将签名的IEEE 754浮点数转换为十六进制表示

Question

我使用的Lua,前端是不幸的是过时的,所以我坚持与5.1版本在这里,这意味着bit32库是遥不可及的(这我可能已经习惯了这种转换).

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,我能够提出的最好的是十进制到二进制/十六进制函数......

Answer 1

以下函数使用 Fran\xc3\xa7ois Perrad\ 的一些代码lua-MessagePack。非常感谢他。

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function float2hex (n)\n    if n == 0.0 then return 0.0 end\n\n    local sign = 0\n    if n < 0.0 then\n        sign = 0x80\n        n = -n\n    end\n\n    local mant, expo = math.frexp(n)\n    local hext = {}\n\n    if mant ~= mant then\n        hext[#hext+1] = string.char(0xFF, 0x88, 0x00, 0x00)\n\n    elseif mant == math.huge or expo > 0x80 then\n        if sign == 0 then\n            hext[#hext+1] = string.char(0x7F, 0x80, 0x00, 0x00)\n        else\n            hext[#hext+1] = string.char(0xFF, 0x80, 0x00, 0x00)\n        end\n\n    elseif (mant == 0.0 and expo == 0) or expo < -0x7E then\n        hext[#hext+1] = string.char(sign, 0x00, 0x00, 0x00)\n\n    else\n        expo = expo + 0x7E\n        mant = (mant * 2.0 - 1.0) * math.ldexp(0.5, 24)\n        hext[#hext+1] = string.char(sign + math.floor(expo / 0x2),\n                                    (expo % 0x2) * 0x80 + math.floor(mant / 0x10000),\n                                    math.floor(mant / 0x100) % 0x100,\n                                    mant % 0x100)\n    end\n\n    return tonumber(string.gsub(table.concat(hext),"(.)",\n                                function (c) return string.format("%02X%s",string.byte(c),"") end), 16)\nend\n\n\nfunction hex2float (c)\n    if c == 0 then return 0.0 end\n    local c = string.gsub(string.format("%X", c),"(..)",function (x) return string.char(tonumber(x, 16)) end)\n    local b1,b2,b3,b4 = string.byte(c, 1, 4)\n    local sign = b1 > 0x7F\n    local expo = (b1 % 0x80) * 0x2 + math.floor(b2 / 0x80)\n    local mant = ((b2 % 0x80) * 0x100 + b3) * 0x100 + b4\n\n    if sign then\n        sign = -1\n    else\n        sign = 1\n    end\n\n    local n\n\n    if mant == 0 and expo == 0 then\n        n = sign * 0.0\n    elseif expo == 0xFF then\n        if mant == 0 then\n            n = sign * math.huge\n        else\n            n = 0.0/0.0\n        end\n    else\n        n = sign * math.ldexp(1.0 + mant / 0x800000, expo - 0x7F)\n    end\n\n    return n\nend\n

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