如何编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))

我在执行该命令时会收到它：

I receive this when I execute it:

Sum = 3
Add any two digits.

为什么加任意两位数字在总和= 3 ？我猜这是因为回调函数先执行。在 main（）中的所有其他代码执行之后，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

在此代码中

if callback != None:
    callback

回调本身无济于事；它接受参数- def callback（a，b）：

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

您做了 callback（1，2）首先将调用该函数，从而打印 Sum = 3 。

The fact that you did callback(1, 2) first will call that function, thereby printing Sum = 3.

由于回调不返回任何显式值，因此返回为 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()

---

解决方案

您可以尝试不首先调用该函数，而只是传递其句柄。

---

Solution

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

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)

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