问题描述
所以我想我明白这一点，但我没有得到我预期的输出，所以很明显，我不明白。
So I thought I understood this, but I'm not getting the output I expected, so obviously I don't understand it.
在红宝石（2.0.0）
In Ruby (2.0.0)
a = [1,2,3,4]
a.each do |e|
a.delete(e)
end
a = [2,4]
它似乎没有通过阵列中的每个项目被循环。然而，当我简单地输出项目，它遍历每个项目。有a.delete（E）的某些机制是影响迭代。
It doesn't seem to be looping through each item in the array. However, when I simply output the item, it loops through each item. There's some mechanism of a.delete(e) that is affecting the iteration.
a = [1,2,3,4]
a.each do |e|
puts e
end
=> 1
=> 2
=> 3
=> 4
最后，我想提出一个条件进入死循环，如：
Ultimately, I want to put a conditional into the loop, such as:
a = [1,2,3,4]
a.each do |e|
if e < 3
a.delete(e)
end
end
我怎样才能得到这个循环通过每个项目迭代并删除它吗？谢谢！
How can I get this loop it iterate through each item and delete it? Thank you!
推荐答案
通过
a = [1,2,3,4]
a.each do |e|
a.delete(e)
end
a # => [2, 4]
第一次迭代是在指数 0 与电子是 1 。这被删除， A 变成 [2,3,4] 和下一个迭代是指数 1 与电子是 3 。这被删除， A 变成 [2,4] 。下一次迭代将处于指数 2 ，但由于 A 是不是长了，它会停止，返回 A 的值 [2,4] 。
The first iteration was at index 0 with e being 1. That being deleted, a becomes [2,3,4] and the next iteration is at index 1, with e being 3. That being deleted, a becomes [2,4]. The next iteration would be at index 2, but since a is not that long anymore, it stops, returning a's value as [2, 4].
为了通过每个项目迭代并删除它，因为没有重复的，常见的方式是向后遍历。
In order to iterate through each item and delete it, given that there is no duplicate, a common way is to iterate backwards.
a = [1,2,3,4]
a.reverse_each do |e|
a.delete(e)
end
a # => []
a = [1,2,3,4]
a.reverse_each do |e|
if e < 3
a.delete(e)
end
end
a # => [3, 4]
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