为什么+ =在列表上表现异常? [英] Why does += behave unexpectedly on lists?

问题描述

python中的+=运算符似乎在列表上意外运行.谁能告诉我这是怎么回事?

The += operator in python seems to be operating unexpectedly on lists. Can anyone tell me what is going on here?

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 

输出

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

foo += bar似乎影响该类的每个实例，而foo = foo + bar似乎以我希望事情表现的方式表现.

foo += bar seems to affect every instance of the class, whereas foo = foo + bar seems to behave in the way I would expect things to behave.

+=运算符称为化合物赋值运算符".

The += operator is called a "compound assignment operator".

推荐答案

通常的答案是+=尝试调用__iadd__特殊方法，如果该方法不可用，它将尝试使用__add__ .因此，问题在于这些特殊方法之间的差异.

The general answer is that += tries to call the __iadd__ special method, and if that isn't available it tries to use __add__ instead. So the issue is with the difference between these special methods.

__iadd__特殊方法用于就地加法，即，它会改变作用于其上的对象. __add__特殊方法返回一个新对象，也用于标准+运算符.

The __iadd__ special method is for an in-place addition, that is it mutates the object that it acts on. The __add__ special method returns a new object and is also used for the standard + operator.

因此，当在定义了__iadd__的对象上使用+=运算符时，该对象将被修改.否则，它将尝试使用普通的__add__并返回一个新对象.

So when the += operator is used on an object which has an __iadd__ defined the object is modified in place. Otherwise it will instead try to use the plain __add__ and return a new object.

这就是为什么对于诸如列表+=之类的可变类型会更改对象的值，而对于诸如元组，字符串和整数之类的不可变类型则会返回一个新对象的原因(a += b等同于a = a + b).

That is why for mutable types like lists += changes the object's value, whereas for immutable types like tuples, strings and integers a new object is returned instead (a += b becomes equivalent to a = a + b).

对于同时支持__iadd__和__add__的类型，您必须小心使用哪种类型. a += b将调用__iadd__并对a进行突变，而a = a + b将创建一个新对象并将其分配给a.它们是不同的操作！

For types that support both __iadd__ and __add__ you therefore have to be careful which one you use. a += b will call __iadd__ and mutate a, whereas a = a + b will create a new object and assign it to a. They are not the same operation!

>>> a1 = a2 = [1, 2]
>>> b1 = b2 = [1, 2]
>>> a1 += [3]          # Uses __iadd__, modifies a1 in-place
>>> b1 = b1 + [3]      # Uses __add__, creates new list, assigns it to b1
>>> a2
[1, 2, 3]              # a1 and a2 are still the same list
>>> b2
[1, 2]                 # whereas only b1 was changed

对于不可变类型(没有__iadd__的类型)，a += b和a = a + b是等效的.这就是让您在不可变类型上使用+=的原因，这似乎是一个奇怪的设计决定，除非您考虑到否则无法在数字等不可变类型上使用+=！

For immutable types (where you don't have an __iadd__) a += b and a = a + b are equivalent. This is what lets you use += on immutable types, which might seem a strange design decision until you consider that otherwise you couldn't use += on immutable types like numbers!

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