Python recursive function value

When I was building a recursive function, I verified it because I couldn't understand how the value outside the function was rewritten inside the function in python.

Prerequisite knowledge

Check how variables defined inside and outside the function are handled.

a=10
def replace1():
    a=15
replace1()
print(a)

>10

In this case, a changes only within the function.

al=[10]
def replace3():
    al=[15]
replace3()
print(al)

>[10]

Again, al changes only within the function.

al=[10]
def replace2():
    al[0]=15
replace2()
print(al)

>[15]

In this case, the value changed inside the function is also changing outside the function. ~~ This is because the list is a mutable object. ~~ It was a question of whether I was assigning to a local variable or referencing a global variable.

a=10
def replace1():
    global a
    a=15
replace1()
print(a)

>15

In the comment, it was pointed out that if you define a global variable, you can assign it. You can actually assign to a variable outside the function from inside the function. There seems to be a way to define it as a class variable. (See comments)

Verification of variable handling with recursive function

From here, we will try to handle variables with recursive functions. The recursive function used here counts how many times it can be calculated until 3-1 becomes 0.

ex1

count=0
a=3
def saiki1(a):
    if(a==0): return
    a=a-1
    count += 1
saiki1(a)
print(a, count)

>UnboundLocalError: local variable 'count' referenced before assignment

I tried it just in case, but an error occurs because count is not defined in the function. If you define `` `global count``` in the function, you will get the correct answer for count without an error.

ex2

count=0
a=3
def saiki2(a):
    count=0
    if(a==0): return
    a=a-1
    count += 1
saiki2(a)
print(a, count)

>2 1
>1 1
>0 1
>3 0

In this case, it changes inside the function, but outside the function it is the value originally defined. In this case, the answer is not correct because count = 0 is reset in the recursion. ex3

count=0
a=3
def saiki3(a, count):
    if(a==0): return
    a=a-1
    count += 1
    print(a, count)
    saiki3(a,count)
saiki3(a, count)
print(a, count)

>2 1
>1 2
>0 3
>3 0

If you add count to the argument of the recursive function, the answer is correct inside the function, but outside the function it is not received as a return value, so it is the value defined first. ex4

count=0
a=3
def saiki4(a, count):
    if(a==0): return a, count
    a=a-1
    count += 1
    print(a, count)
    saiki4(a, count)
    return a, count
a, count= saiki4(a, count)
print(a, count)

>2 1
>1 2
>0 3
>2 1

Next, when a and count are set as return values. In this case, it's complicated, but since we didn't receive a return value when we ran saiki4 on line 8, we're getting 21 of the first calculation in the function.

ex5 (legitimate)

count=0
a=3
def saiki5(a, count):
    if(a==0): return a, count
    a=a-1
    count += 1
    print(a, count)
    a,count= saiki5(a, count)
    return a, count
a, count= saiki5(a, count)
print(a, count)

>2 1
>1 2
>0 3
>0 3

In this case, since a and count are received in the function, the validity can be output. The same result can be obtained by writing `` `return saiki5 (a, count) ``` together on the 8th and 9th lines.

ex6 (legitimate)

count=[0]
a=3
def saiki6(a, count):
    if(a==0): return a
    a=a-1
    count[0] += 1
    print(a, count)
    a = saiki6(a, count)
    return a
a= saiki6(a, count)
print(a, count)

>2 [1]
>1 [2]
>0 [3]
>3 [3]

Finally, list count and check if only that element is rewritten. In this case, count is not the return value, but the value defined outside the function is rewritten inside the function. If you set `t = count [0] + 1 count = [t] ``` etc., the objects in the list will change, so 3 [0] `` will be output as the final result.

Postscript In the comments, I learned how to do it with global variables.

reference

https://excel-ubara.com/python/python016.html