Recursive Py Program Returning j=none when j==1?

Question

Recursive Py Program Returning j=None when j==1? This does not make sense as base case specified j must equal 1 and does not call function again.

import sys
y=10
def decrease(j):
    if j==1:
        print('j =' + str(j) + '(1)')
        print('returning j')
        return j
    else:
        print('j =' + str(j) + '(not 1)')
        print('decreasing j')
        j = j-1
        print('calling decrease j')
        decrease(j)
y=decrease(y)
print('complete')
print(y)

The `else` clause doesn't `return` anything so you will get `None`. Perhaps you meant `return decrease(j)` — AChampion, Sep 14 '17 at 01:53

score 1 · Answer 1 · answered Sep 14 '17 at 01:56

1

You forget to return decrease(j) at the end of the second branch.

Usually when you encounter unexpected None returned from function, check first that all the branches end with a return statement. Without it, the function returns None

answered Sep 14 '17 at 01:56

doraemon

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Recursive Py Program Returning j=none when j==1?

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