为什么这个递归 function 中的第二个条件出现两次

Question

Trying to use recursion for a maze problem and made a conditional test case. Can someone explain to me why the 2nd condition adds 20 bs instead of 10?

def recur_test(count, list1, status):
    if count == 10 and status == 1:
        return (count, list1)
    if count == 10 and status == 0:
        count = 0
        status = 1
        recur_test(count, list1, status)

    if count < 10 and status == 0:
        count += 1
        list1.append("a")
        print("A thread", count)
        recur_test(count, list1, status)

    if count < 10 and status == 1:
        count += 1
        list1.append('b')
        print("B Thread", count)
        recur_test(count, list1, status)

print(recur_test(0, [], 0))

Answer 1

Consider a simpler example:

def recur_test(value):
    if value:
        print("in first condition")
        recur_test(False)

    if not value:
        print("in second condition")

recur_test(True)

The way to understand this most easily is to substitute in another function that does the same thing, and call that instead of making the recursive call:

def recur_test_2(value):
    if value:
        print("in first condition")
        value = False
        recur_test_2(value)

    if not value:
        print("in second condition")

def recur_test_1(value):
    if value:
        print("in first condition")
        value = False
        recur_test_2(value)

    if not value:
        print("in second condition")

recur_test_1(True)

Now there should be no confusion: when the call to recur_test_2 returns, recur_test_1 keeps going, and prints the in second condition message a second time - since value has been changed.

The same fundamental thing happens with recursion - each call to the function is "separate", it just happens that the function has the same name each time.

Your own code has the same issue - on the call where count == 10 and status == 0 , the one where you flip status to 1 , on the way "in" this switches from appending a s to appending b s, but on the way "out" those calls continue - and since count was reset to 0 in that call , it is now 0 again, and status is similarly 1 , so the b -appending branch is entered again.

There is a further complication, though: in your code, the reassignments to count and status in further-on recursive calls don't matter to the current call, but the .append to list1 does - because it is the same list object , being modified in-place.

Answer 2

Your if statement unexpectedly falls through to each of the others even when it is true. When the second condition if count == 10 and status == 0: is true, you set count = 0 and status = 1, then make the recursive call. The recursive call goes all the way through the "B Thread" and returns control to where you made the recursive call in that second if body. So now count and status are 0 and 1 respectively. This makes the last condition if count < 10 and status == 1: true, so it continues to go all the way through the "B Thread" again.

I would suggest changing the if statements to if else.