void*) arr[44]);
    printf("foo_4:  %p\n", (void*) &arr[44]);
}

int main(void)
{
    foo(my_arr);
    printf("my_arr[%03d][%03d] is %c\n", 44, 23, my_arr[44][23]);
    /* my_arr[44][23] is now 'a', (also here)  */

    printf("main_1: %p\n", (void*) my_arr);
    printf("main_2: %p\n", (void*) &my_arr);
    printf("main_3: %p\n", (void*) my_arr[44]);
    printf("main_4: %p\n", (void*) &my_arr[44]);

    return 0;
}

Example output:

foo_1:  0x804a040  <---+
foo_2:  0xbece91f0     |
foo_3:  0x804b19c  <--------+
foo_4:  0x804b19c  <--------+
my_arr[044][023] is a  |    |
main_1: 0x804a040 <----+    |
main_2: 0x804a040 <----+    |
main_3: 0x804b19c <---------+
main_4: 0x804b19c <---------+