Python 3.x变量赋值与绑定的区别及绑定语法咨询

Great question! Let's break this down clearly—Python's variable model works a bit differently than some other languages, so it's totally normal to have this confusion.

为什么你的代码里var2 = var1没让两个变量“绑定”？ #

First, let's get one key concept straight: in Python, variables are just labels that point to objects in memory, not boxes that hold values.

Let's walk through your code line by line:

var1 = 1  # Creates an integer object `1`, then sticks the label `var1` on it
var2 = var1  # Now `var2` is another label stuck to the EXACT SAME `1` object
var2 = var2 + 3  # Here's the critical part: integers are immutable. `var2 +3` creates a NEW integer object `4`, then moves the `var2` label to this new object. `var1` still stays attached to the original `1`.

That's why your print statements output 1 and 4—the two variables are now pointing to entirely separate objects.

怎么让两个变量绑定到同一内存空间（并保持关联）？ #

This only works with mutable objects (like lists, dictionaries, sets, or custom classes you define). Mutable objects can be modified in-place without creating a brand new object, so all labels pointing to it will see the changes.

Here's an example:

var1 = [1]  # Creates a mutable list object `[1]`, labels it `var1`
var2 = var1  # `var2` also gets stuck to the same list object
var2.append(3)  # Modifies the list IN-PLACE—no new object is made
print(var1)  # Outputs [1, 3] (since it's the same list)
print(var2)  # Outputs [1, 3]

Now both variables are "bound" to the same memory space because they reference the same mutable object. Any in-place changes to that object will show up in both variables.

For immutable objects (int, str, tuple, etc.), this kind of binding isn't possible. Any operation that looks like a "modification" actually creates a new object, so reassigning the variable just moves its label to the new object, leaving the original variable pointing to the old one.

如何避免意外绑定两个变量？ #

If you're working with mutable objects and don't want changes to one variable affecting the other, you need to create a copy of the object instead of just reassigning the label.

Some common ways to make copies:

• For lists: var2 = var1.copy() or var2 = var1[:]
• For dictionaries: var2 = var1.copy()
• For nested or complex objects: use the copy module's copy.deepcopy() function

This way, var2 points to a brand new, separate object in memory—modifying it won't have any impact on var1.

内容的提问来源于stack exchange，提问作者James Bell