Linked list practice problem help

[violet bluff]

Can someone explain the tracing of this code and how it gets to the list after the function call?

[worldly sonnet]

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[pastel basalt]

presumably you would pass in the head node and the number of elements in the linked list. So, once the function is called, you could exam the linked list from the head node after calling the function to know the remaining content.

[orchid whale]

An iteration through the loop:

current = head
other = head->next

3 → 8 → 12 → 5 → 1 → ...
↑   ↑
c   o

current->next = other->next

3 → 8 → 12 → 5 → 1 → ...
↑ ↑ ↑
c o c.n

free(other)

3 → 12 → 5 → 1 → ...
↑   ↑
c  c.n

current = current->next

3 → 12 → 5 → 1 → ...
↑
c

if (check)
    current = current->next
    other = current->next

3 → 12 → 5 → 1 → ...
         ↑   ↑
         c   o
```The head of the list (in the example `3` rather than `head`) isn't modified, so it remains at the same address before and after the function call

[worldly sonnet]

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