list assign in python

List Assign with different strategies in Python

Here is a code block

def run_infer(self, g_blocks: List[TraceBlock], max_o_num: int = THREADS_NUM) -> Iterable[B_PAIRS]:
    self.model.eval()
    _g_blocks = [*g_blocks]
    _default_inst = ArmInst(opcode='mov', destination='reg1', source='reg0')
    _trace_inst = TraceInst.init_from_inst(_default_inst)
    _default_block = TraceBlock([_trace_inst], line_idx=-1)
    diff = self.batch_size - len(_g_blocks)

Questions:

• What is this kind of copy?
  • A shallow copy of the list g_blocks
  • If a deep copy is wanted, should use copy.deepcopy() instead.
• If it is a shallow copy, why not directly use _g_blocks = g_blocks?
  • Using _g_blocks = g_blocks directly links _g_blocks and g_blocks to the same list object. This means any changes made to the list through _g_blocks will also reflect in g_blocks, and vice versa, because both variables point to the same memory location.
  • In contrast, _g_blocks = [*g_blocks] creates a new list object with the same elements. Here's the difference:
    • Direct Assignment (_g_blocks = g_blocks):
      • Both _g_blocks and g_blocks reference the same list.
      • Changes to the list via either variable will be visible through the other.
      • Example: Adding an item to _g_blocks will also show up when accessing g_blocks.
    • Shallow Copy (_g_blocks = [*g_blocks]):
    • A new list object is created and assigned to _g_blocks.
    • Both _g_blocks and g_blocks contain the same elements, but they are independent lists.
    • Changes to the structure of the list (like adding or removing items) in _g_blocks won't affect g_blocks. However, changes to the elements themselves will reflect in both lists if those elements are mutable objects.