Python，通过引用传递但参数返回不变

Question

Python新手，有一些C＆amp;的JavaScript。

我有以下两个函数和测试用例：

def switch(a, b):
    print("switch__You gave me (%d, %d)." % (a,b))
    y = a
    a = b 
    b = y
    print("switch__The order is now (%d, %d)" % (a, b))

def sortHi_Low(c, d):
    print("SORT You gave me (%d, %d)." % (c,d))
    if (c < d):
        switch(c, d)
        print("SORT The order is now (%d, %d)" % (c, d))
    else:
        print("SORT The order did not change (%d, %d)" % (c, d))

print("Case 1 \n")
sortHi_Low(10, 3)

print("\nCase 2 \n")
sortHi_Low(4,5)

switch()按预期工作。 sortHi_Low()运行switch()，但始终按照传递的顺序返回数字。

我对Python的理解是对象总是通过引用传递。当switch将sort的变量作为参数接收时，它应该通过引用更改它们的值 - 然后sort应返回这些新值。但是，这是在解释器中打印的内容。

Case 1

SORT You gave me (10, 3).
SORT The order did not change (10, 3)

Case 2

SORT You gave me (4, 5).
switch__You gave me (4, 5).
switch__The order is now (5, 4)
SORT The order is now (4, 5)

排序应该是Hi to Low，但不起作用。