python - 尝试并在 Python 中除外，意外输出

Question

编辑：改变条件..谢谢

我正在尝试学习尝试/异常。我没有得到我应该得到的输出。它通常只做一杯或不做。理想情况下，它应该是 9 或 10。

指示：

创建一个 NoCoffee 类，然后编写一个名为 make_coffee 的函数，该函数执行以下操作：使用 random 模块以 95% 的概率通过打印一条消息并正常返回来创建一壶咖啡。有 5% 的机会引发 NoCoffee 错误。

接下来，编写一个函数 attempt_make_ten_pots，它使用一个 try 块和一个 for 循环，通过调用 make_coffee 来尝试制作十个罐子。函数 attempt_make_ten_pots 必须使用 try 块处理 NoCoffee 异常，并且应该返回一个整数来表示实际制作的罐子的数量。

import random

# First, a custom exception
class NoCoffee(Exception):
    def __init__(self):
        super(NoCoffee, self).__init__()
        self.msg = "There is no coffee!"

def make_coffee():
    try:
        if random.random() <= .95:
            print "A pot of coffee has been made"

    except NoCoffee as e:
        print e.msg

def attempt_make_ten_pots():
    cupsMade = 0

    try:
        for x in range(10):
            make_coffee()
            cupsMade += 1

    except:
        return cupsMade


print attempt_make_ten_pots()

score 4 · Accepted Answer

If you want to allow 95% then the condition should have been
```
if random.random() <= .95:
```

Now, to make your program throw an error and return the number of coffees made, you need to raise an exception when the random value is greater than .95 and it should be excepted in the attempt_make_ten_pots function not in make_coffee itself.

import random

# First, a custom exception
class NoCoffee(Exception):
    def __init__(self):
    super(NoCoffee, self).__init__()
    self.msg = "There is no coffee!"

def make_coffee():
    if random.random() <= .95:
        print "A pot of coffee has been made"
    else:
        raise NoCoffee

def attempt_make_ten_pots():
    cupsMade = 0
    try:
        for x in range(10):
            make_coffee()
            cupsMade += 1
    except NoCoffee as e:
        print e.msg
    return cupsMade

print attempt_make_ten_pots()

python - 尝试并在 Python 中除外，意外输出

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