在 Lambda 中捕获值而不是引用 [英] Capturing Value instead of Reference in Lambdas

问题描述

Eli Bendersky 给出的这个例子让我有点惊讶(http://eli.thegreenplace.net/2015/the-scope-of-index-variables-in-pythons-for-loops/)

<预><代码>>>>定义 foo():... lst = []...对于范围内的我(4):... lst.append(lambda: i)... 打印([f() for f in lst])...>>>富()[3, 3, 3, 3]

但是当我考虑它时，它是有道理的——lambda 正在捕获对 i 的引用，而不是 i 的值.

解决这个问题的方法如下:

<预><代码>>>>定义 foo():... lst = []...对于范围内的我(4):... lst.append((lambda a: lambda: a)(i))...打印([f() for f in lst])...>>>富()[0, 1, 2, 3]

看起来这样做的原因是当 i 被提供给外部 lambda 时，外部 lambda 会创建一个范围并取消引用 i，将 a 设置为 i.然后，返回的内部 lambda 持有对 a 的引用.

这是正确的解释吗?

解决方案

默认参数是另一种捕获值的方法:

lst.append(lambda i=i: i)

I was slightly surprised by this example given by Eli Bendersky (http://eli.thegreenplace.net/2015/the-scope-of-index-variables-in-pythons-for-loops/)

>>> def foo():
...     lst = []
...     for i in range(4):
...         lst.append(lambda: i)
...     print([f() for f in lst])
...
>>> foo()
[3, 3, 3, 3]

But when I thought about it, it made some sense — the lambda is capturing a reference to i rather than i's value.

So a way to get around this is the following:

>>> def foo():
...     lst = []
...     for i in range(4):
...         lst.append((lambda a: lambda: a)(i))
...     print([f() for f in lst])
...
>>> foo()
[0, 1, 2, 3]

It appears that the reason that this works is that when i is provided to the outer lambda, the outer lambda creates a scope and dereferences i, setting a to i. Then, the inner lambda, which is returned, holds a reference to a.

Is this a correct explanation?

解决方案

Default param is an another way to catch a value:

lst.append(lambda i=i: i)

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