Why not getting return on line #16?

[deanhystad]

Yes. That is the purpose of my code. To demonstrate why your convert function returns None when called with a non-zero value for n. Is the reason clear now?

Your convert function, as you wrote it, has two potential outcomes. If the value of n == 0 the function returns new. If n is any value other than 0, commit does not return a value, so the value of the function call is None. I tried to highlight this by writing convert0() to return new, and convert6(), convert2() and convert1() not returning a value. This is exactly how your code works. convert(0) returns new. convert(6), convert(2) and convert(1) do not return a value.

When your main code calls convert(4562) it does not return a value. 4562 is not 0, so the code executes the branch that executes the recursion, not the branch that returns a value. It operates just like convert(6) or convert6().

[jahuja73]

Yes. That is the purpose of my code. To demonstrate why your convert function returns None when called with a non-zero value for n. Is the reason clear now?

Your convert function, as you wrote it, has two potential outcomes. If the value of n == 0 the function returns new. If n is any value other than 0, commit does not return a value, so the value of the function call is None. I tried to highlight this by writing convert0() to return new, and convert6(), convert2() and convert1() not returning a value. This is exactly how your code works. convert(0) returns new. convert(6), convert(2) and convert(1) do not return a value.

When your main code calls convert(4562) it does not return a value. 4562 is not 0, so the code executes the branch that executes the recursion, not the branch that returns a value. It operates just like convert(6) or convert6().

I seem very slow learner, as unable to grasp still.
For the final attempt have modified my code to see that output is 'None' many times.

#https://ideone.com/XKyhyR

n = 4562; 
rev = 0
new= []
m = 0
  
def convert(n,m):
    print "round #:", m,"n :", n
    a = ((n+1) % 3)-1
    if n:
        new.append(a); print 'new :', new
        print 'call convert(', (n+1)//3,',',m+1,'):', convert((n+1)//3,m+1)
        print 'after convert(' , n, ',' ,m,  ')','\n'
    else:
        print 'n==',n, 'new: ::: ', new
        return new
  
print  convert(n,m), "<<---- None returned ? "
  
sum = 0
  
def value(l):
    sum = 0
    for i in range(0, len(new),1):
        sum += new[i]*(3**i)
        print "sum :" , sum
    return sum
  
print value(new), 

Output:

round #: 0 n : 4562
new : [-1]
call convert( 1521 , 1 ): round #: 1 n : 1521
new : [-1, 0]
call convert( 507 , 2 ): round #: 2 n : 507
new : [-1, 0, 0]
call convert( 169 , 3 ): round #: 3 n : 169
new : [-1, 0, 0, 1]
call convert( 56 , 4 ): round #: 4 n : 56
new : [-1, 0, 0, 1, -1]
call convert( 19 , 5 ): round #: 5 n : 19
new : [-1, 0, 0, 1, -1, 1]
call convert( 6 , 6 ): round #: 6 n : 6
new : [-1, 0, 0, 1, -1, 1, 0]
call convert( 2 , 7 ): round #: 7 n : 2
new : [-1, 0, 0, 1, -1, 1, 0, -1]
call convert( 1 , 8 ): round #: 8 n : 1
new : [-1, 0, 0, 1, -1, 1, 0, -1, 1]
call convert( 0 , 9 ): round #: 9 n : 0
n== 0 new: ::: [-1, 0, 0, 1, -1, 1, 0, -1, 1]
[-1, 0, 0, 1, -1, 1, 0, -1, 1]
after convert( 1 , 8 )

None
after convert( 2 , 7 )

None
after convert( 6 , 6 )

None
after convert( 19 , 5 )

None
after convert( 56 , 4 )

None
after convert( 169 , 3 )

None
after convert( 507 , 2 )

None
after convert( 1521 , 1 )

None
after convert( 4562 , 0 )

None <<---- None returned ?
sum : -1
sum : -1
sum : -1
sum : 26
sum : -55
sum : 188
sum : 188
sum : -1999
sum : 4562
4562

[deanhystad]

Yes. The if statement in your function makes it behave in one of two ways. Either the function will recursively call itself and not return anything, or the function doesn't recursively call itself and returns new. I think maybe you are understanding this now.

The value of your function, when called by the main code, is going to depend on which branch of the if statement is executed IN THAT PARICULAR CALL. If n==0 your function will return new. If n != 0, your function doesn't return a value.

Since it is unlikely you will always call you function with n=0, and since you always want your function to return new, you have a bug in you logic. The bug is fixed by moving the return statement from the conditional code to the end of the function.

def convert(n):
    a = ((n+1) % 3)-1
    if n:
        new.append(a)
        convert((n+1)//3)
    return new

Now the function always returns new regardless of the value of n.