我希望将一个函数的 kwargs 转发到另一个函数,但添加一个进一步的参数。然而,以下内容不起作用...
def A(**kwargs):
B(type="moose", kwargs=kwargs)
def B(**kwargs):
print(kwargs) # I'd like to see {"call": True, "seven": 8, "type": "moose" }
A(call=True, seven=8)
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只需B()致电**kwargs:
def A(**kwargs):
B(type="moose", **kwargs)
def B(**kwargs):
print(kwargs)
A(call=True, seven=8)
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印刷:
{'type': 'moose', 'call': True, 'seven': 8}
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