Question

有人可以解释一下吗？

def digit_block(size = 1)
  col = 2 + 1*size
  row = 1 + 2*size
  r = []
  for i in 0...col
    r.push ' '
  end
  a = []
  for i in 0...row
    a.push r
  end
  a
end

block = digit_block
puts block.inspect
block[1][2] = 'x'
puts block.inspect

输出：

[[" ", " ", " "], [" ", " ", " "], [" ", " ", " "]]
[[" ", " ", "x"], [" ", " ", "x"], [" ", " ", "x"]]

我的理解是块[1] [2]只更改第1行第2列的单元格，但为什么它会更改第2列中的所有单元格？

Answer 1

  for i in 0...row
    # you are pushing the same array object to an array
    a.push r
  end

因此block中的每个元素都是同一个对象。

block[0] === block[1]  # true
block[1] === block[2]  # true

的更新 的

你需要为每个元素创建一个新数组，你的代码可以重写如下：

def digit_block(size = 1) Array.new(1 + 2*size){Array.new(2 + size){' '}} end

Answer 2

您只生成一个数组r。即使你在多个地方使用它，它们的身份也是一样的。如果在一个位置更改它，它会影响其他位置的同一对象。要回答标题中的问题，您需要为每一行创建一个不同的数组。

def digit_block(size = 1)
  col = 2 + 1*size
  row = 1 + 2*size
  a = []
  for i in 0...row
    # For every iteration of the loop, the code here is evaluated,
    #  which means that the r is newly created for each row.
    r = []
    for i in 0...col
      r.push ' '
    end
    a.push r
  end
  a
end

如何更改ruby 2d阵列中的一个单元格？

2 个答案: