Python中函数的异常处理

Suppose I have a function definiton:

假设我有一个函数定义:

def test():
    print 'hi'

I get a TypeError whenever I gives an argument.

每当我给出一个参数时,我得到一个TypeError。

Now, I want to put the def statement in try. How do I do this?

现在,我想把def语句放在try中。我该怎么做呢?

6 个解决方案

#1

try: 
    test()
except TypeError:
    print "error"

#2

In [1]: def test():
     ...:     print 'hi'
     ...:

In [2]: try:
     ...:     test(1)
     ...: except:
     ...:     print 'exception'
     ...:
exception

Here is the relevant section in the tutorial

这是教程中的相关部分

By the way. to fix this error, you should not wrap the function call in a try-except. Instead call it with the right number of arguments!

顺便说说。要修复此错误,您不应该在try-except中包装函数调用。而是使用正确数量的参数调用它!

#3

You said

Now, I want to put the def statement in try. How to do this.

现在,我想把def语句放在try中。这该怎么做。

The def statement is correct, it is not raising any exceptions. So putting it in a try won't do anything.

def语句是正确的,它不会引发任何异常。所以试一试不会做任何事情。

What raises the exception is the actual call to the function. So that should be put in the try instead:

引发异常的是对函数的实际调用。所以应该把它放在try中:

try: 
    test()
except TypeError:
    print "error"

#4

If you want to throw the error at call-time, which it sounds like you might want, you could try this aproach:

如果你想在通话时抛出错误,听起来你可能想要,你可以试试这个方法:

def test(*args):
    if args:
        raise
    print 'hi'

This will shift the error from the calling location to the function. It accepts any number of parameters via the *args list. Not that I know why you'd want to do that.

这会将错误从调用位置转移到函数。它通过* args列表接受任意数量的参数。不是我知道你为什么要那样做。

#5

A better way to handle a variable number of arguments in Python is as follows:

在Python中处理可变数量的参数的更好方法如下:

def foo(*args, **kwargs):
    # args will hold the positional arguments
    print args

    # kwargs will hold the named arguments
    print kwargs


# Now, all of these work
foo(1)
foo(1,2)
foo(1,2,third=3)

#6

This is valid:

这是有效的:

try:
  def test():
    print 'hi'
except:
  print 'error'


test()

#1

try: 
    test()
except TypeError:
    print "error"

#2

In [1]: def test():
     ...:     print 'hi'
     ...:

In [2]: try:
     ...:     test(1)
     ...: except:
     ...:     print 'exception'
     ...:
exception