How to shift byte array with Golang?

Here simple working code to left shift first bit of a byte

package main

import (
	"fmt"
)

type Byte byte

func SL(b Byte) Byte {
	if b&0x80 == 0x80 {
		b <<= 1
		b ^= 0x01
	} else {
		b <<= 1
	}
	return b
}

func main() {
	var b Byte
	b = 0xD3
	fmt.Printf("old byte %#08b\n", b) // 11010011
	c := SL(b)
	fmt.Printf("new byte %#08b", c)   // 10100111
}

What should I do to shift array of bytes, like
type Byte [2]byte?

Thanks for advance!

You appear to want to rotate, not shift. Any particular reason you aren't using a uint16 type instead of [2]byte?

Anyway, if you really want [2]byte, this is simpler and doesn't branch:

func rol(v [2]byte) [2]byte {
	x := int(v[0])<<8 | int(v[1])
	x <<= 1
	v[0] = byte(x >> 8)
	v[1] = byte((x & 0xff) | x>>16)
	return v
}

If you want to do such operations on an arbitrary large number of bits you could use math/big.

A solution to shift left 1 bit.

func shiftBytesLeft(a []byte) (dst []byte) {
	n := len(a)
	dst = make([]byte, n)
	for i := 0; i < n-1; i++ {
		dst[i] = a[i] << 1
		dst[i] = (dst[i] & 0xfe) | (a[i+1] >> 7)
	}
	dst[n-1] = a[n-1] << 1
	return dst
}

Here's an implementation that can do both left and right shifts:

// ShiftLeft performs a left bit shift operation on the provided bytes.
// If the bits count is negative, a right bit shift is performed.
func ShiftLeft(data []byte, bits int) {
	n := len(data)
	if bits < 0 {
		bits = -bits
		for i := n - 1; i > 0; i-- {
			data[i] = data[i]>>bits | data[i-1]<<(8-bits)
		}
		data[0] >>= bits
	} else {
		for i := 0; i < n-1; i++ {
			data[i] = data[i]<<bits | data[i+1]>>(8-bits)
		}
		data[n-1] <<= bits
	}
}

Yep! I found a solution.

package main

import (
	"fmt"
)

type Byte [2]byte

//shift left
func SL(b Byte) Byte {
	if b[0]&0x80 == 0x80 {
		b[0] <<= 1
		if b[1]&0x80 == 0x80 {
			b[0] ^= 1
			b[1] <<= 1
		} else {
			b[1] <<= 1
		}
		b[1] ^= 0x01
	} else {
		b[0] <<= 1
		if b[1]&0x80 == 0x80 {
			b[0] ^= 1
			b[1] <<= 1
		} else {
			b[1] <<= 1
		}
	}
	return b
}

func main() {
	//var b Byte
	b := Byte{0x23, 0x86}
	fmt.Printf("old byte %#08b %#08b\n", b[0], b[1]) // 00100011 10000110
	c := SL(b)
	fmt.Printf("new byte %#08b %#08b", c[0], c[1]) // 01000111 00001100
}

shift left by multiple bits, expanding qin's answer:

func shiftBytesLeft(a []byte, byBits int) (dst []byte) {
    n := len(a)
    dst = make([]byte, n)
    for i := 0; i < n-1; i++ {
        dst[i] = a[i] << byBits
        dst[i] = (dst[i] & (0xff - byte(byBits))) | (a[i+1] >> (8 - byte(byBits)))
    }
    dst[n-1] = a[n-1] << byBits
    return dst
}