問題描述

為什么

s = [1, 2, 3, 4, 5, 6]i = 0i = s[i] = 3 結果是: [1, 2, 3, 3, 5, 6] 而不是 [3, 2, 3, 4, 5, 6]

問題解答

可以參考下寫過的一篇文章: Python: 鏈式賦值的坑

根據(jù) Assignment statements 可知:

a = b = c = d = ... = E

等同于

a = Eb = Ec = Ed = E...

所以: i=s[i]=3等同于:

i = 3s[i] = 3

Python 中的賦值是一個語句,并不是一個操作符,所以表達式(a=b)會產(chǎn)生語法錯誤,賦值語句沒有值.

參考 https://stackoverflow.com/que...

等同于

s = [1, 2, 3, 4, 5, 6]i = 0temp_value = 3i = temp_values[i] = temp_value

首先是i變成了3，然后才賦值s[i]

看結果反推，是i=3 在 s[i] = 3之前執(zhí)行了。

你就分開兩句寫不行嗎？

可使用 PythonTutor.com i = s[i] = 3 那一行基本上就是先後執(zhí)行i=3 及s[i]=3

i = s[i] = 3 等價于 i = 3; s[i] = 3

用dis模塊來解析執(zhí)行過程：

>>> def f(): s = [1, 2, 3, 4, 5, 6] i = 0 i = s[i] = 3 >>> import dis>>> dis.dis(f) 2 0 LOAD_CONST 1 (1) 3 LOAD_CONST 2 (2) 6 LOAD_CONST 3 (3) 9 LOAD_CONST 4 (4) 12 LOAD_CONST 5 (5) 15 LOAD_CONST 6 (6) 18 BUILD_LIST 6 21 STORE_FAST 0 (s) # s = [1, 2, 3, 4, 5, 6] 3 24 LOAD_CONST 7 (0) 27 STORE_FAST 1 (i) # i = 0 4 30 LOAD_CONST 3 (3) # 常量3 入棧 33 DUP_TOP# 復制棧頂，也就是 常量3 34 STORE_FAST 1 (i) # i = 3 37 LOAD_FAST0 (s) 40 LOAD_FAST1 (i) 43 STORE_SUBSCR # s[i] = 3 44 LOAD_CONST 0 (None) # 返回 None 47 RETURN_VALUE

分開寫的例子

>>> def f2(): s = [1, 2, 3, 4, 5, 6] i = 0 i = 3 s[i] = 3 >>> dis.dis(f2) 2 0 LOAD_CONST 1 (1) 3 LOAD_CONST 2 (2) 6 LOAD_CONST 3 (3) 9 LOAD_CONST 4 (4) 12 LOAD_CONST 5 (5) 15 LOAD_CONST 6 (6) 18 BUILD_LIST 6 21 STORE_FAST 0 (s) # s = [1, 2, 3, 4, 5, 6] 3 24 LOAD_CONST 7 (0) 27 STORE_FAST 1 (i) # i = 0 4 30 LOAD_CONST 3 (3) 33 STORE_FAST 1 (i) # i = 3 5 36 LOAD_CONST 3 (3) 39 LOAD_FAST0 (s) 42 LOAD_FAST1 (i) 45 STORE_SUBSCR # s[i] = 3 46 LOAD_CONST 0 (None) 49 RETURN_VALUE>>>