Python尝试/除了Collat​​z序列

Question

我正在尝试编写一个Collat​​z序列，同时添加一个try和except语句来检测用户是否输入非整数字符串，而我似乎无法弄清楚如何这样做。如果阅读Try/Except in Python: How do you properly ignore Exceptions?，但仍然感到茫然。到目前为止我的代码：

def collatz(y):
    try:
        if y % 2 == 0:
            print(int(y/2))
            return int(y/2)

        else:
            print(int(y*3+1))
            return int(y*3 +1)

    except ValueError:  
        print('Error: Invalid Value, the program should take in integers.')

print('Please enter a number and the Collatz sequence will be printed')
x = int(input())

while x != 1:
    x = collatz(x)

Answer 1

在非整数输入上，程序将在您编写代码时使用代码调用collatz()之前失败，这意味着您的try块不会捕获异常：

def collatz(y):
    try:
        if y % 2 == 0:
            print(int(y/2))
            return int(y/2)

        else:
            print(int(y*3+1))
            return int(y*3 +1)

    except ValueError:  
        print('Error: Invalid Value, the program should take in integers.')

print('Please enter a number and the Collatz sequence will be printed')
x = int(input())  # <-- THIS throws ValueError if input is non integer...

while x != 1:
    x = collatz(x)  # <-- ...but your try/except is in here.

而是将输入转换包装在同一个try / except中。然后在collatz()函数中没有必要：

def collatz(y):
    if y % 2 == 0:
        print(int(y/2))
        return int(y/2)
    else:
        print(int(y*3+1))
        return int(y*3 +1)


print('Please enter a number and the Collatz sequence will be printed')
try:
    x = int(input())
except ValueError:  
    print('Error: Invalid Value, the program should take in integers.')
    exit()

while x != 1:
    x = collatz(x)