【Python】求问Python的一个列表赋值问题
为什么
s = [1, 2, 3, 4, 5, 6]
i = 0
i = s[i] = 3
结果是: [1, 2, 3, 3, 5, 6] 而不是 [3, 2, 3, 4, 5, 6]
回答
可以参考下写过的一篇文章: Python: 链式赋值的坑
根据 python.org/release/3.0.1/reference/simple_stmts.html#assignment-statements" rel="nofollow noreferrer">Assignment statements 可知:
a = b = c = d = ... = E
等同于
a = Eb = E
c = E
d = E
...
所以: i=s[i]=3
等同于:
i = 3s[i] = 3
Python 中的赋值是一个语句,并不是一个操作符,所以表达式(a=b)
会产生语法错误,赋值语句没有值.
参考 https://stackoverflow.com/que...
等同于
s = [1, 2, 3, 4, 5, 6]i = 0
temp_value = 3
i = temp_value
s[i] = temp_value
首先是i变成了3,然后才赋值s[i]
看结果反推,是i=3 在 s[i] = 3之前执行了。
你就分开两句写不行吗?
可使用 PythonTutor.com
i = s[i] = 3 那一行基本上就是先後執行i=3 及s[i]=3
i = s[i] = 3
等价于 i = 3; s[i] = 3
用dis模块来解析执行过程:
>>> 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 入栈
33 DUP_TOP # 复制栈顶,也就是 常量3
34 STORE_FAST 1 (i) # i = 3
37 LOAD_FAST 0 (s)
40 LOAD_FAST 1 (i)
43 STORE_SUBSCR # s[i] = 3
44 LOAD_CONST 0 (None) # 返回 None
47 RETURN_VALUE
分开写的例子
>>> 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_FAST 0 (s)
42 LOAD_FAST 1 (i)
45 STORE_SUBSCR # s[i] = 3
46 LOAD_CONST 0 (None)
49 RETURN_VALUE
>>>
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