有人分享了一些关于他们如何完成挑战的Ruby代码，您必须将数字转换为单词。这个解决方案是迄今为止最简单的解决方案，但我并不完全理解它的递归方面。具体来说，在下面的行中......
elsif int.to_s.length == 1 && int/num > 0
return str + "#{name}"
如果一个return语句将代码从循环中分解出来，为什么这个代码不会执行并在一旦为真时中断循环？例如，当int = 4且num = 1时，这将被证明是真的并触发return语句，但代码一直持续到num = 4。
对代码的这一方面的任何澄清，以及理解此解决方案的任何一般性建议都会有所帮助。仍在努力理解递归。感谢
def in_words(int)
numbers_to_name = {
1000000 => "million",
1000 => "thousand",
100 => "hundred",
90 => "ninety",
80 => "eighty",
70 => "seventy",
60 => "sixty",
50 => "fifty",
40 => "forty",
30 => "thirty",
20 => "twenty",
19=>"nineteen",
18=>"eighteen",
17=>"seventeen",
16=>"sixteen",
15=>"fifteen",
14=>"fourteen",
13=>"thirteen",
12=>"twelve",
11 => "eleven",
10 => "ten",
9 => "nine",
8 => "eight",
7 => "seven",
6 => "six",
5 => "five",
4 => "four",
3 => "three",
2 => "two",
1 => "one"
}
str = ""
numbers_to_name.each do |num, name|
if int == 0
return str
elsif int.to_s.length == 1 && int/num > 0
return str + "#{name}"
elsif int < 100 && int/num > 0
return str + "#{name}" if int%num == 0
return str + "#{name} " + in_words(int%num)
elsif int/num > 0
return str + in_words(int/num) + " #{name} " + in_words(int%num)
end
end
end
puts in_words(4) == "four"
puts in_words(27) == "twenty seven"
puts in_words(102) == "one hundred two"
puts in_words(38_079) == "thirty eight thousand seventy nine"
puts in_words(82102713) == "eighty two million one hundred two thousand seven hundred thirteen"
答案 0 :(得分：1)
这是一个修改版本，每个分支都有调试行：
def in_words(int, indent="")
puts "#{indent}in_words(#{int})"
numbers_to_name = {
1000000 => "million",
1000 => "thousand",
100 => "hundred",
90 => "ninety",
80 => "eighty",
70 => "seventy",
60 => "sixty",
50 => "fifty",
40 => "forty",
30 => "thirty",
20 => "twenty",
19=>"nineteen",
18=>"eighteen",
17=>"seventeen",
16=>"sixteen",
15=>"fifteen",
14=>"fourteen",
13=>"thirteen",
12=>"twelve",
11 => "eleven",
10 => "ten",
9 => "nine",
8 => "eight",
7 => "seven",
6 => "six",
5 => "five",
4 => "four",
3 => "three",
2 => "two",
1 => "one"
}
str = ""
numbers_to_name.each do |num, name|
puts "#{indent} testing #{num}"
if int == 0
puts "#{indent} Empty string"
return str
elsif int.to_s.length == 1 && int/num > 0
puts "#{indent} Single digit found (#{name})!"
return str + "#{name}"
elsif int < 100 && int/num > 0
puts "#{indent} Double digits found (#{name})!"
return str + "#{name}" if int%num == 0
puts "#{indent} Recursive call with #{int%num} :"
return str + "#{name} " + in_words(int%num, indent+" ")
elsif int/num > 0
puts "#{indent} Recursive call with #{int/num} and #{int%num} :"
indent += " "
return str + in_words(int/num, indent) + " #{name} " + in_words(int%num,indent)
end
end
end
对于in_words(4)，输出：
in_words(4)
testing 1000000
testing 1000
testing 100
testing 90
testing 80
testing 70
testing 60
testing 50
testing 40
testing 30
testing 20
testing 19
testing 18
testing 17
testing 16
testing 15
testing 14
testing 13
testing 12
testing 11
testing 10
testing 9
testing 8
testing 7
testing 6
testing 5
testing 4
Single digit found (four)!
four
对于27：
in_words(27)
testing 1000000
testing 1000
testing 100
testing 90
testing 80
testing 70
testing 60
testing 50
testing 40
testing 30
testing 20
Double digits found (twenty)!
Recursive call with 7 :
in_words(7)
testing 1000000
testing 1000
testing 100
testing 90
testing 80
testing 70
testing 60
testing 50
testing 40
testing 30
testing 20
testing 19
testing 18
testing 17
testing 16
testing 15
testing 14
testing 13
testing 12
testing 11
testing 10
testing 9
testing 8
testing 7
Single digit found (seven)!
twenty seven
例如，当int = 4且num = 1时，这将证明是真的 触发return语句，但代码一直持续到num = 4。
以4作为输入，num = 1永远不会发生。 numbers_to_name按递减顺序排列，因此返回语句将发生在num=4，并且该方法将不会执行任何其他代码。