Question

在python3中，当我尝试执行以下行时：

$ python3
Python 3.3.5 (default, Mar 18 2014, 02:00:02) 
[GCC 4.2.1 20070831 patched [FreeBSD]] on freebsd9
Type "help", "copyright", "credits" or "license" for more information.
>>> def fail(arg):
...   raise Exception
... 
>>> def fail_type(arg):
...   raise TypeError
... 
>>> identity = lambda x: x
>>> m = [0,1]
>>> print("{:8.2f}{:8.2f}".format(*map(identity, m)))
    0.00    1.00
>>> print("{:8.2f}{:8.2f}".format(*map(fail, m)))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in fail
Exception
>>> print("{:8.2f}{:8.2f}".format(*map(fail_type, m)))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: format() argument after * must be a sequence, not map
>>> list(map(identity, m))
[0, 1]
>>> list(map(fail, m))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in fail
Exception
>>> list(map(fail_type, m))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in fail_type
TypeError
>>>

为什么python会在我看来极为误导TypeError: format() argument after * must be a sequence, not map这个内置函数？如果发现其他raise的明确Exception被捕获，则会更加令人困惑。在*错误处理中是否存在假设TypeError必须归因于参数列表不是序列？

Answer 1

TypeError异常匹配发生在ext_do_call（Python / ceval.c，第4507行）中，调用它来处理操作码CALL_FUNCTION_VAR，CALL_FUNCTION_KW和{ {1}}。它假设PySequence_Tuple提出CALL_FUNCTION_VAR_KW，因为参数不可迭代，正如PyObject_GetIter提出的那样。

在补丁评论＆＃34;中有一个已打开的Python错误。阶段3年：Function calls taking a generator as star argument can mask TypeErrors in the generator

有时，地图是一个序列，有时不是？

1 个答案: