Check first character of a pointer to an array

A pointer is a variable that holds the address to something else as its value. To access the value at the address held by the pointer (e.g. the something else), you dereference the pointer with the unary '*'.

Since you have an array of pointers, you will need to dereference the first element of the array, e.g. mystrings[0]. With C - operator precedence, the [..] has greater precedence than '*', so you can simply look at the first character of the string pointed to by mystrings[0] by dereferencing the pointer, e.g. *mystrings[0]

A short example will help:

#include <stdio.h>

#define MAX 40

int main (void) {

    char *string = "example",       /* string literal */
        *arrptrs[MAX] = { NULL };   /* array of pointers */

    arrptrs[0] = string;    /* set 1st pointer to point to string */

    if (*arrptrs[0] == 'e')         /* test 1st char is 'e' */
        printf ("found : %s\n", arrptrs[0]);

    return 0;
}

Since you are looking at the first element of your array of pointers, you can also print the string "example" by dereferecing the array, e.g.

        printf ("found : %s\n", *arrptrs);

Example Use/Output

$ ./bin/arrptrs
found : example

(the output is the same regardless whether you access the first element with arrptrs[0] or *arrptrs, they are equivalent -- do you understand why?)

note: always pay attention to C-Operator Precedence. Let me know if you have further questions.