STL linked lists

Hi all,
need to ask quick question about stl linked lists.

When i push objects into a list am i correct in saying that on each push
"new" objects are created i.e. the linker fixes addresses for those objects in the container.
While the address for those objects in the container are different the value they hold are "copied" from the object that was pushed in.

So if i have

Code:

---

class myClass
{
    int a ;
  myClass()
  {
      a =10 ;

  }
}
myClass a ;
list <myClass> ;

myClass.push_back(a) ;

---

the member variable a for the new objects in the list would have the value 10 unless changed.
So is my logic flawed?

Sorry for this ridiculous question. I had to use linked lists all of a sudden, so have to read it on my own.
The main question though is the erase function using an iterator.

Code:

---


while(it!=myClass.end())
{
    if(....)
    {

                it=  myClass.erase() ;
          }
}

---

why do we need to initialize it to myClass.erase()
what does erase() return?
would not do if we just increment it?
so

Code:

---

      myClass.erase()
      it++ ;

---

in a hurry, sorry if formatting is bad.
Thanks in advance

When you push an object into a list, the object is copied. That is, a new object is created inside the list and the old value is copied over. Otherwise it wouldn't take a reference.
About erase... consider the list (below) and an iterator to the "5".

Code:

---

[ 1 2 3 4 5 6 7 8 9 ]
          ^

---

If we now "erase" that 5, where does the iterator point? If we're lucky, it might point to 6. But then again, it might not.
Why not? Because there is no pointer "next to" 5 in memory. All the items are linked together with pointers to different places in memory.
So to ensure we have a valid iterator, we must reassign it after calling erase.

Quote:

---

Originally Posted by Elysia

When you push an object into a list, the object is copied. That is, a new object is created inside the list and the old value is copied over. Otherwise it wouldn't take a reference.
About erase... consider the list (below) and an iterator to the "5".

Code:

---

[ 1 2 3 4 5 6 7 8 9 ]
          ^

---

If we now "erase" that 5, where does the iterator point? If we're lucky, it might point to 6. But then again, it might not.
Why not? Because there is no pointer "next to" 5 in memory. All the items are linked together with pointers to different places in memory.
So to ensure we have a valid iterator, we must reassign it after calling erase.

---

Oh right, excellent :P
i assumed stl linked lists were structured sequentially in memory.
your explanation makes sense, ty much.

Quote:

---

Originally Posted by Eman

i assumed stl linked lists were structured sequentially in memory.

---

That's a vector.
But the same principle applies to vectors, as well. If we delete 5, then the iterator will point to 6. So if you then increment it, it will point to 7, which you'll delete instead.
Take care.

Quote:

---

Originally Posted by Elysia

That's a vector.
But the same principle applies to vectors, as well. If we delete 5, then the iterator will point to 6. So if you then increment it, it will point to 7, which you'll delete instead.
Take care.

---

em I haven't done vectors yet. In fact we didn't even cover linked lists just found it was better to use it.

but if the same concept apply to vectors, but i think i can see why it delete 7.

Ah, well, I don't suppose it is that important to understand them now, but later you are definitely going to need to learn what different containers are and do.
Each has their advantages and weaknesses.

yeah true, i assume it is much more difficult to implement linked lists in C :(.