假设a = 2, b = 3

那么，||= b将得到a的值，即2。

当a计算为某个值时，结果不是false或nil..这就是为什么它不计算b的值。

现在假设a = nil, b = 3。

那么||= b将得到3，即b的值。

因为它首先尝试评估a的值，结果是nil..它求出b的值。

ror app中使用的最佳示例是:

#To get currently logged in iser
def current_user
  @current_user ||= User.find_by_id(session[:user_id])
end

# Make current_user available in templates as a helper
helper_method :current_user

其中，User.find_by_id(session[:user_id])当且仅当@current_user之前未初始化时触发。

x ||= y

is

x || x = y

"如果x为假或未定义，则x指向y"

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。

还请记住||=不是原子操作，因此，它不是线程安全的。根据经验，不要将它用于类方法。

ruby-lang语法。正确的答案是查看ruby-lang文档。所有其他解释都令人困惑。

谷歌

"ruby-lang文档缩写赋值"。

Ruby-lang文档

https://docs.ruby-lang.org/en/2.4.0/syntax/assignment_rdoc.html#label-Abbreviated+Assignment