CODEWITHSUNDEEP
cpointersstructlinked-list
iugubhk
iugubhk

Reputation: 31

Linked Lists Struct in C. Altering shared element in only one list

I have these structs in C:

typedef struct team{
    char *name;
    int wins;
} *pTeam;

typedef struct node_team{
    pTeam team;
    struct node_team *next;
} *link_team;

So I have various of these teams, and I'm putting them on a linked list, each element of the linked list is a link_team, so having 2 link_team's, A and B, if I want to put B after A, I do A->next=B. However, I have 2 lists, and both lists share some of the same elements, and so if I have A in both of them and want B after A in one list, I don't want it in the other list. However, since we're talking about pointers, and both lists have the pointer pointing to the struct A, by doing A->next=B in one list, B is linked to A automatically in the other list as well. How can I prevent this?

Upvotes: 1

Views: 43

Answers (1)

Daniel Walker
Daniel Walker

Reputation: 6740

The next pointers are what make the linked list a linked list. So, you couldn't mix the lists the way you're doing it. However, the only data a node contains besides the pointer to the next object is the pTeam object. That can easily be shared across different lists (though you need to be careful).

... // Suppose you already have five pTeam pointers: p1, ..., p5

link_team *head1, *head2, *node1, *node2, *node3, *node4;

head1=malloc(sizeof(*link_team));
head2=malloc(sizeof(*link_team)); // I should be checking these return values but oh well.

head1->team=p1;
head2->team=p2;

node1=malloc(sizeof(*link_team));
node1->team=p3;
head1->next=node1;

node2=malloc(sizeof(*link_team));
node2->team=p3;
head2->next=node2;

node3=malloc(sizeof(*link_team));
node3->team=p4;
node3->next=NULL;
node1->next=node3;

node4=malloc(sizeof(*link_team));
node4->team=p5;
node4->next=NULL;
node2->next=node4;

Upvotes: 0

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