how to call a function inside function itself in python

Question

do we have any way to assign a function inside function itself in python? I tried to call out a nth Fibonacci as below:

memo={0:0,1:1}
def fib(n):
    if n<=1:
        return n
    if memo[n] is None:
        fib(n)=fib(n-1)+fib(n-2)
    return memo[n]

print(fib(5))

You're trying to implement function which return n-th element of fibonacci sequence and remember all calculated items for next searches? — Olvin Roght
– Olvin Roght, Commented Jul 2, 2021 at 11:08
You call a function inside itself the same way you would call it anywhere else, but fib(n)=... wouldn't be valid anywhere, inside or outside fib. Assigning to fib(n) is not what needs to happen here. — user2357112
– user2357112, Commented Jul 2, 2021 at 11:08
You're trying to implement function which return n-th element of fibonacci sequence and remember all calculated items for next searches? --> yes , correct — Hoang Minh Nguyen
– Hoang Minh Nguyen, Commented Jul 2, 2021 at 11:13

Yoav Glazner · Accepted Answer · 2021-07-02 11:08:25Z

2

two fixes

memo={0:0,1:1}
def fib(n):
    if n not in memo: # test
        memo[n] = fib(n-1)+fib(n-2)   # <-- set memo
    return memo[n]

print(fib(5))

answered Jul 2, 2021 at 11:08

Yoav Glazner

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Comments

ShlomiF · Accepted Answer · 2021-07-02 11:15:13Z

2

Of course you can. It's called recursion. However, you have some errors in your code...
Try -

memo={0:0,1:1}
def fib(n):
    if n not in memo:
        memo[n] = fib(n - 1) + fib(n - 2)
    return memo[n]
        
print(fib(5))

answered Jul 2, 2021 at 11:11

ShlomiF

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Actually, simple list will be enough for data container (demo)