有没有办法在 Python 中传递块参数而不先将其定义为函数? [英] Is there any way to pass a block argument in Python without defining it as a function first?

问题描述

在 Ruby 中，块参数的工作方式如下:

in Ruby, a block argument works like this:

def foo_bar (&block)
    block.(4)
end

foo_bar do |x|
    puts x
    puts x * 2
end
=begin
4
8
=end

我在 Python 中看到了以下等价物，但我觉得它很不令人满意，因为它需要定义函数然后才将它作为参数传递:

I've seen the following equivalent in Python, but I find it quite unsatisfactory, because it requires defining the function and only then passing it as an argument:

def foo_bar(block):
    block(4)

def callback(x):
    print(x)
    print(x * 2)

foo_bar(callback) 
'''
4
8
'''

在 Python 中是否有任何替代方法，不需要先定义函数?

is there any alternative to it in Python, that doesn't require the function to be defined first?

推荐答案

不，python 不允许这样的语法糖.它没有匿名函数；它提供的最接近的是 lambdas，但它们有许多限制，即它们只能有一个表达式，即一行代码.

Nope, python doesn't allow such syntax sugar.It doesn't have anonymous functions; the closest it offers is lambdas, but those have a number of restrictions, namely, they can only have one expression, i.e, one "line" of code.

使用def 定义函数是创建可重用代码块的pythonic 方式.

Defining functions with def is the pythonic way to create a reusable block of code.

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