Different structs in the same node Linked Lists C

Question

I am having a bit of difficulty with my assignment and was wondering if someone can point me in the right path.

I wish to create a linked list which supports the use of different nodes, as I need different data sets to be used.

At the moment I am using three structs:

struct equipment_data {
    char EquipID[4+1]; /* 4 Max */
    char EquipName[31+1]; /* 31 Max */
    unsigned TotalNumber;
};

struct member_data {
    unsigned MemberID;
    char LastName[31+1]; /* 31 Max */
    char FirstName[31+1]; /* 31 Max */
};

struct loan_data {
    unsigned MemberID;
    char EquipID[4+1]; /* 4 Max */
    unsigned Number;
};

I need to somehow use this within the same node.

struct ets {
    struct node *equipment_data;
    struct node *member_data;
    struct node *loan_data;
    unsigned equip_count;
    unsigned member_count;
    unsigned loan_count;
};

struct node {
    void *data;
    struct node *next;
};

Looks like I need to create an ADT Linked List. Could you please help? Thanks!

Answer 1

I would make structures for the types that you need to support, and then a union with a type indicator in the linked list node:

typedef struct { int a, b, c;    } Type1;
typedef struct { char buf[80];   } Type2;
typedef struct { int a; float b; } Type3;

typedef union {
    Type1 t1,
    Type2 t2,
    Type3 t3;
} Anytype;

typedef struct node {
    int thistype;   // 1 for type1, 2 for type2 etc.
    Anytype data;
    struct node *next;
} Listnode;

Just make sure that you set the thistype right in each Listnode.

Listnode *node = malloc(sizeof(Listnode));
node->thistype = 1;  // example: Type1, the 3 ints
node->t1.a = 1;
node->t1.b = 2;
node->t1.c = 3;
node->next = someothernode;

The you can use a switch to access the data:

Listnode *node;
switch (node->thistype) {
    case 1: 
        // do stuff with node->t1.a, node->t1.b, node->t1.c
        break
    case 2:
        // do stuff with node->t2.buf
        break;
    case 3:
        // do stuff with node->t3.a, node.t3.b
        break
}