当输入是十的倍数时，Lua 函数失败

Question

我有一个 Lua 函数可以将两位整数转换为书面文字，例如“12”变成“十二”。我以为我已经完成了所有工作，除非我输入的数字是十的直接倍数（10、20、30 等），但代码会失败。

我得到的具体错误是

./.lua/num2word:4: attempt to perform arithmetic on local 'number' (a nil value)
stack traceback:
    ./.lua/num2word:4: in function 'num2wordint'
    ./.lua/num2word:117: in main chunk
    [C]: in ?

但我在任何其他数字上都得到了正确的输出。

我尝试重新定义 number 变量的范围，但它要么在所有情况下都破坏了代码，要么对问题情况没有帮助。

我已经看了几个小时的代码和一堆 lua 支持文档，这超出了我的能力。

#! /usr/local/bin/lua

function num2wordint(number)
    local number = number + 0
    local outstring=""
    if number / 10 >= 9 then
        local remainder = number % 10
        if remainder == 0 then
            outstring= outstring.."Ninety"
        else
            outstring= outstring.."Ninety-"
            return outstring, remainder
        end
    elseif number / 10 >= 8 then
        local remainder = number % 10
        if remainder == 0 then
            outstring= outstring.."Eighty"
        else
            outstring= outstring.."Eighty-"
            return outstring, remainder
        end
    elseif number / 10 >= 7 then
        local remainder = number % 10
        if remainder == 0 then
            outstring= outstring.."Seventy"
        else
            outstring= outstring.."Seventy-"
            return outstring, remainder
        end
    elseif number / 10 >= 6 then
        local remainder = number % 10
        if remainder == 0 then
            outstring= outstring.."Sixty"
        else
            outstring= outstring.."Sixty-"
            return outstring, remainder
        end
    elseif number / 10 >= 5 then
        local remainder = number % 10
        if remainder == 0 then
            outstring= outstring.."Fifty"
        else
            outstring= outstring.."Fifty-"
            return outstring, remainder
        end
    elseif number / 10 >= 4 then
        local remainder = number % 10
        if remainder == 0 then
            outstring= outstring.."Forty"
        else
            outstring= outstring.."Forty-"
            return outstring, remainder
        end
    elseif number / 10 >= 3 then
        local remainder = number % 10
        if remainder == 0 then
            outstring= outstring.."Thirty"
        else
            outstring= outstring.."Thirty-"
            return outstring, remainder
        end
    elseif number / 10 >= 2 then
        local remainder = number % 10
        if remainder == 0 then
            outstring= outstring.."Twenty"
        else
            outstring= outstring.."Twenty-"
            return outstring, remainder
        end
    else
        if number == 19 then
            outstring=outstring.."Nineteen"
        elseif number == 18 then
            outstring=outstring.."Eighteen"
        elseif number == 17 then
            outstring=outstring.."Seventeen"
        elseif number == 16 then
            outstring=outstring.."Sixteen"
        elseif number == 15 then
            outstring=outstring.."Fifteen"
        elseif number == 14 then
            outstring=outstring.."Fourteen"
        elseif number == 13 then
            outstring=outstring.."Thirteen"
        elseif number == 12 then
            outstring=outstring.."Twelve"
        elseif number == 11 then
            outstring=outstring.."Eleven"
        elseif number == 10 then
            outstring=outstring.."Ten"
        elseif number == 9 then
            outstring=outstring.."Nine"
        elseif number == 8 then
            outstring=outstring.."Eight"
        elseif number == 7 then
            outstring=outstring.."Seven"
        elseif number == 6 then
            outstring=outstring.."Six"
        elseif number == 5 then
            outstring= outstring .. "Five"
        elseif number == 4 then
            outstring=outstring.."Four"
        elseif number == 3 then
            outstring=outstring.."Three"
        elseif number == 2 then
            outstring=outstring.."Two"
        elseif number == 1 then
            outstring=outstring.."One"
        end
        return outstring, 0
    end
end

local words, leftOver = num2wordint(arg[1])
local i = 1
while i ~= 0 do
    local inword, inrem = num2wordint(leftOver)
    words = words .. inword
    if inrem == 0 then
        i = 0
    else
        leftOver = inrem
    end
end

print(words)

Answer 1

当remainder 为0 时，函数num2wordint 不会返回任何内容。

肮脏而简单的解决方法是将每个 sn-p 转换为：

    if remainder == 0 then
        outstring= outstring.."Ninety"
    else
        outstring= outstring.."Ninety-"
        return outstring, remainder
    end

到

    if remainder == 0 then
        outstring= outstring.."Ninety"
    else
        outstring= outstring.."Ninety-"
    end
    return outstring, remainder

Answer 2

我对此的解决方案是忘记 if 语句，只做一些表索引。不过，我节省了一些空间，只是用一些基本值制定了一些简单的规则：

local num2str
do
   local bases = {[0]="","one","two","three","four","five","six","seven","eight","nine",[10]="ten",[11]="eleven",[12]="twelve",[13]="thirteen",[15]="fifteen",[20]="twenty",[30]="thirty",[40]="fourty",[50]="fifty",[60]="sixty",[70]="seventy",[80]="eighty",[90]="ninety"}

   function num2str(n)
      local s = tostring(n)
      if n < 20 then
         return bases[n] or bases[tonumber(s:sub(2,2))].."teen"
      else
         return bases[tonumber(s:sub(1,1))*10]..bases[tonumber(s:sub(2,2))]
      end
   end
end

for i = 1,99 do
   print(num2str(i))
end