元组作为函数参数 [英] tuples as function arguments

问题描述

tuple 在 python 中(在代码块中)由逗号定义；括号不是强制性的(在以下情况下).所以这三个都是等价的:

a tuple in python (in a code block) is defined by the commas; the parentheses are not mandatory (in the cases below). so these three are all equivalent:

a, b = 1, 2
a, b = (1, 2)
(a, b) = 1, 2

如果我定义了一个函数

def f(a, b):
    print(a, b)

这样调用就行了:

f(2, 3)

这不会:

f((2, 3))
# TypeError: f() missing 1 required positional argument: 'b'

当元组是函数参数时，python 如何区别对待元组?这里括号是必要的(我理解为什么这是这种情况，我很高兴 Python 以这种方式工作！).

how does python treat tuples differently when they are function arguments? here the parentheses are necessary (i understand why this is the case and i am happy python works this way!).

我的问题是:当元组是函数参数时，python 如何以不同的方式对待它们.

my question is: how does python treat tuples differently when they are function arguments.

推荐答案

为方便起见，Python 会根据赋值语句的需要构造一个临时元组.因此，您的所有三个赋值语句在到达数据移动时都完全相同.

For convenience, Python constructs a temporary tuple as needed for an assignment statement. Thus, all three of your assignment statements are exactly the same once they reach data movement.

函数调用不是赋值语句；这是一个参考映射.因此，语义是不同的.

A function call is not an assignment statement; it's a reference mapping. Therefore, the semantics are different.

如果您希望 Python 将元组解包为两个单独的参数，请使用 * 运算符:

If you want Python to unpack your tuple into two separate arguments, use the * operator:

f(*(2, 3))

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