一行中的 Python 多个赋值语句 [英] Python Multiple Assignment Statements In One Line

问题描述

(别担心，这不是关于解包元组的另一个问题.)

在python中，像foo = bar = baz = 5这样的语句将变量foo、bar和baz赋值为5.它从左到右将这些变量赋值，正如nastier所证明的像

这样的例子<预><代码>>>>foo[0] = foo = [0]回溯(最近一次调用最后一次):文件<stdin>"，第 1 行，在 <module> 中NameError: 名称 'foo' 未定义>>>foo = foo[0] = [0]>>>富[[...]]>>>富[0][[...]]>>>foo 是 foo[0]真的

但是 python 语言参考指出赋值语句的形式为

(target_list "=")+ (expression_list | yield_expression)

在赋值时，expression_list 首先被求值，然后赋值发生.

既然 bar = 5 不是 expression_list，那么行 foo = bar = 5 怎么可能有效?如何解析和评估一行上的这些多个分配?我读错了语言参考吗?

解决方案

感谢@MarkDickinson，他在评论中回答了这个问题:

<块引用>

注意(target_list "=")+中的+，表示一份或多份.在foo = bar = 5中，有两个(target_list "=")产生式，而expression_list部分就是5

赋值语句中的所有 target_list 产生式(即看起来像 foo = 的东西)从左到右分配给 expression_list 在语句的右端，在 expression_list 被求值之后.

当然，通常的元组解包"赋值语法也适用于这种语法，让您可以执行诸如

<预><代码>>>>foo, boo, moo = boo[0], moo[0], foo[0] = moo[0], foo[0], boo[0] = [0], [0], [0]>>>富[[[[[...]]]]]>>>foo[0] 是嘘真的>>>foo[0][0] 是 moo真的>>>foo[0][0][0] 是 foo真的

(Don't worry, this isn't another question about unpacking tuples.)

In python, a statement like foo = bar = baz = 5 assigns the variables foo, bar, and baz to 5. It assigns these variables from left to right, as can be proved by nastier examples like

>>> foo[0] = foo = [0]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'foo' is not defined
>>> foo = foo[0] = [0]
>>> foo
[[...]]
>>> foo[0]
[[...]]
>>> foo is foo[0]
True

But the python language reference states that assignment statements have the form

(target_list "=")+ (expression_list | yield_expression)

and on assignment the expression_list is evaluated first and then the assigning happens.

So how can the line foo = bar = 5 be valid, given that bar = 5 isn't an expression_list? How are these multiple assignments on one line getting parsed and evaluated? Am I reading the language reference wrong?

解决方案

All credit goes to @MarkDickinson, who answered this in a comment:

Notice the + in (target_list "=")+, which means one or more copies. In foo = bar = 5, there are two (target_list "=") productions, and the expression_list part is just 5

All target_list productions (i.e. things that look like foo =) in an assignment statement get assigned, from left to right, to the expression_list on the right end of the statement, after the expression_list gets evaluated.

And of course the usual 'tuple-unpacking' assignment syntax works within this syntax, letting you do things like

>>> foo, boo, moo = boo[0], moo[0], foo[0] = moo[0], foo[0], boo[0] = [0], [0], [0]
>>> foo
[[[[...]]]]
>>> foo[0] is boo
True
>>> foo[0][0] is moo
True
>>> foo[0][0][0] is foo
True

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