Convert a bitstring into a byte array

Question

I can't seem to convert a bit string of 1s and 0s to a byte array.

Here is the code I have so far:

package main

import (
    "strconv"
    "encoding/binary"
    "fmt"
)

func main() {
    /* goal: convert a bit string (ex "10110110") to a byte array */
    bitString := "00000000000000000000000100111000100001100000000000000000000000000000000000000000000000000000000000000000000000000000000000000011"

    bitNum, err := strconv.ParseUint(bitString, 2, 128) // turn my 128-bit bitstring into an int

    if err != nil {
        panic(err) // currently panics with "value out of range"
    }

    // convert my integer to a byte array
    // code from https://stackoverflow.com/questions/16888357/convert-an-integer-to-a-byte-array
    bs := make([]byte, 128)                   // allocate memory for my byte array
    binary.LittleEndian.PutUint64(bs, bitNum) // convert my bitnum to a byte array
    fmt.Println(bs)

}

I am clearly missing something, but I can't seem to convert a bit string of that size to a byte array.

edit I got passed the first error with this:

package main

import (
    "fmt"
    "strconv"
)

func main() {
    /* goal: convert a bit string (ex "10110110") to a byte array */
    bitString := "00000000000000000000000100111000100001100000000000000000000000000000000000000000000000000000000000000000000000000000000000000011"

    myBytes := make([]byte, 16)
    for len(bitString) > 7 {
        currentByteStr := bitString[:8]
        bitString = bitString[8:]
        currentByteInt, _ := strconv.ParseUint(currentByteStr, 2, 8)
        currentByte := byte(currentByteInt)
        myBytes = append(myBytes, currentByte)
    }

    fmt.Println(myBytes)

}

but it doesn't output what I would expect a byte array to look like:

[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 56 134 0 0 0 0 0 0 0 0 0 0 3]

I would have expected it to be hex? is that not what a byte array looks like in golang?

You will probably need to use big integers: golang.org/pkg/math/big — Haroldo_OK
– Haroldo_OK, Commented Feb 6, 2018 at 15:36
I just clicked that link. I'll look at the sizes of those. And yes, I should do it byte by byte i suppose. — Jeff
– Jeff, Commented Feb 6, 2018 at 15:38
Do you want each sequence of 8 bits to be stored as a byte, the whole number as something that can be used as a number, or one byte for each bit? Your sample code sounds like one big number, requiring math/big — Marc
– Marc, Commented Feb 6, 2018 at 15:51
Sorry @Marc, that was unclear. I am wanting an array of bytes. — Jeff
– Jeff, Commented Feb 6, 2018 at 15:54

leaf bebop · Accepted Answer · 2018-02-06 15:59:08Z

3

I think in case of this very specified tasks, it is good to construct the array just by hands.

func main() {
    /* goal: convert a bit string (ex "10110110") to a byte array */
    bitString := "00000000000000000000000100111000100001100000000000000000000000000000000000000000000000000000000000000000000000000000000000000011"

    lenB := len(bitString) / 8 + 1
    bs:=make([]byte,lenB)

    count,i := 0,0
    var now byte
    for _,v:=range bitString {
        if count == 8 {
            bs[i]=now
            i++
            now,count = 0,0
        }
        now = now << 1 + byte(v-'0')
        count++
    }
    if count!=0 {
        bs[i]=now << (8-byte(count))
        i++
    }

    bs=bs[:i:i]
    fmt.Println(bs)

}

The code uses a count to count the numbers of digit consumed since last "flush" and "flush" when there is 8. variable i keeps info on the bytes area and after the loop there might be a final flush.

byte(v-'0') is the converting from rune to a bit.

Playground: https://play.golang.org/p/eB1mc_FjiQc

answered Feb 6, 2018 at 15:59

leaf bebop

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2 Comments

Jeff Over a year ago

Your answer got me the same (except I just wanted my byte array to be a bit larger).

leaf bebop Over a year ago

@Jeff waht do you mean by larger?

hbogert · Accepted Answer · 2021-05-12 15:28:18Z

2

I would opt for making a new type, bitString, with methods for converting to a []byte and []string (for hex if you like). This should also guard against an input string not evenly divisible by 8.

package main

import (
    "encoding/hex"
    "fmt"
    "strconv"
)

type bitString string

func (b bitString) AsByteSlice() []byte {
    var out []byte
    var str string

    for i := len(b); i > 0; i -= 8 {
        if i-8 < 0 {
            str = string(b[0:i])
        } else {
            str = string(b[i-8 : i])
        }
        v, err := strconv.ParseUint(str, 2, 8)
        if err != nil {
            panic(err)
        }
        out = append([]byte{byte(v)}, out...)
    }
    return out
}

func (b bitString) AsHexSlice() []string {
    var out []string
    byteSlice := b.AsByteSlice()
    for _, b := range byteSlice {
        out = append(out, "0x" + hex.EncodeToString([]byte{b}))
    }
    return out
}

func main() {
    x := bitString("00000000000000000000000100111000100001100000000000000000000000000000000000000000000000000000000000000000000000000000000000000011")
    fmt.Println(x.AsByteSlice())
    fmt.Println(x.AsHexSlice())
}

OUTPUT

[64 0 1 56 134 0 0 0 0 0 0 0 0 0 0 3]
[0x00 0x00 0x01 0x38 0x86 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x00 0x03]

Playgroud

If speed is a concern, then optimization can be achieved by allocating the slice with the appropriate capacity and tracking the index yourself.

edited May 12, 2021 at 15:28

hbogert

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answered Feb 6, 2018 at 16:29

sberry

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3 Comments

leaf bebop Over a year ago

The most overhead is in strconv.ParseUint. Not memory allocation.

sberry Over a year ago

@leafbebop fair enough... but that's what pprof is for :).

leaf bebop Over a year ago

And that is auctually what pprof told me before (I can't find the exact code but basically the same).

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