ruby - 防止 Ruby 中的副作用

Question

a = [1,2,3,4]
b = a << 5
a == [1,2,3,4] # returns false

如何在不修改自身的情况下分配给附加b到末尾？a5a

Answer 1

只需对两个数组求和：

a = [1,2,3,4]
b = a + [5]

# b == [1, 2, 3, 4, 5]
# a == [1, 2, 3, 4]

Answer 2

Ruby 变量保存对对象的引用=，操作员复制这些引用。

您似乎希望：clone a

irb(main):001:0> a = [1,2,3,4]
=> [1, 2, 3, 4]
irb(main):002:0> b = a.clone << 5
=> [1, 2, 3, 4, 5]
irb(main):003:0> a
=> [1, 2, 3, 4]
irb(main):004:0> b
=> [1, 2, 3, 4, 5]

Answer 3

我通常这样做：

b = [*a, 5]