Question

有方法：

possible_queries=[]
variant="tenra"
for i in 0...variant.length
  variant="tenra"
  if variant[i]=~/[\w]/
    letter=variant[i]
    variant[i]="."
    for k in 0...variant.length
      if variant[k]=~/[\w]/ && i!=k
        letter_two=variant[k]
        variant[k]="."
        possible_queries.push(variant)
        print variant+", "
        variant[k]=letter_two
      end
    end
  end
end
print "\n"
print possible_queries.inspect

所以我将variant发送到每个内部循环的数组，并在此处打印，例如，但实际发送到数组的variant项与实际数组项不匹配。

印刷版本：

..nra, .e.ra, .en.a, .enr., ..nra, t..ra, t.n.a, t.nr., .e.ra, t..ra, te..a, te.r., .en.a, t.n.a, te..a, ten.., .enr., t.nr., te.r., ten..,

且possbile_queries：

[".enra", ".enra", ".enra", ".enra", "t.nra", "t.nra", "t.nra", "t.nra", "te.ra", "te.ra", "te.ra", "te.ra", "ten.a", "ten.a", "ten.a", "ten.a", "tenr.", "tenr.", "tenr.", "tenr."]

为什么会这样？

Answer 1

我不确定输出与您的预期有何不同，但我会猜测推送.dup ed字符串会有所帮助：

possible_queries.push(variant.dup)

这样，你在推送它之后对variant所做的更改不会影响你已推送的内容，这似乎是你所期望的。

Answer 2

错误比你想象的要简单。你在推送中添加到数组中的是指向文本的指针..然后我的朋友你改变了指针的内容（在你打印之后）。

  possible_queries.push(variant)   #reference to the object (aka pointer) variant
  print variant+", "               #prints the object (staticly)
  variant[k]=letter_two            #you changed the object variant..  
                                   #the value in the array is still pointing to the object and you've changed its contents.

简单的解决方案..当您推送到数组时..添加dup ..

variant.dup

数组中的内容与循环时的输出不匹配

2 个答案: