How to prevent conflicting parameters?-CodePudding

I got a function that takes two lists. For example, list1 = [1, 2] and list2 = [3, 4], the function merges them together into list1 = [1, 2, 3 ,4], leaving list2 empty. But now, I'm trying to make the function do nothing when it takes in the same list. For example, both *a and *b are passed as list1. How I do that?

typedef struct node {
    ElemType val;
    struct node *next;
} NODE;

struct list_struct {
    NODE *front;
    NODE *back;
};

void merge(LIST *a, LIST *b) {
    if (a->front != NULL || b->front != NULL) {
        a->back->next = b->front;
        a->back = b->back;
        b->front = NULL;
    }
}

CodePudding user response：

Just check the parameters and act accordingly, for example like this:

void merge(LIST *a, LIST *b) {
    if (a == b) {
      // do nothing, or whatever...
    }
    else
    {
        if (a->front != NULL || b->front != NULL) {
            a->back->next = b->front;
            a->back = b->back;
            b->front = NULL;
        }
    }
}

But beware: with your code, a will contain all elements from b after a merge, but if you now remove elements from b, these now inexistent elements will still be referenced by a and you have a problem.

IOW: your merge function works, as long as you never remove elements from lists you have merged from.

Are you sure this is what you want?

BTW: you forgot the ytepdef for LIST:

typedef struct list_struct {
    NODE *front;
    NODE *back;
} LIST;