为什么我的递归函数不接受第二个参数？

Question

"""

def permNums(inp,sec):
    newInp = []
    for i in inp:
        for j in sec:
            if j not in i: newInp.append( i+j )      #I put the print after this line
    return newInp


b = permNums(permNums(permNums(inp='word',sec='word')))

print b
"""


def permNums(inp):
    newInp = []
    for i in inp:
        for j in 'word':
            if j not in i: newInp.append( i+j )
    return newInp


b = permNums(permNums(permNums(inp='word')))

print b

The way I see it the code that has been commented out and the one that has not, should be equivalent. The commented out code throws me an error saying i've only given 1 argument instead of 2. Where am I going wrong?

I tried putting a print function right after

 if j not in i: newInp.append( i+j )

,in the commented out code and noticed that the innermost function does get called but then gives an error at the 2nd recursion possibly because it doesnt take 'sec' as an argument.. Can someone clear this out for me please.

Answer 1

The problem comes from the second call. Let me illustrate it. When the result of the inner call is returned, Python then has to do this one:

permNums(result_of_previous_call)  # sec won't implicitly carry through.

which is obviously only one argument. Therefore to fix the commented-out one, you need to supply the extra sec argument for each call:

b = permNums(permNums(permNums(inp='word', sec='word'), sec='word'), sec='word')