How Does Python's = Operator Differentially Affect Lists Depending on __iadd__ and __add__?

How the = Operator Operates on Lists in Python

The = operator in Python exhibits unexpected behavior when applied to lists. This behavior is attributed to the distinction between the iadd and add special methods.

iadd vs add

• __iadd__': In-place addition; mutates the object being acted upon.
• __add__': Returns a new object; used for the standard operator.

Behavior on Lists

When = is used on a list with an iadd method, the list is modified in place. If iadd is not defined, add is invoked, resulting in a new list.

Example

Consider the following code:

class foo:
    bar = []
    def __init__(self, x):
        self.bar += [x]

class foo2:
    bar = []
    def __init__(self, x):
        self.bar = self.bar + [x]

f = foo(1)
g = foo(2)
print(f.bar)
print(g.bar)

f.bar += [3]
print(f.bar)
print(g.bar)

f.bar = f.bar + [4]
print(f.bar)
print(g.bar)

f = foo2(1)
g = foo2(2)
print(f.bar)
print(g.bar)

Output

[1, 2]
[1, 2]
[1, 2, 3]
[1, 2, 3]
[1, 2, 3, 4]
[1, 2, 3]
[1]
[2]

Explanation

• foo implements __iadd__, so = modifies f.bar, causing it to affect all instances of foo.
• foo2 lacks __iadd__, so = triggers __add__, returning a new list that replaces f.bar.
• foo2 preserves its original bar list, while foo modifies it globally.

Conclusion

The behavior of = on lists depends on whether it calls iadd or add__. In-place modification occurs with __iadd__, while __add creates a new list.

The above is the detailed content of How Does Python's = Operator Differentially Affect Lists Depending on __iadd__ and __add__?. For more information, please follow other related articles on the PHP Chinese website!