Question

def iglob(pathname, *, recursive=False):
"""Return an iterator which yields the paths matching a pathname pattern.

The pattern may contain simple shell-style wildcards a la
fnmatch. However, unlike fnmatch, filenames starting with a
dot are special cases that are not matched by '*' and '?'
patterns.

If recursive is true, the pattern '**' will match any files and
zero or more directories and subdirectories.
"""
it = _iglob(pathname, recursive)
if recursive and _isrecursive(pathname):
    s = next(it)  # skip empty string
    assert not s
return it

当我浏览python3.5.1中的glob代码时，这里定义的函数，为什么函数参数列表中有*。如果我将三个参数传递给此函数，则引发TypeError，*的效果是什么？先谢谢。

Answer 1

在python 3中，您可以指定*几乎不会强制参数作为关键字参数：

>>>def fn(arg1, arg2, *, kwarg1, kwarg2):
...     print(arg1, arg2, kwarg1, kwarg2)
... 
>>> fn(1, 2, 3, 4)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: fn() takes 2 positional arguments but 4 were given
>>> fn(1, 2, kwarg1=3, kwarg2=4)
1 2 3 4
>>>

在此示例中，它强制kwarg1和kwarg2仅作为关键字参数发送。

python函数定义中*的用法是什么？

1 个答案: