如果它不在范围内则放“nil”-puts“nil”ifitisnotinrange

As pointed out by @YuHao, it's an infinite loop when your element isn't found as: low = high = 0 (on the lower end) which makes the loop run indefinitely.

正如@YuHao指出的那样,当你的元素找不到时,它是一个无限循环:low = high = 0(在低端),这使循环无限期地运行。

def binary_search(num, test_array)
  counter = 0
  low = 0
  high = test_array.length

  while (low <= high) do
    i = ((low + high) / 2).floor;
    if num == test_array[i]
      return i
    elsif num  test_array[i]
      low = i
    end
    counter += 1

    break if counter > test_array.length / 2 # could probably be more efficient
  end

  nil
end

test_array = [13, 19, 24, 29, 32, 37, 43]

binary_search(11, test_array) # => nil
binary_search(35, test_array) # => nil

As Yu Hao mentioned, when nothing is found, it goes into an infinite loop. Here is an example of checking for the infinite loop case and return nil.

正如余浩所说,当什么也没发现时,它会进入无限循环。这是检查无限循环情况并返回nil的示例。

def binary_search(num, test_array)
  counter = 0
  low = 0
  high = test_array.length

  while (low <= high) do
    i = ((low + high) / 2).floor;
    if num == test_array[i]
      return i
      break
    elsif num  test_array[i]
      low = i
    end
    counter += 1

    # check for value not found
    if (high - low) <= 1
      return nil
    end

  end
end