Python，如何传递参数到函数指针参数？ [英] Python, how to pass an argument to a function pointer parameter?

问题描述

我刚刚开始学习Python，发现我可以传递一个函数作为另一个函数的参数。现在如果我调用 foo（bar（））它不会作为一个函数指针，但使用的函数的返回值。调用 foo（bar）将传递函数，但是这种方式我不能传递任何额外的参数。如果我想传递一个函数指针调用 bar（42）？

I only just started learning Python and found out that I can pass a function as the parameter of another function. Now if I call foo(bar()) it will not pass as a function pointer but the return value of the used function. Calling foo(bar) will pass the function, but this way I am not able to pass any additional arguments. What if I want to pass a function pointer that calls bar(42)?

def repeat(function, times):
    for calls in range(times):
        function()

def foo(s):
        print s

repeat(foo("test"), 4)

在这种情况下，函数 foo（test）应该连续调用4次。有没有办法实现这一点，而不必传递测试到重复而不是 foo ？

In this case the function foo("test") is supposed to be called 4 times in a row. Is there a way to accomplish this without having to pass "test" to repeat instead of foo?

推荐答案

您可以使用 lambda ：

repeat(lambda: bar(42))

或 functools.partial ：

from functools import partial
repeat(partial(bar, 42))

：

def repeat(times, f, *args):
    for _ in range(times):
        f(*args)

这种最终样式在标准库和主要的Python工具中很常见。 * args 表示可变数量的参数，因此您可以将此函数用作

This final style is quite common in the standard library and major Python tools. *args denotes a variable number of arguments, so you can use this function as

repeat(4, foo, "test")

或

def inquisition(weapon1, weapon2, weapon3):
    print("Our weapons are {}, {} and {}".format(weapon1, weapon2, weapon3))

repeat(10, inquisition, "surprise", "fear", "ruthless efficiency")

请注意，为方便起见，我将重复次数放在前面。如果要使用 * args 构造，它不能是最后一个参数。

Note that I put the number of repetitions up front for convenience. It can't be the last argument if you want to use the *args construct.

您可以添加关键字参数以及 ** kwargs 。）

(For completeness, you could add keyword arguments as well with **kwargs.)

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