Hex or binary to float

General
1

Is there someone sitting on a conversion snippert that could perform the above on a 8 byte array ?

For example 32-bit hex 79 E9 F6 42 should convert to float 123.456

Little or big endian set properly of course.

I can do it using Processing but not been able to in Lua yet.

Re Peter

2

This probably isn’t what you want, but this is an easy way to convert regular hex to decimal.


function setup()
    print(tonumber("ab", 16))
    print(tonumber("abcd", 16))
    print(tonumber("abcdef", 16))
    print(tonumber("79E9F642",16))   
end
3

There’s a lua binary-decimal-hexadecimal library on the web that works in Codea. I put it in my library: http://www.math.ntnu.no/~stacey/code/CodeaLibrary

4

Here’s the routine. The conversion is dependent upon the precision of Codea which is only about 6 to 7 digits.

EDIT: changed code to one function.


-- convert hex to floating point.

function setup()
    tab={["0"]="0000",["1"]="0001",["2"]="0010",["3"]="0011",
         ["4"]="0100",["5"]="0101",["6"]="0110",["7"]="0111",
         ["8"]="1000",["9"]="1001",["a"]="1010",["b"]="1011",
         ["c"]="1100",["d"]="1101",["e"]="1110",["f"]="1111",
         ["A"]="1010",["B"]="1011",["C"]="1100",["D"]="1101",["E"]="1110",["F"]="1111"}
        
    print(hexToFloat("C58EBF11"))    -- -4567.89
    print(hexToFloat("42f6e979"))    -- 123.456
    print(hexToFloat("3f9e0610"))    -- 1.23456
end

function hexToFloat(str)
    local str1=""
    local a,z
    for z=1,string.len(str) do
        a=string.sub(str,z,z)
        str1=str1..tab[a]
    end
    local pm=string.sub(str1,1,1)
    local exp=string.sub(str1,2,9)
    local c=tonumber(exp,2)-127
    local p=math.pow(2,c)
    local man="1"..string.sub(str1,10,32)
    local x=0
    for z=1,string.len(man) do
        if string.sub(man,z,z)=="1" then
            x=x+p
        end
        p=p/2
    end
    if pm=="1" then
        x= -x
    end    
    return(x)
end
5

.@macflyerdk I’m writing a routine now. It kind of works so far I think, but I need to add code to look for error conditions and out of range numbers. I converted these 2 hex numbers. be580000 = .210938 44361000 = 728.25 . As for your 79e9f642 I get 1.5185e+35. Do you know what the value is supposed to be.

6

Nice work and thanks for taking the time dave1707 :slight_smile:
Sorry for my late reply, but Easter came in the way

First it did not work in my setup, but when I reversed the byte array to little endian format instead, all went well.

I have added this line to your function and one more parameter passed (endian)

if endian == "little" then -- big endian or little endian (reversed array)
str = string.sub(str,7,8)..string.sub(str,5,6)..string.sub(str,3,4)..string.sub(str,1,2) end

Do we have a place for code snippets like this ?
It can be a huge time and frustration saver, if these things could be collected and saved somewhere.

Re Peter

Edit: Nearly forgot. My number 79e9f642 was in the little endian format so reversing it to 42f6e979 results the number in my original post. Hence my added line to your function.

7

. @macflyerdk Thanks for the original post. I learned something about floating point hex and the code turned out easier than I thought it would. I never thought about little and big endian when I was doing the code. I looked on the Internet for the hex floating point format and went from there. One of my examples is 123.456 and I didn’t realize it was the reverse of what you had in your example. I like writing code for different things, it keeps me thinking.