Python通过引用混淆传递对象

Question

Hi I seem to having issues understanding of passing object by reference in Python. I understand example 1 output, but shouldn't example 2 behave in the similar way and not change the A matrix?

Example 1:

def reassign(list):
  list = [0, 1, 2]

list = [3]
reassign(list)
print(list)

Returns: [3]

Example 2:

import numpy as np

A = np.ones((4,4))

def xyz(A):
    for i in range(4):
        A[i,i] = 0    
    return None

x = xyz(A)
print(A)

# Returns

[[0. 1. 1. 1.]
 [1. 0. 1. 1.]
 [1. 1. 0. 1.]
 [1. 1. 1. 0.]]

Answer 1

Passing by reference means that, when you pass a variable into the function, you don't pass the variable itself , you pass the pointer to the variable, which is copied from outside to inside the function. In Example 1, you pass list into the function, which is a pointer to a list that contains the elements [3] . But then, immediately after, you take that variable holding the pointer to the list, and put a new pointer in it, to a new list that contains the elements [0, 1, 2] . Note that you haven't changed the list you started with - you changed what the variable referring to it referred to. And when you get back out of the function, the variable you passed into the function (still a pointer to the first list) hasn't changed - it's still pointing to a list that contains the elements [3] .

In Example 2, you pass A into xyz() . Whereas in Example 1 you did something along the lines of

A = something_else

here, you're doing

A[i] = something_else

This is an entirely different operation - instead of changing what the variable holding the list is pointing to, you're changing the list itself - by changing one of its elements. Instead of making A point to something else, you're changing the value that A points to by dereferencing it.