爲什麼可以寫或不寫||？

Question

我知道有在another answer所示優先級的差別：

p foo = false || true 
# => true 

p foo = false or true 
# => false 

但似乎有更多的東西那是or和||不同。

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) 

但它是一個語法錯誤。那麼發生了什麼？

Answer 1

理論：

下面是一個Ruby precedence table。

從該表中不清楚，但沒有括號的Ruby方法調用的優先級低於||和=，但高於or。請參閱question。

因此，對於你的代碼，從最高到最低的優先級：

• ||
• =
• raise "something"
• or

表達與or

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

首先是=：

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

然後raise：

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

然後or：

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

表達與||

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

首先是||：

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

這是你的語法錯誤！

你想：

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

或

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

或只是

foo = 42 || raise 

警告！

當您遇到優先級問題時，您應該小心添加另一個不帶括號的puts或p！

例如：

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

輸出：

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

即使你可能會認爲：

[2, 4, 6] 

Answer 2

||和or是不一樣的操作。

第一個等同於方法調用，後者是控制流關鍵字。您可能總是希望使用||以避免與優先級混淆。大多數Ruby的樣式指南都有一條禁止使用and和or的條款。

那麼，

A or B 

# can be considered equivalent to 

if A then A else B end 

而

A || B 

# can be considered equivalent to 

A.or { B } # given a hypothetical "logical or" method 

現在，讓我們看看你的or例如

p foo = false or true 

相當於

temp = p(foo = false) # => nil 
if temp 
    temp 
else 
    true 
end 

，從而執行打印false時，返回true

[1] pry(main)> p foo = false or true 
false 
=> true 
[2] pry(main)> foo 
=> false 

而

p foo = false || true 

相當於（粉飾布爾和邏輯之間，或者現在因爲你的例子，無論如何處理布爾值的差異）

p(foo = false.|(true)) 

因此當執行打印true並返回true

[1] pry(main)> p foo = false || true 
true 
=> true 
[2] pry(main)> foo 
=> true