Python: pass-by-reference

再次提出这个问题是因为遇到了一个 “想在 function 中修改 list 内容” 的场景，而正确的写法应该是用 slice-then-assign:

def test1(lst):
    lst = [999, 1000]

def test2(lst):
    lst[:] = [999, 1000]

if __name__ == "__main__":
    lst = [1, 2]
    
    test1(lst)
    print(lst)

    test2(lst)
    print(lst)

# Output:
#   [1, 2]
#   [999, 1000]

函数体内的 lst 只是 argument 的一个 copy，而这里 argument 是一个 reference，指向一个 list；而你对这个 reference copy 赋值，相当于让它指向一个新的 list，原来的 reference 没有变动。所以 test1 的逻辑大约是：

if __name__ == "__main__":
    lst = [1, 2]

    ENTER test1:
        lst_copy = lst  # => [1, 2]
        lst_copy = [999, 1000]
    EXIT test1

    print(lst)

# Output
#   [1, 2]

而 test2 的先 slice 再 assignment 实际上是调用了 __setitem__。根据 How assignment works with python list slice:

1. If the left side of assignment is subscription, Python will call __setitem__ on that object. a[i] = x is equivalent to a.__setitem__(i, x).
2. If the left side of assignment is slice, Python will also call __setitem__, but with different arguments: a[1:4]=[1,2,3] is equivalent to a.__setitem__(slice(1,4,None), [1,2,3])

所以即是 caller 是 reference copy，但它能实际影响 reference 指向的值。test2 的逻辑大约是：

if __name__ == "__main__":
    lst = [1, 2]

    ENTER test2:
        lst_copy = lst  # => [1, 2]
        lst_copy.__setitem__(slice(None, None, None), [999, 1000])
    EXIT test2

    print(lst)

# Output
#   [999, 1000]