将有符号的 IEEE 754 浮点数转换为十六进制表示 [英] Convert signed IEEE 754 float to hexadecimal representation

问题描述

我正在使用 Lua 的前端，不幸的是它已经过时了，所以我在这里坚持使用 5.1 版，这意味着 bit32 库是遥不可及的(我可能已经使用了转换这个).

I'm using a front-end of Lua which is unfortunately outdated, so I'm stuck with version 5.1 here, meaning the bit32 library is out of reach (which I probably could have used to convert this).

所以我想知道是否有人知道我可以实现浮点到二进制(数字)函数的方法，或者更好的是，浮点到十六进制.到目前为止我能想出的最好的是十进制到二进制/十六进制函数......

So I'm wondering if anyone knows of a way I could implement either a floating-point to binary (digits) function or, better yet, floating-point to hex. The best I've been able to come up with so far is a decimal to binary/hex function...

推荐答案

以下函数使用了 François Perrad 的 lua-MessagePack 中的一些代码.非常感谢他.

The following functions uses some code from François Perrad's lua-MessagePack. A big thank you goes to him.

function float2hex (n)
    if n == 0.0 then return 0.0 end

    local sign = 0
    if n < 0.0 then
        sign = 0x80
        n = -n
    end

    local mant, expo = math.frexp(n)
    local hext = {}

    if mant ~= mant then
        hext[#hext+1] = string.char(0xFF, 0x88, 0x00, 0x00)

    elseif mant == math.huge or expo > 0x80 then
        if sign == 0 then
            hext[#hext+1] = string.char(0x7F, 0x80, 0x00, 0x00)
        else
            hext[#hext+1] = string.char(0xFF, 0x80, 0x00, 0x00)
        end

    elseif (mant == 0.0 and expo == 0) or expo < -0x7E then
        hext[#hext+1] = string.char(sign, 0x00, 0x00, 0x00)

    else
        expo = expo + 0x7E
        mant = (mant * 2.0 - 1.0) * math.ldexp(0.5, 24)
        hext[#hext+1] = string.char(sign + math.floor(expo / 0x2),
                                    (expo % 0x2) * 0x80 + math.floor(mant / 0x10000),
                                    math.floor(mant / 0x100) % 0x100,
                                    mant % 0x100)
    end

    return tonumber(string.gsub(table.concat(hext),"(.)",
                                function (c) return string.format("%02X%s",string.byte(c),"") end), 16)
end


function hex2float (c)
    if c == 0 then return 0.0 end
    local c = string.gsub(string.format("%X", c),"(..)",function (x) return string.char(tonumber(x, 16)) end)
    local b1,b2,b3,b4 = string.byte(c, 1, 4)
    local sign = b1 > 0x7F
    local expo = (b1 % 0x80) * 0x2 + math.floor(b2 / 0x80)
    local mant = ((b2 % 0x80) * 0x100 + b3) * 0x100 + b4

    if sign then
        sign = -1
    else
        sign = 1
    end

    local n

    if mant == 0 and expo == 0 then
        n = sign * 0.0
    elseif expo == 0xFF then
        if mant == 0 then
            n = sign * math.huge
        else
            n = 0.0/0.0
        end
    else
        n = sign * math.ldexp(1.0 + mant / 0x800000, expo - 0x7F)
    end

    return n
end

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