Question

我正在研究某个问题时，我发现扔掉所有内容的是我在while循环之外声明了一个变量。有问题的变量是sub_string = 1.似乎我需要在第一个while循环之后声明它，就在它实际在第二个while循环中发挥作用之前。通过早期声明它，循环以某种方式只能拾取从字符串[0]开始的回文，然而，可以是任何长度。对于我的生活，我不确定它为什么会以这种方式表现，因为我宣称切片的长度（而不是起始位置）。

如果我错过了一些非常基本的东西，请原谅我，因为我对编码很新。

这是有问题的代码（省略了回文的定义？检查方法）。

def longest_palindrome(string)
  idx = 0
  longest = nil
  #substring_string = 1  ; declaring it early seems to cause weird issues to show up
  while idx < string.length
    substring_length = 1 #I have to declare the substring_length here

    while (idx + substring_length) <= string.length
      substring = string.slice(idx, substring_length)
      if palindrome?(substring) && (longest == nil || substring.length > longest.length)
    longest = substring
    print substring_length
  end

  substring_length += 1
end

  idx += 1
  end

  return longest
end

Answer 1

场景1：while循环中定义的 substring_length

def longest_palindrome(string)
  idx = 0
  longest = nil
  while idx < string.length
    substring_length = 1
    while (idx + substring_length) <= string.length
      # palindrome
      substring_length += 1 
    end
    idx += 1
  end
  longest
end

在对象上调用此方法时，执行控制流将移至第一个while循环。 while idx < string.length
在该循环中，您初始化 substring_length至1
然后代码进入嵌套循环while (idx + substring_length) <= string.length
在嵌套循环中，调用回文方法，然后substring_length 递增1 。此代码将一直运行，直到其条件返回false (idx + substring_length) < string.length。这意味着，一旦这个内部循环结束，substring_length 总是将成为大于1的值
一旦这个内部循环结束，控制就会回到外部while循环。现在代码的第一行重新分配 substring_length到1。实质上，无论内部循环结束时substring_length的值是什么，一旦控制流回到外部循环，当您将substring_length设置为 1

场景2： substring_length在while循环中定义

def longest_palindrome(string) idx = 0 longest = nil substring_length = 1 while idx < string.length while (idx + substring_length) <= string.length # palindrome substring_length += 1 end idx += 1 end longest end

在对象上调用此方法时，substring_length 初始化为1 ，然后执行控制流移至第一个while循环。 while idx < string.length

如果条件为真，则控制流将移至第二个循环inner循环

在内环中，调用回文方法，然后substring_length 增加1 。此代码将一直运行，直到其条件返回false (idx + substring_length) < string.length。就像在方案1 中一样，这意味着一旦这个内部循环结束，substring_length 始终将是大于1的值

一旦这个内循环结束，控制就会回到外部while 环。这次没有初始化语句，因此内部循环结束时substring_length的值是控制返回外部循环时使用的相同值。

让我们尝试使用一个例子

情景1

longest_palindrome('him') idx = 0; substring_length = 1; string.length = 3 while (0 < 3) # the outer loop condition is true, substring_length = 1 # while (idx + substring_length) < string.length while (0 + 1) < 3 # inner loop condition is true # run code block, increment substring_length by 1, substring_length is NOW 2 while (0 + 2) < 3 # inner loop condition is still true # run code block, increment substring_length by 1, substring_length is NOW 3 while (0 + 3) < 3 # inner loop condition fails, substring_length = 3 # Go to line after end of inner loop increase idx by 1 # idx = 1 # Go back to outer loop set substring_length to 1 # substring_length is NO MORE 3 # while (idx + substring_length) < string.length while (1 + 1) < 3 # run code block, increment substring_length by 1, substring_length is NOW 2 while (1 + 2) < 3 # condition fails ..... and so on

场景2

longest_palindrome('him') idx = 0; substring_length = 1; string.length = 3 while (0 < 3) # the outer loop condition is true, # while (idx + substring_length) < string.length while (0 + 1) < 3 # inner loop condition is true # run code block, increment substring_length by 1, substring_length is NOW 2 while (0 + 2) < 3 # inner loop condition is still true # run code block, increment substring_length by 1, substring_length is NOW 3 while (0 + 3) < 3 # inner loop condition fails, substring_length = 3 # Go to line after end of inner loop increase idx by 1 # idx = 1 # Go back to outer loop. Remember substring_length =3 # while (idx + substring_length) < string.length while (1 + 3) < 3 # condition is false. Inner loop is NOT EXECUTED # Go to line after end of inner loop increase idx by 1 # idx = 1 # Go back to outer loop while (idx < string.length) while (1 < 3) # outer loop condition is true # Go to inner loop while (idx + substring_length) < string.length while (1 + 3) < 3 # Once again condition is false and inner loop is NOT EXECUTED # Go to line after end of inner loop increase idx by 1 # idx = 1

正如您在此场景中所看到的，内部循环停止执行，因为substring_length未被重新分配为1并且条件不断失败。

Ruby：在while循环之前和之后声明变量之间的区别？

1 个答案: