简洁完整的回答

a ||= b

计算方法与下面每一行相同

a || a = b
a ? a : a = b
if a then a else a = b end

-

另一方面，

a = a || b

计算方法与下面每一行相同

a = a ? a : b
if a then a = a else a = b end

-

编辑:正如AJedi32在评论中指出的那样，这只在以下情况下成立:A是一个已定义变量。2. 计算一次时间和两次时间不会导致程序或系统状态的差异。

简洁完整的回答

a ||= b

计算方法与下面每一行相同

a || a = b
a ? a : a = b
if a then a else a = b end

-

另一方面，

a = a || b

计算方法与下面每一行相同

a = a ? a : b
if a then a = a else a = b end

-

编辑:正如AJedi32在评论中指出的那样，这只在以下情况下成立:A是一个已定义变量。2. 计算一次时间和两次时间不会导致程序或系统状态的差异。

准确地说，||= b意味着“如果a是未定义的或假的(false或nil)，将a设置为b并求值为(即返回)b，否则求值为a”。

其他人经常试图用||= b等价于|| a = b或a = a || b来说明这一点。这些等价有助于理解这个概念，但要注意，它们并非在所有条件下都是准确的。请允许我解释:

a ||= b ⇔ a || a = b? The behavior of these statements differs when a is an undefined local variable. In that case, a ||= b will set a to b (and evaluate to b), whereas a || a = b will raise NameError: undefined local variable or method 'a' for main:Object. a ||= b ⇔ a = a || b? The equivalency of these statements are often assumed, since a similar equivalence is true for other abbreviated assignment operators (i.e. +=,-=,*=,/=,%=,**=,&=,|=,^=,<<=, and >>=). However, for ||= the behavior of these statements may differ when a= is a method on an object and a is truthy. In that case, a ||= b will do nothing (other than evaluate to a), whereas a = a || b will call a=(a) on a's receiver. As others have pointed out, this can make a difference when calling a=a has side effects, such as adding keys to a hash. a ||= b ⇔ a = b unless a?? The behavior of these statements differs only in what they evaluate to when a is truthy. In that case, a = b unless a will evaluate to nil (though a will still not be set, as expected), whereas a ||= b will evaluate to a. a ||= b ⇔ defined?(a) ? (a || a = b) : (a = b)???? Still no. These statements can differ when a method_missing method exists which returns a truthy value for a. In this case, a ||= b will evaluate to whatever method_missing returns, and not attempt to set a, whereas defined?(a) ? (a || a = b) : (a = b) will set a to b and evaluate to b.

好的，好的，那么||= b等于什么?有办法在Ruby中表达这一点吗?

好吧，假设我没有忽略任何东西，我相信||= b在功能上等价于……(击鼓声)

begin
  a = nil if false
  a || a = b
end

坚持住!这不是第一个前面有noop的例子吗?嗯，不完全是。还记得我之前说的||= b只有当a是一个未定义的局部变量时，||= b才不等于|| a = b吗?如果为false，则a = nil确保a永远没有定义，即使这一行永远不会执行。Ruby中的局部变量是词法范围的。

x ||= y

is

x || x = y

"如果x为假或未定义，则x指向y"

||= b是一个条件赋值运算符。它的意思是:

如果a是未定义的或假的，则计算b并将a设置为结果。 否则(如果定义了a并计算为真)，则不计算b，并且不进行赋值。

例如:

a ||= nil # => nil
a ||= 0 # => 0
a ||= 2 # => 0

foo = false # => false
foo ||= true # => true
foo ||= false # => true

令人困惑的是，它看起来类似于其他赋值操作符(例如+=)，但行为不同。

A += b转换为A = A + b ||= b大致可以转换为|| A = b

它是|| a = b的近似缩写。不同之处在于，当a未定义时，|| a = b将引发NameError，而||= b将a设置为b。如果a和b都是局部变量，这种区别不重要，但如果其中一个是类的getter/setter方法，则很重要。

进一步阅读:

http://www.rubyinside.com/what-rubys-double-pipe-or-equals-really-does-5488.html

这就像延迟实例化。 如果变量已经定义，它将获取该值，而不是重新创建该值。