如何在lua中将32字节字符缓冲区转换为uint64?
我正在编写一种机制来生成从sha256派生的唯一ID。我使用:如何在lua中将32字节字符缓冲区转换为uint64?,lua,char,buffer,uint64,Lua,Char,Buffer,Uint64,我正在编写一种机制来生成从sha256派生的唯一ID。我使用: require 'sha2' function getUint64ID( callId ) local minimumValue = 100000000000000000 local maximumValue = 999999999999999999 outputHash = sha2.sha256hex(callId) print(outputHash .. "\n") local c1 = ""
require 'sha2'
function getUint64ID( callId )
local minimumValue = 100000000000000000
local maximumValue = 999999999999999999
outputHash = sha2.sha256hex(callId)
print(outputHash .. "\n")
local c1 = ""
for a, b in outputHash:gmatch"(%x)(%x)" do
hexTuple = tostring(a) .. tostring(b)
intVal = tonumber(hexTuple, 16)
c1 = c1 .. string.char(intVal)
end
uint64ID = ( minimumValue + ( tonumber(c1) % ( maximumValue - minimumValue + 1 ) ) )
print('uint64ID:')
print(string.format("%18.0f",uint64ID))
end
getUint64ID("test")
我在上面的代码中得到以下错误:
stdin:17: attempt to perform arithmetic on a nil value
stack traceback:
stdin:17: in function 'getUint64ID'
stdin:1: in main chunk
[C]: ?
如何在lua中将32字节字符缓冲区(c1)转换为uint64数字?如何只执行
uint64ID=minimumValue+tonumber(输出:sub(1,14),16)
?tonumber(c1)
是nil
,因为c1
包含任意符号(而不是十进制数字)