ruby - 为什么可以写or却不能写||？

Question

我知道优先级有所不同，如 another answer 所示。 :

p foo = false || true
# => true

p foo = false or true
# => false

但是 or 和 || 之间似乎还有更多不同之处。

例如:

p foo = 42 or raise "Something went wrong with foo"
# => 42
p foo = nil or raise "Something went wrong with foo"
# => Something went wrong with foo (RuntimeError)
p foo = 42 || raise "Something went wrong with foo"
# => syntax error, unexpected tOP_ASGN, expecting end-of-input

我期望得到:

p foo = 42 or raise "Something went wrong with foo"
# => 42
p foo = nil or raise "Something went wrong with foo"
# => Something went wrong with foo (RuntimeError)
p foo = 42 || raise "Something went wrong with foo"
# => Something went wrong with foo (RuntimeError)

但这是一个语法错误。那么发生了什么？

Answer 1

理论:

这是一个 precedence table用于 ruby 。

从这个表中看不出来，但是不带括号的Ruby方法调用的优先级低于||和=，但高于或 .看这个question .

因此对于您的代码，从最高到最低的优先级:

• ||
• =
• 提出“某事”
• 或

用或表达

foo = 42 or raise "Something went wrong with foo"

先到= :

( foo = 42 ) or raise "Something went wrong with foo"

然后加注:

( foo = 42 ) or ( raise "Something went wrong with foo" )

然后或:

( ( foo = 42 ) or ( raise "Something went wrong with foo" ) )

用||表达

foo = 42 || raise "Something went wrong with foo"

先到||:

foo = ( 42 || raise ) "Something went wrong with foo"

这是你的语法错误!

你想要:

foo = 42 || (raise "Something went wrong with foo") #=> 42

或

foo = 42 || raise("Something went wrong with foo")  #=> 42

或者只是

foo = 42 || raise 

警告!

当你遇到优先级问题时，你应该小心添加另一个没有括号的 puts 或 p !

例如:

p [1,2,3].map do |i|
  i*2
end

输出:

#<Enumerator: [1, 2, 3]:map>

即使您可能已经预料到:

[2, 4, 6]