Question

I’m working on a program and can’t understand the compiler errors.

arrays.c:12:21: error: flexible array member not at end of struct
arrays.c: In function 'insert':
arrays.c:22:4: warning: statement with no effect

I know these are very simple things, but I cant fix them. I am not sure if I my insert function is correct. Can anyone please help me? Thanks

#include <stdio.h>

#define AMOUNT 7




    return 0;
}

Answer 1

A struct member declared with empty square brackets [] is called a flexible array member. This is explained in the comp.lang.c FAQ, question 2.6 (scroll all the way to the bottom). A flexible array member may appear only as the last declared member of a struct, which is what your compiler is complaining about.

But moving it to the end of the struct is not the best solution; you probably don’t want to use a flexible array member at all. (You could, but it introduces unnecessary complexity.)

As dasblinkenlight says, you can make the name member a pointer:

struct person {
    char *name;
    int age;
};

name would then point to a string representing the person’s name — which means you’ll need to allocate memory to hold the string.

Or you can make name an array of char, with a fixed maximum size:

#define MAX_NAME 50
struct person {
    char name[MAX_NAME];
    int age;
};

You can then use the strcpy function to copy a value into name — being careful that you don’t try to copy more than MAX_NAME bytes.

In your insert function, these statements:

people[nextcell].names;
people[nextcell].ages;

were presumably intended to initialize the names and ages members (better names: name and age), but in fact they do nothing; they refer to those members, but they don’t do anything with them. For age, you can simply assign the value of the age parameter (which you’re currently ignoring) to the age member:

people[nextcell].age = age;

The way to assign a value to name depends on whether you decide to define it as a pointer or as an array.