一个常见的误解是，||= b并不等同于a = a || b，但它的行为就像|| a = b。

但这里有一个棘手的情况。如果a未定义，则|| a = 42将引发NameError，而||= 42将返回42。所以，它们似乎不是等价的表达式。

irb(main):001:0> a = 1
=> 1
irb(main):002:0> a ||= 2
=> 1

因为a已经被设为1了

irb(main):003:0> a = nil
=> nil
irb(main):004:0> a ||= 2
=> 2

因为a是nil

这是默认的赋值表示法

例如:x ||= 1 这将检查x是否为nil。如果x确实为nil，那么它就会赋给它一个新值(在我们的例子中是1)

更明确的表示: 如果x == nil X = 1 结束

一个常见的误解是，||= b并不等同于a = a || b，但它的行为就像|| a = b。

但这里有一个棘手的情况。如果a未定义，则|| a = 42将引发NameError，而||= 42将返回42。所以，它们似乎不是等价的表达式。

X ||= y表示

如果x有任何值，不要改变它的值，否则 将x设为y

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。