防止副作用紅寶石

Question

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

如何分配b到a與5追加到末尾，而無需修改a本身？

Answer 1

紅寶石變量持有引用對象和=操作複製引用。

看來你想clonea：

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 2

算了筆兩個數組：

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

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

Answer 3

我通常做這種方式：

b = [*a, 5]