No, this copies the value from next into the memory occupied by head. For example if the linked list was: &next: (100, 0xFFFF) 0xFFFF: (101, 0xDEFF) 0XDEFF:

No, this copies the value from next into the memory occupied by head. For example if the linked list was: &next: (100, 0xFFFF) 0xFFFF: (101, 0xDEFF) 0XDEFF: (102, 0) Then calling remove on head would leave us with: &next: (101, 0xDEFF) 0xFFFF: (101, 0xDEFF) 0XDEFF: (102, 0) (It would be on the caller to free the memory address 0XFFFF)