lua - lua 5.1中的uint64到十六进制

Question

嘿，我无法正确进行此转换。我使用下面的代码

RawGUID=17379524724484210731 --It's impossible to store variable this way, it will always convert into test3 eventually. Stored as userdata I cannot change
test1="17379524724484210731"

test2="1.7379524724484e+019"
test3=1.7379524724484e+019

function tohex(num)
    local charset = {"0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"}
    local tmp = {}
    repeat
        table.insert(tmp,1,charset[num%16+1])
        num = math.floor(num/16)
    until num==0
    return table.concat(tmp)
end

RawGUID 是我需要转换为十六进制字符串的示例，其余变量只是相同数字的转换。该代码适用于 64 位以下的任何内容。

老实说，为了实现我的目标，我需要一个十六进制字符串来使我的其余代码正常工作

我得到的结果：F13079A800000000

我想要的结果：F13079A80000002B