*args represents any multiple unnamed parameters, which is a tuple; **kwargs represents keyword parameters, which is a dict.
def fun(*args, **kwargs):
print 'args = ', args
print 'kwargs = ', kwargs
print '###'
if __name__ == '__main__':
foo(1,2,3,4)
foo(a=1,b=2,c=3)
foo(1,2,3,4, a=1,b=2,c=3)
foo('a', 1, None, a=1, b='2', c=3)The output result is as follows:
args = (1, 2, 3, 4)
kwargs = {}
###
args = ()
kwargs = {'a': 1, 'c': 3, 'b': 2}
###
args = (1, 2, 3, 4)
kwargs = {'a': 1, 'c': 3, 'b': 2}
###
args = ('a', 1, None)
kwargs = {'a': 1, 'c': 3, 'b': '2'}
###As you can see, these two are variable parameters in python.
Note: When using *args and **kwargs at the same time, the *args parameter must be listed before **kwargs, like foo(a=1, b='2', c=3, a', 1, None, ) If called in this way, the syntax error "SyntaxError: non-keyword arg after keyword arg" will be prompted.
def fun2(param1, *args, **kwargs):
print 'param1 = ', param1
print 'args = ', args
print 'kwargs = ', kwargs
print '###'
fun2(1, 2, 3, 4, a=1,b=2,c=3)Output result:
param1 = 1
args = (2,3,4)
kwargs = {'a': 1, 'c': 3, 'b': 2}
###1 is assigned to param1, the remaining 2, 3, and 4 are assigned to *args, and the others are assigned to **kwargs
There is also a very beautiful usage , is to create a dictionary:
def kw_dict(**kwargs):
return kwargs
print kw_dict(a=1,b=2,c=3)Result:
{'a':1, 'b':2, 'c':3}In fact, there is a dict class in python. You can create a dictionary using dict(a=1,b=2,c=3).
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