防止Ruby中的副作用

Question

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

如何将b分配到a，并在不修改5的情况下将a添加到最后？{/ p>

Answer 1

只需将两个数组相加：

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

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

Answer 2

Ruby变量将引用保存到对象，=运算符复制引用。

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]