Python 如何写一个简单的回调函数？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/40843039/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to write a simple callback function?
提问by srgbnd
Python 2.7.10
蟒蛇 2.7.10
I wrote the following code to test a simple callback function.
我编写了以下代码来测试一个简单的回调函数。
def callback(a, b):
print('Sum = {0}'.format(a+b))
def main(callback=None):
print('Add any two digits.')
if callback != None:
callback
main(callback(1, 2))
I receive this when I execute it:
我在执行它时收到这个：
Sum = 3
Add any two digits.
Why Add any two digitsis after Sum = 3? I guess it is because the callback function executes first. How to execute the callback function after all other code in main()executed?
为什么Add any two digits是后Sum = 3？我猜是因为回调函数先执行。执行完所有其他代码后如何执行回调函数main()？
回答by OneCricketeer
In this code
在这段代码中
if callback != None:
callback
callbackon its own doesn't do anything; it accepts parameters - def callback(a, b):
callback它自己不做任何事情；它接受参数 -def callback(a, b):
The fact that you did callback(1, 2)first will call that function, thereby printing Sum = 3.
您callback(1, 2)首先执行的事实将调用该函数，从而打印Sum = 3.
Since callbackreturns no explicit value, it is returned as None.
由于不callback返回显式值，因此返回为None。
Thus, your code is equivalent to
因此，您的代码等效于
callback(1, 2)
main()
Solution
解决方案
You could try not calling the function at first and just passing its handle.
您可以尝试先不调用该函数，而只是传递其句柄。
def callback(sum):
print("Sum = {}".format(sum))
def main(a, b, _callback = None):
print("adding {} + {}".format(a, b))
if _callback:
_callback(a+b)
main(1, 2, callback)
回答by erip
As mentioned in the comments, your callback is called whenever it's suffixed with open and close parens; thus it's called when you pass it.
正如评论中提到的，只要以 open 和 close 括号为后缀，就会调用您的回调；因此当你通过它时它会被调用。
You might want to use a lambda and pass in the values.
您可能想要使用 lambda 并传入值。
#!/usr/bin/env python3
def main(callback=None, x=None, y=None):
print('Add any two digits.')
if callback != None and x != None and y != None:
print("Result of callback is {0}".format(callback(x,y)))
else:
print("Missing values...")
if __name__ == "__main__":
main(lambda x, y: x+y, 1, 2)
回答by Eric Duminil
Here's what you wanted to do :
这是你想要做的：
def callback(a, b):
print('Sum = {0}'.format(a+b))
def main(a,b,f=None):
print('Add any two digits.')
if f != None:
f(a,b)
main(1, 2, callback)
回答by ettanany
Your code is executed as follows:
您的代码执行如下：
main(callback(1, 2))
callbackfunction is called with (1, 2)and it returns None(Without return statement, your function prints Sum = 3and returns None)
callback函数被调用(1, 2)并返回None（没有返回语句，你的函数打印Sum = 3并返回None）
mainfunction is called with Noneas argument (So callback != Nonewill always be False)
main函数被None作为参数调用（所以callback != None将永远是False）
回答by farsil
The problem is that you're evaluating the callback before you pass it as a callable. One flexible way to solve the problem would be this:
问题是您在将回调作为可调用对象传递之前对其进行评估。解决问题的一种灵活方法是：
def callback1(a, b):
print('Sum = {0}'.format(a+b))
def callback2(a):
print('Square = {0}'.format(a**2))
def callback3():
print('Hello, world!')
def main(callback=None, cargs=()):
print('Calling callback.')
if callback != None:
callback(*cargs)
main(callback1, cargs=(1, 2))
main(callback2, cargs=(2,))
main(callback3)
Optionally you may want to include a way to support keyword arguments.
（可选）您可能希望包含一种支持关键字参数的方法。
