Question

这是一个测试用例。

try:
    targ = raw_input("Please enter target: ")
except KeyboardInterrupt:
    print "Cancelled"
print targ

当我按ctrl + c -

时，我的输出如下

NameError: name 'targ' is not defined

我的意图是输出为“已取消”。当我在raw_input期间尝试捕获KeyboardInterrupt时，有什么想法会发生这种情况吗？

谢谢！

Answer 1

在上面的代码中，当引发异常时，没有定义targ。只有在未引发异常时才应打印。

try:
    targ = raw_input("Please enter target: ")
    print targ
except KeyboardInterrupt:
    print "Cancelled"

Answer 2

发生错误是因为如果引发KeyboardInterrupt，变量targ永远不会被初始化。

try:
    targ = raw_input("Please enter target: ")
except KeyboardInterrupt:
    print "Cancelled"

Please enter target: 
Cancelled

>>> targ

Traceback (most recent call last):
  File "<pyshell#19>", line 1, in <module>
    targ
NameError: name 'targ' is not defined

何时，它不会发生，

try:
    targ = raw_input("Please enter target: ")
except KeyboardInterrupt:
    print "Cancelled"


Please enter target: abc

>>> targ
'abc'

如果未引发异常，您可以将代码更改为打印targ，方法是在try语句中打印，请参阅以下演示。

try:
    targ = raw_input("Please enter target: ")
    print targ
except KeyboardInterrupt:
    print "Cancelled"


Please enter target: abc
abc

try:
    targ = raw_input("Please enter target: ")
    print targ
except KeyboardInterrupt:
    print "Cancelled"


Please enter target: 
Cancelled

为什么我不能在raw_input期间捕获KeyboardInterrupt？

2 个答案: