ruby – 为什么写或者不是||？

我知道 another answer中显示的优先级有所不同：

p foo = false || true
# => true

p foo = false or true
# => false

但似乎在和|和|之间存在更多不同的东西.

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

我期待得到：

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？

解决方法

理论：

这是Ruby的precedence table.

从这个表中不清楚,但没有括号的Ruby方法调用的优先级低于||和=,但高于或.见question.

因此,对于您的代码,从最高优先级到最低优先级：

> ||
> =
>提出“某事”
>或

表达与或

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

先来=：

( 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" ) )

表达式为||

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

首先是|| ：

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

这是你的语法错误！

你要 ：

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

要么

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

要不就

foo = 42 || raise

警告！

当你遇到优先事项时,你应该小心添加另一个没有括号的put或p！

例如 ：

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

产出：

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

即使你可能已经预料到：

[2,4,6]