Make a local copy of data going into a linked-list

Question

this one is not homework, it just the fact that i've been out of school for more than 20 years and I NEVER had to use linked lists for anything at all. So i'm out of my element here. anyway, I have

struct Data_Struct {
    char *Name;
    char *Task;
    char *Pos;
    struct Data_Struct *Next;
};
typedef struct Data_Struct MyData;

Which is a pretty simple struct for a linked list.

As you can see the data is stored in char* and when I populate the list I copy the pointers there. All is well. The problem begins when those pointers get overwritten and I lose the original data. I tried copying the data into local char* malloc'ed on every "list_add()" but I keep on either crashing or losing data. Yes, I do allocate enough and yes, I do copy the data and check that it's copied correctly. I tried with memcpy, strcpy, etc ,etc.

I'm out of things to try so my question is, how do I make sure that the data I will be adding to the list is local. how do I copy those char* going into the list into local variables.

while an explanation on how to do this and why my code is bad is OK, I prefer code with the explanation. Thank you.

Jess.

Answer 1

You don't want to "copy into local", you want to do as you say you've done; allocate fresh memory using malloc(), store the data there, and save the pointers in the list node.

If you have it, you can use strdup() to combine the malloc()+strcpy() into one call.

It's hard to understand under what circumstances you experience that the data is overwritten, what are you doing to the list node(s) that cause this?

A basic prepend ought to look like this, for your node definition:

MyData * list_prepend(MyData* head, const char *name, const char *task,
                  const char *pos)
{
  MyData *node = malloc(sizeof *node);
  node->Name = strdup(name);
  node->Task = strdup(task);
  node->Pos = strdup(pos);
  node->Next = head;
  return node;
}

Obviously this lacks error handling (malloc() and strdup() can both fail). Note that it's a prepend, and that the new head of the list is returned.