Getting warning message when trying to initialize “employee” struct

Question

Getting the following warning message:

database.c:15:19: warning: assignment makes integer from pointer without a cast [-Wint-conversion]
  ptr->LastName[0] = NULL;
                   ^
database.c:16:26: warning: assignment makes integer from pointer without a cast [-Wint-conversion]
  ptr->FirstMiddleName[0] = NULL;

                      ^

I tried using pointers a few different ways but I don't understand them very well and can't figure out how to avoid this.

#include <stdio.h>

int main() {
        struct employee {
                char LastName[30];
                char FirstMiddleName[35];
                float Salary;
                int YearHired;
        };

        struct employee employees[20];
        struct employee *ptr, person;
        ptr = &person;

        ptr->LastName[0] = NULL;
        ptr->FirstMiddleName[0] = NULL;
        ptr->Salary = -1;
        ptr->YearHired = -1;

        printf("%i", person.YearHired);
        printf("%s", person.LastName[0]);

        for(int i = 0; i < 20; i++) {
                employees[i] = person;
                //printf("%i\n", i);
        }

        printf("%c", employees[3].LastName[0]);
}

I would like to initialize an array of 20 "employees" with the initial values such that numerical values are set to -1, and the strings contain the null character as the zeroth character. Instead I get the above warning, and if I replace the NULL assignment with a letter it says "Segmentation fault (core dumped)".

Answer 1

NULL is a null pointer constant, not the null character. Use:

ptr->LastName[0] = '\0';

Answer 2

Compiler warning message is clear enough to tell what you are doing the wrong. Here

ptr->LastName[0]  = NULL; /* NULL is equivalent of (void*)0 not \0 */

LastName[0] is character not character pointer . You might want

ptr->LastName[0]  = '\0'; /* now here \0 and Lastname[0] both are of char type */

Answer 3

NULL is a pointer constant, and you're trying to assign that to an element of the LastName or FirstMiddleName fields. Instead, assign 0 to the first character of each, which makes them empty strings.

    ptr->LastName[0] = 0;
    ptr->FirstMiddleName[0] = 0;

This also isn't valid:

printf("%s", person.LastName[0]);

Because person.LastName[0] is a single character, not a string. You instead want:

printf("%s", person.LastName);

Answer 4

To fix the warning replace NULL with the null character '\\0'. NULL is a pointer, not what you want.

ptr->LastName[0] = '\0';

To fix the segmentation fault replace person.LastName[0] with person.LastName in your printf. person.LastName[0] is a single character. printf %s expects the address of a null-terminated string.

printf("%s", person.LastName);

That printf statement may still fail if you don't put a null termination somewhere in your string:

 ptr->LastName[0] = 'a';   //may still fail in printf without termination.

vs

ptr->LastName[0] = 'a';   
ptr->LastName[1] = '\0';  

Answer 5

The macro NULL is an invalid pointer not the ASCII NUL character, you should not use it as a character constant. The NUL character is \\0 :

    ptr->LastName[0] = `\0` ;
    ptr->FirstMiddleName[0] = `\0` ;

Or it is also valid to simply use a literal zero:

    ptr->LastName[0] = 0 ;
    ptr->FirstMiddleName[0] = 0 ;