Why are pointers expected in structures?

Question

I was just testing structures using C. The code is:

#include <stdio.h>

struct {
    int age;
    char name[]; //just an array 
    char gender; //just a character (M for male and F for female)
}person;

int main()
{
    person.age=10;
    person.name="John";
    person.gender="M";
    printf("Person's age: %d", person.age);
    printf("Person's name: %s", person.name);
    printf("Person's gender: %c", person.gender);
    return 0;
}

However, what the compiler returns is:

warning: initialization of 'char' from 'char *' makes integer from pointer without a cast [-Wint-conversion]|

I have no idea what this means or why it appears.

Answer 1

This code has multiple issues.

First:

person.name="John";

You cannot assign a char* to a char array. The two types are not compatible.

You must either change your struct to:

char* name; //just an char pointer

Or use strcpy() to copy the name to this array.

strcpy(person.name, "John");

Also, you need to specify the size of this array in your struct. Depending on your compiler, funny things can happen with an undefined array size like this.

Second:

person.gender="M";

You also cannot assign a char* to a char.

Either change gender to be of type char*, or change your assignment to use a char.

person.gender = 'M';

Answer 2

"Why are pointers expected in structures?"

They are not. They are also not unexpected.

This construction is however:

struct {
    int age;
    char name[]; //just an array                <- A BOMB
    char gender; //just a character (M for male and F for female)
} person;

You here have an instance (person) of an anonymous struct with a flexible array. Old-school structs with this setup had their last member declared as type identifier[1];.

[1] was historically used because a zero element array has never been valid in standard C (although accepted as a language extension by some compilers).

The C99 standard made char name[] legal by allowing indexing such an array if it's made the last element in a struct. The name[] member in your struct is not.

The purpose of the flexible array was to let writing to such an array be legal to let the programmer (or a called function) pre-allocate memory and have functions return an unknown amount of elements by writing their result in that memory.

If your struct is accepted by a compiler, a funcion using it will overwrite gender when writing more than zero chars to name as an answer.

This would be a definition (and typedef) with the potential of being used properly:

typedef struct person_ {
    int age;
    char gender;
    struct person_ *next;  // points somewhere after the '\0' in name[] or NULL
    char name[];           // Pre C99 "char name[1];" was often seen
} person;                  // a typedef, not an instance here

Answer 3

Why are pointers expected in structures?

Pointers are not expected, nor are they dis-allowed.

"warning: initialization of 'char' from 'char *' makes integer from pointer without a cast [-Wint-conversion]|"

person.gender="M"; errantly attempts to take a pointer (the address of the stirng literal "M") and assign it to a char (.gender). Pointers do not fit in a char.

---

OP's code is invalid. OP is relying on a code extension.

A member like char name[]; must be the last member in a struct. There it is a flexible array member.

struct {
    int age;
    // char name[]; // move to the end
    char gender; 
    char name[];
}person;

There are no pointers in OP's struct. With repair, member .name[] is an array.

---

Example usage

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

// Define a type
typedef struct {
  int age;
  char gender;
  char name[];  // last member
} person;

// Allocate and populate data
person* person_initialize(int age, const char *name, char gender) {
  size_t len = strlen(name);
  // Size needed is for the `person` and the name as a _string_.
  person *p = malloc(sizeof *p + len + 1);
  // If successful ....
  if (p) {
    p->age = age;
    p->gender = gender;
    strcpy(p->name, name);
  }
  return p;
}

int main(void) {
  person *p = person_initialize(10, "John", 'M');  // Note 'M', not "M".
  if (p) {
    printf("Person's age: %d", p->age);
    printf("Person's name: %s", p->name);
    printf("Person's gender: %c", p->gender);
    free(p);
  }
  return 0;
}