Passing array of character strings to function

Question

I am trying to pass an array of character strings (C style strings) to a function. However, I don't want to place a maximum size on length of each string coming into the function, nor do I want to allocate the arrays dynamically. Here is the code I wrote first:

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

void fun(char *s[])
{
    printf("Entering Fun\n");
    printf("s[1]=%s\n",(char *)s[1]);
}

int main(void)
{
    char myStrings[2][12];

    strcpy(myStrings[0],"7/2/2010");
    strcpy(myStrings[1],"hello");
    fun(myStrings);

    return(0);
}

I got a seg fault when run and the following warning from the compiler: stackov.c: In function ‘main’: stackov.c:17: warning: passing argument 1 of ‘fun’ from incompatible pointer type stackov.c:5: note: expected ‘char **’ but argument is of type ‘char (*)[12]’

However, when I change the main() to the following it works:

int main(void)
{
    char myStrings[2][12];
    char *newStrings[2];

    strcpy(myStrings[0],"7/2/2010");
    strcpy(myStrings[1],"hello");
    newStrings[0]=myStrings[0];
    newStrings[1]=myStrings[1];
    fun(newStrings);

    return(0);
}

Isn't array[2][12] the same thing as an array of character pointers when it is passed to a function?

Answer 1

No, char array[2][12] is a two-dimensional array (array of arrays). char *array[2] is an array of pointers.

char array[2][12] looks like:

7/2/2010\0\x\x\xhello\0\x\x\x\x\x\x

where \0 is NUL and \x is indeterminate.

while

char *array[2] is:

0x CAFEBABEDEADBEEF

(assuming 32-bit)

The first has 24 contiguous characters, the second has two pointers (to the beginnings of strings elsewhere).

Answer 2

Try

void fun(char s[][12]) { ...}

Read also from the c-faq: My compiler complained when I passed a two-dimensional array to a function expecting a pointer to a pointer