在 Ruby 中赋值是否在逻辑运算符之前? [英] Does assignment precede logical operator in Ruby?

问题描述

我在 Ruby 的 if 条件中遇到了一个奇怪的情况.这是重现这种情况的示例代码.

I have encountered a strange situation in the if condition of Ruby. Here is the sample code to reproduce the situation.

p1 = /hello/
p2 = /world/

s = "hello, world"

if m1 = s.match(p1) && m2 = s.match(p2)
    puts "m1=#{m1}"
    puts "m2=#{m2}"
end

输出将是

m1=world
m2=world

但我期望 m1=hello 因为 && 操作符具有 更高的优先级 Ruby 运算符.

But I expected m1=hello because the && operator has higher precedence in Ruby operators.

此代码

if m1 = s.match(p1) && m2 = s.match(p2)

似乎被解释为

if m1 = (s.match(p1) && m2 = s.match(p2))

为什么逻辑 AND 运算符 && 在赋值运算符 = 之前?

Why is the logical AND operator && preceded over the assignment operator =?

推荐答案

在 ruby​​ 中，几乎所有东西(见评论)都会返回一个值.

In ruby, pretty much everything (see comments) returns a value.

运算符 && 返回其右侧的最后一个表达式.所以 1 &&3 产生 3. && 将在第一个 falsey 值上短路.它返回该值，或最后计算的真值表达式.

The operator && returns the last expression to its the right. So 1 && 3 yields 3. && will short circuit on the first falsey value. It returns either that value, or the last evaluated truthy expression.

|| 返回其左侧的第一个表达式 - 所以 1 ||3 产生 1.|| 将短路第一个真值，返回它.

|| returns the first expression to the its left - so 1 || 3 yields 1. || will short circuit on the first truthy value, returning it.

检查这个差异:

1 + 5 * 3 + 1
# => 17 

1 + 5 && 3 + 1
# => 4

1 + 5 || 3 + 1
# => 6 

---

这是m1 = s.match(p1) && 中的求值顺序m2 = s.match(p2)

1. s.match(p1) =>你好"
2. && =>正确评估一切
3. m2 = s.match(p2) =>世界"
4. m1 = "你好";&&世界" =>世界"

1. s.match(p1) => "hello"
2. && => evaluate everything to its right
3. m2 = s.match(p2) => "world"
4. m1 = "hello" && "world" => "world"

您对 m2 的赋值返回用于 && 表达式world"的第二部分的值.ruby 中的赋值也返回一个值！

Your assignment to m2 returns the value which is used for the second part of the && expression, "world". Assignments in ruby also return a value!

所以你将有 m1 和 m2 的值都为world".

So you will have m1 and m2 both with the value "world".

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