排列生成器代码中的条件和递归

Question

def genperm(perm,n):
    if len(perm)==n:
3         print('output',perm)
          return
    for i in range(n):
          print('i',i)
7         if i not in perm:
                perm.append(i)
                print('appended',perm)
                genperm(perm,n)
11            print('to pop',perm)
                    perm.pop()
genperm([],2)O

So I have this code that outputs permutations with out repetitions. It works but I don't understand how. So I put some unnecessary prints to see how they are generated.

For the sake of this example n=2 which are values (0,1) and so the output should be 0,1 and 1,0. Once run, it prints:

i 0 
appended [0]
i 0
i 1
appended [0, 1]
3  output [0, 1]
11 to pop [0, 1]
to pop [0]
i 1
appended [1]
i 0
appended [1, 0]
output [1, 0]
to pop [1, 0]
i 1
to pop [1]

So let's say we get to a state where perm=[0,1] and so line 3 prints it, but then it jumps to line 11. My question is how does it enters the conditional in line 7 if at that point, all possibles values of i are in perm and so it is false.

Answer 1

It goes into the recursive function at line 10. This means it runs the function again from the start at this point. So it is jumping from line 10 to line 2, not jumping to line 11. It then prints at line 3 (Where perm=[0,1] )and returns , which ends the recursive function at line 10. So it continues on to line 11, where it pops 1 before going to line 7 again for the next i , 1. Which is True.