为什么在使用raw_input时无法捕获KeyboardInterrupt？

Question

这是一个测试用例。

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

当我按下ctrl+c时，我的输出如下所示-

NameError: name 'targ' is not defined

我的意图是输出“已取消”。当我尝试在输入时捕获KeyboardInterrupt时，为什么会发生这种情况？

谢谢！

- 0xhughes

Answer 1

在上面的代码中，当出现异常时，targ未定义。您应该仅在未出现异常时打印。

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

- falsetru

Answer 2

错误发生的原因是，如果出现 KeyboardInterrupt 异常，变量 targ 就不会被初始化。

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

当它没有发生时，

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


Please enter target: abc

>>> targ
'abc'

您可以在try语句中打印targ，以确保没有异常抛出时进行打印。请参考以下演示进行操作。

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

- Sukrit Kalra