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]

this link (eskimo.com/~scs/cclass/int/sx9a.html) should help you. — Kunal
– Kunal, Commented Apr 10, 2012 at 1:55

Mahmoud Al-Qudsi · Accepted Answer · 2012-04-10 01:51:33Z

3

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).

answered Apr 10, 2012 at 1:51

Mahmoud Al-Qudsi

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3 Comments

ggkmath Over a year ago

Thanks, If I change the prototype in main() to be void my_function( char *); and change the function definition to void my_function( char *my_array) I get this error:

warning: passing argument 30 of 'fn_call_master_v1p0' from incompatible pointer type ../Project_Files/myheader.h:59: note: expected 'char *' but argument is of type 'char (*)[101]'

ggkmath Over a year ago

Thanks, If I change the prototype in main() to be void my_function( char *my_array[][COLS]); and change the function definition to void my_function( char *my_array[][COLS]) I get this warning:

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

Mahmoud Al-Qudsi Over a year ago

I didn't say char *my_array[][COLS] but char my_array[][cols] (notice the extra * in your code). You can also use char (*my_array)[COLS], but there's no reason to do so.

bta · Accepted Answer · 2012-04-10 01:56:39Z

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.

Morpfh · Accepted Answer · 2012-04-10 03:03:49Z

You can have a function signature with multi dimensional arrays, i.e.:

my_fun(char my_array[][COLS]);

You might get some out of this:

A Tutorial on Pointers and Arrays in C, see I.e. 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 <---------+

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