词汇闭包如何工作? [英] How do lexical closures work?

问题描述

虽然我正在研究Java代码中的词法闭包问题，但我在Python中遇到了这个问题:

While I was investigating a problem I had with lexical closures in Javascript code, I came along this problem in Python:

flist = []

for i in xrange(3):
    def func(x): return x * i
    flist.append(func)

for f in flist:
    print f(2)

请注意，此示例应避免使用lambda.它打印"4 4 4"，这是令人惊讶的.我希望"0 2 4".

Note that this example mindfully avoids lambda. It prints "4 4 4", which is surprising. I'd expect "0 2 4".

此等效的Perl代码可以正确执行此操作:

This equivalent Perl code does it right:

my @flist = ();

foreach my $i (0 .. 2)
{
    push(@flist, sub {$i * $_[0]});
}

foreach my $f (@flist)
{
    print $f->(2), "\n";
}

打印"0 2 4".

您能解释一下其中的区别吗?

Can you please explain the difference ?

更新:

i具有全局性，问题不是.这将显示相​​同的行为:

The problem is not with i being global. This displays the same behavior:

flist = []

def outer():
    for i in xrange(3):
        def inner(x): return x * i
        flist.append(inner)

outer()
#~ print i   # commented because it causes an error

for f in flist:
    print f(2)

如注释行所示，此时i是未知的.仍然会打印"4 4 4".

As the commented line shows, i is unknown at that point. Still, it prints "4 4 4".

推荐答案

Python实际上按照定义运行. 创建了三个单独的函数，但是每个函数都具有所定义的环境的封闭状态-在这种情况下，是全局环境(如果是外部函数，则为外部环境)循环放在另一个函数中).不过，这确实是问题所在-在这种环境下， i被突变了，并且所有闭包都引用了相同的i .

Python is actually behaving as defined. Three separate functions are created, but they each have the closure of the environment they're defined in - in this case, the global environment (or the outer function's environment if the loop is placed inside another function). This is exactly the problem, though - in this environment, i is mutated, and the closures all refer to the same i.

这是我能想到的最好的解决方案-创建一个函数创建器并调用那个.这将为所创建的每个函数强制不同的环境，每个函数中都具有不同的i .

Here is the best solution I can come up with - create a function creater and invoke that instead. This will force different environments for each of the functions created, with a different i in each one.

flist = []

for i in xrange(3):
    def funcC(j):
        def func(x): return x * j
        return func
    flist.append(funcC(i))

for f in flist:
    print f(2)

这是将副作用和功能编程混合使用时发生的情况.

This is what happens when you mix side effects and functional programming.

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