如何写一个简单的回调函数? [英] How to write a simple callback function?

问题描述

Python 2.7.10

我编写了以下代码来测试一个简单的回调函数.

I wrote the following code to test a simple callback function.

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))

我在执行时收到:

Sum = 3
Add any two digits.

为什么任意两位数要在Sum = 3之后?我猜是因为回调函数先执行.main()中的所有其他代码执行完后如何执行回调函数?

Why Add any two digits is 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?

推荐答案

在这段代码中

if callback != None:
    callback

callback 本身不做任何事情；它接受参数 - def callback(a, b):

callback on its own doesn't do anything; it accepts parameters - def callback(a, b):

您执行 callback(1, 2) 的事实将首先调用该函数，从而打印 Sum = 3，然后打印 main() 使用 callback 函数的结果被调用，该函数正在打印第二行

The fact that you did callback(1, 2) first will call that function, thereby printing Sum = 3, and then main() gets called with the result of the callback function, which is printing the second line

由于 callback 没有返回显式值，所以返回为 None.

Since callback returns no explicit value, it is returned as None.

因此，您的代码等效于

callback(1, 2)
main()

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解决方案

您可以尝试先不调用该函数，而仅传递其句柄.

---

Solution

You could try not calling the function at first and just passing its handle.

def callback(n):
    print("Sum = {}".format(n))

def main(a, b, _callback = None):
    print("adding {} + {}".format(a, b))
    if _callback:
        _callback(a+b)

main(1, 2, callback)

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