Clean list function

okay..so basically what i have is two linked lists(link1, link2) and link2 contains all of the values stored in link1. What i need to do is use string compare to clean the big list free of everything that is in link1. I believe i am on the right track but i am stuck as to how i should make this work. here is the code for the clean list function:

Code:

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struct info_node * clean_list(struct info_node * link1, struct info_node * link2)
{
                struct info_node * top = NULL ;
                struct info_node * pre_pntr = NULL;
                struct info_node* current_pntr = NULL;

        While (current_pntr != NULL)
        {
        if (strcmp(link1->fname, link2->fname) && (strcmp(link1->lname, link2->lname) != 0)
        {
                link1 = link1->next ;
                link2 = link2->next ;
        }
        else
        {



        }

        }


return(0) ;
} //end of func clean_list()

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You might want:

if ((strcmp(link1->fname, link2->fname)!=0) && (strcmp(link1->lname, link2->lname != 0)))

instead of

if (strcmp(link1->fname, link2->fname) && (strcmp(link1->lname, link2->lname) != 0)

because even if you correct the missing/extra parantheses, you will still only be testing the second strcmp.

Code:

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if (strcmp(link1->fname, link2->fname) && !strcmp(link1->lname, link2->lname))

---

Looks so much cleaner.

Quote:

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Originally Posted by master5001

Code:

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if (strcmp(link1->fname, link2->fname) && !strcmp(link1->lname, link2->lname))

---

Looks so much cleaner.

---

But that will be true when link1->fname is NOT equal to link2->fname and at the same time link1->lname IS equal to link2->lname.

You probably want to have a condition that is true when both are equal?

--
Mats

Until these OP's start throwing down cash or something, I will not be guaranteed to follow all of the logic embedded in sample code. This post included.