Convert an integer to a byte array

I have a function which receives a []byte but what I have is an int, what is the best way to go about this conversion ?

err = a.Write([]byte(myInt))

I guess I could go the long way and get it into a string and put that into bytes, but it sounds ugly and I guess there are better ways to do it.

I agree with Brainstorm's approach: assuming that you're passing a machine-friendly binary representation, use the encoding/binary library. The OP suggests that binary.Write() might have some overhead. Looking at the source for the implementation of Write(), I see that it does some runtime decisions for maximum flexibility.

func Write(w io.Writer, order ByteOrder, data interface{}) error {
    // Fast path for basic types.
    var b [8]byte
    var bs []byte
    switch v := data.(type) {
    case *int8:
        bs = b[:1]
        b[0] = byte(*v)
    case int8:
        bs = b[:1]
        b[0] = byte(v)
    case *uint8:
        bs = b[:1]
        b[0] = *v
    ...

Right? Write() takes in a very generic data third argument, and that's imposing some overhead as the Go runtime then is forced into encoding type information. Since Write() is doing some runtime decisions here that you simply don't need in your situation, maybe you can just directly call the encoding functions and see if it performs better.

Something like this:

package main

import (
	"encoding/binary"
	"fmt"
)

func main() {
	bs := make([]byte, 4)
	binary.LittleEndian.PutUint32(bs, 31415926)
	fmt.Println(bs)
}

Let us know how this performs.

Otherwise, if you're just trying to get an ASCII representation of the integer, you can get the string representation (probably with strconv.Itoa) and cast that string to the []byte type.

package main

import (
	"fmt"
	"strconv"
)

func main() {
	bs := []byte(strconv.Itoa(31415926))
	fmt.Println(bs)
}

Check out the "encoding/binary" package. Particularly the Read and Write functions:

binary.Write(a, binary.LittleEndian, myInt)

Sorry, this might be a bit late. But I think I found a better implementation on the go docs.

buf := new(bytes.Buffer)
var num uint16 = 1234
err := binary.Write(buf, binary.LittleEndian, num)
if err != nil {
	fmt.Println("binary.Write failed:", err)
}
fmt.Printf("% x", buf.Bytes())

i thought int type has any method for getting int hash to bytes, but first i find math / big method for this
https://golang.org/pkg/math/big/

var f int = 52452356235;          // int
var s = big.NewInt(int64(f))      // int to big Int
var b = s.Bytes()                 // big Int to bytes
// b - byte slise
var r = big.NewInt(0).SetBytes(b) // bytes to big Int
var i int = int(r.Int64())        // big Int to int

https://play.golang.org/p/VAKSGw8XNQq

However, this method uses an absolute value.
If you spend 1 byte more, you can transfer the sign

func IntToBytes(i int) []byte{
	if i > 0 {
		return append(big.NewInt(int64(i)).Bytes(), byte(1))
	}
	return append(big.NewInt(int64(i)).Bytes(), byte(0))
}
func BytesToInt(b []byte) int{
	if b[len(b)-1]==0 {
		return -int(big.NewInt(0).SetBytes(b[:len(b)-1]).Int64())
	}
	return int(big.NewInt(0).SetBytes(b[:len(b)-1]).Int64())
}

https://play.golang.org/p/mR5Sp5hu4jk

or new(https://play.golang.org/p/7ZAK4QL96FO)

(The package also provides functions for fill into an existing slice)

https://golang.org/pkg/math/big/#Int.FillBytes

Adding this option for dealing with basic uint8 to byte[] conversion

foo := 255 // 1 - 255
ufoo := uint16(foo) 
far := []byte{0,0}
binary.LittleEndian.PutUint16(far, ufoo)
bar := int(far[0]) // back to int
fmt.Println("foo, far, bar : ",foo,far,bar)

output :
foo, far, bar : 255 [255 0] 255

Here is another option, based on the Go source code 1:

package main

import (
   "encoding/binary"
   "fmt"
   "math/bits"
)

func encodeUint(x uint64) []byte {
   buf := make([]byte, 8)
   binary.BigEndian.PutUint64(buf, x)
   return buf[bits.LeadingZeros64(x) >> 3:]
}

func main() {
   for x := 0; x <= 64; x += 8 {
      buf := encodeUint(1<<x-1)
      fmt.Println(buf)
   }
}

Result:

[]
[255]
[255 255]
[255 255 255]
[255 255 255 255]
[255 255 255 255 255]
[255 255 255 255 255 255]
[255 255 255 255 255 255 255]
[255 255 255 255 255 255 255 255]

Much faster than math/big:

BenchmarkBig-12         28348621                40.62 ns/op
BenchmarkBit-12         731601145                1.641 ns/op

1. https://github.com/golang/go/blob/go1.16.5/src/encoding/gob/encode.go#L113-L117

Convert Integer to byte slice.

import (
	"bytes"
	"encoding/binary"
	"log"
)

func IntToBytes(num int64) []byte {
	buff := new(bytes.Buffer)
	bigOrLittleEndian := binary.BigEndian
	err := binary.Write(buff, bigOrLittleEndian, num)
	if err != nil {
		log.Panic(err)
	}

	return buff.Bytes()
}

Maybe the simple way is using protobuf, see the Protocol Buffer Basics: Go

define message like

message MyData {
  int32 id = 1;
}

get more in Defining your protocol format

// Write
out, err := proto.Marshal(mydata)

read more in Writing a Message

Try math/big package to convert bytes array to int and to convert int to bytes array.

package main

import (
	"fmt"
	"math/big"
)

func main() {
	// Convert int to []byte
	var int_to_encode int64 = 65535
	var bytes_array []byte = big.NewInt(int_to_encode).Bytes()
	fmt.Println("bytes array", bytes_array)

	// Convert []byte to int
	var decoded_int int64 = new(big.Int).SetBytes(bytes_array).Int64()
	fmt.Println("decoded int", decoded_int)
}




</details>



# 答案10
**得分**: 0

这是最直接（也是最短（也是最安全）（也可能是最高效））的方法：

`buf.Bytes()` 的类型是字节切片。

```go
	var val uint32 = 42
	buf := new(bytes.Buffer)
	err := binary.Write(buf, binary.LittleEndian, val)
	if err != nil {
		fmt.Println("binary.Write 失败：", err)
	}
	fmt.Printf("% x\n", buf.Bytes())

参见 https://stackoverflow.com/a/74819602/589493

	var val uint32 = 42
	buf := new(bytes.Buffer)
	err := binary.Write(buf, binary.LittleEndian, val)
	if err != nil {
		fmt.Println("binary.Write failed:", err)
	}
	fmt.Printf("% x\n", buf.Bytes())

答案11

得分: -6

将其转换为字符串有什么问题吗？

[]byte(fmt.Sprintf("%d", myint))

[]byte(fmt.Sprintf("%d", myint))