I want to assign string to bytes array:
var arr [20]byte
str := "abc"
for k, v := range []byte(str) {
arr[k] = byte(v)
}
Have another method?
Safe and simple:
[]byte("Here is a string....")
For converting from a string to a byte slice, string -> []byte:
[]byte(str)
For converting an array to a slice, [20]byte -> []byte:
arr[:]
For copying a string to an array, string -> [20]byte:
copy(arr[:], str)
Same as above, but explicitly converting the string to a slice first:
copy(arr[:], []byte(str))
The built-in copy function only copies to a slice, from a slice.
Arrays are "the underlying data", while slices are "a viewport into underlying data".
Using [:] makes an array qualify as a slice.
A string does not qualify as a slice that can be copied to, but it qualifies as a slice that can be copied from (strings are immutable).
If the string is too long, copy will only copy the part of the string that fits (and multi-byte runes may then be copied only partly, which will corrupt the last rune of the resulting string).
This code:
var arr [20]byte
copy(arr[:], "abc")
fmt.Printf("array: %v (%T)\n", arr, arr)
...gives the following output:
array: [97 98 99 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0] ([20]uint8)
I also made it available at the Go Playground
For example,
package main
import "fmt"
func main() {
s := "abc"
var a [20]byte
copy(a[:], s)
fmt.Println("s:", []byte(s), "a:", a)
}
Output:
s: [97 98 99] a: [97 98 99 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
Piece of cake:
arr := []byte("That's all folks!!")
I think it's better..
package main
import "fmt"
func main() {
str := "abc"
mySlice := []byte(str)
fmt.Printf("%v -> '%s'",mySlice,mySlice )
}
Check here: http://play.golang.org/p/vpnAWHZZk7
Go, convert a string to a bytes slice
You need a fast way to convert a []string to []byte type. To use in situations such as storing text data into a random access file or other type of data manipulation that requires the input data to be in []byte type.
package main
func main() {
var s string
//...
b := []byte(s)
//...
}
which is useful when using ioutil.WriteFile, which accepts a bytes slice as its data parameter:
WriteFile func(filename string, data []byte, perm os.FileMode) error
Another example
package main
import (
"fmt"
"strings"
)
func main() {
stringSlice := []string{"hello", "world"}
stringByte := strings.Join(stringSlice, " ")
// Byte array value
fmt.Println([]byte(stringByte))
// Corresponding string value
fmt.Println(string([]byte(stringByte)))
}
Output:
[104 101 108 108 111 32 119 111 114 108 100] hello world
Please check the link playground
Besides the methods mentioned above, you can also do a trick as
s := "hello"
b := *(*[]byte)(unsafe.Pointer((*reflect.SliceHeader)(unsafe.Pointer(&s))))
Go Play: http://play.golang.org/p/xASsiSpQmC
You should never use this :-)
Ended up creating array specific methods to do this. Much like the encoding/binary package with specific methods for each int type. For example binary.BigEndian.PutUint16([]byte, uint16).
func byte16PutString(s string) [16]byte {
var a [16]byte
if len(s) > 16 {
copy(a[:], s)
} else {
copy(a[16-len(s):], s)
}
return a
}
var b [16]byte
b = byte16PutString("abc")
fmt.Printf("%v\n", b)
Output:
[0 0 0 0 0 0 0 0 0 0 0 0 0 97 98 99]
Notice how I wanted padding on the left, not the right.
http://play.golang.org/p/7tNumnJaiN
Arrays are values... slices are more like pointers. That is [n]type is not compatible with []type as they are fundamentally two different things. You can get a slice that points to an array by using arr[:] which returns a slice that has arr as it's backing storage.
One way to convert a slice of for example []byte to [20]byte is to actually allocate a [20]byte which you can do by using var [20]byte (as it's a value... no make needed) and then copy data into it:
buf := make([]byte, 10)
var arr [10]byte
copy(arr[:], buf)
Essentially what a lot of other answers get wrong is that []type is NOT an array.
[n]T and []T are completely different things!
When using reflect []T is not of kind Array but of kind Slice and [n]T is of kind Array.
You also can't use map[[]byte]T but you can use map[[n]byte]T.
This can sometimes be cumbersome because a lot of functions operate for example on []byte whereas some functions return [n]byte (most notably the hash functions in crypto/*).
A sha256 hash for example is [32]byte and not []byte so when beginners try to write it to a file for example:
sum := sha256.Sum256(data)
w.Write(sum)
they will get an error. The correct way of is to use
w.Write(sum[:])
However, what is it that you want? Just accessing the string bytewise? You can easily convert a string to []byte using:
bytes := []byte(str)
but this isn't an array, it's a slice. Also, byte != rune. In case you want to operate on "characters" you need to use rune... not byte.
If someone is looking for a quick consider use unsafe conversion between slices, you can refer to the following comparison.
package demo_test
import (
"testing"
"unsafe"
)
var testStr = "hello world"
var testBytes = []byte("hello world")
// Avoid copying the data.
func UnsafeStrToBytes(s string) []byte {
return *(*[]byte)(unsafe.Pointer(&s))
}
// Avoid copying the data.
func UnsafeBytesToStr(b []byte) string {
return *(*string)(unsafe.Pointer(&b))
}
func Benchmark_UnsafeStrToBytes(b *testing.B) {
for i := 0; i < b.N; i++ {
_ = UnsafeStrToBytes(testStr)
}
}
func Benchmark_SafeStrToBytes(b *testing.B) {
for i := 0; i < b.N; i++ {
_ = []byte(testStr)
}
}
func Benchmark_UnSafeBytesToStr(b *testing.B) {
for i := 0; i < b.N; i++ {
_ = UnsafeBytesToStr(testBytes)
}
}
func Benchmark_SafeBytesToStr(b *testing.B) {
for i := 0; i < b.N; i++ {
_ = string(testBytes)
}
}
go test -v -bench="^Benchmark" -run=none
output
cpu: Intel(R) Core(TM) i7-8565U CPU # 1.80GHz
Benchmark_UnsafeStrToBytes
Benchmark_UnsafeStrToBytes-8 1000000000 0.2465 ns/op
Benchmark_SafeStrToBytes
Benchmark_SafeStrToBytes-8 289119562 4.181 ns/op
Benchmark_UnSafeBytesToStr
Benchmark_UnSafeBytesToStr-8 1000000000 0.2530 ns/op
Benchmark_SafeBytesToStr
Benchmark_SafeBytesToStr-8 342842938 3.623 ns/op
PASS
Related
I generated a hash value using sha3 and I need to convert it to a big.Int value. Is it possible ? or is there a method to get the integervalue of the hash ?
the following code throws an error that cannot convert type hash.Hash to type int64 :
package main
import (
"math/big"
"golang.org/x/crypto/sha3"
"fmt"
)
func main(){
chall := "hello word"
b := byte[](chall)
h := sha3.New244()
h.Write(chall)
h.Write(b)
d := make([]byte, 16)
h.Sum(d)
val := big.NewInt(int64(h))
fmt.Println(val)
}
TL;DR;
sha3.New224() cannot be represented in uint64 type.
There are many hash types - and of differing sizes. Go standard library picks a very generic interface to cover all type of hashes: https://golang.org/pkg/hash/#Hash
type Hash interface {
io.Writer
Sum(b []byte) []byte
Reset()
Size() int
BlockSize() int
}
Having said that some Go hash implementations optionally include extra methods like hash.Hash64:
type Hash64 interface {
Hash
Sum64() uint64
}
others may implement encoding.BinaryMarshaler:
type BinaryMarshaler interface {
MarshalBinary() (data []byte, err error)
}
which one can use to preserve a hash state.
sha3.New224() does not implement the above 2 interfaces, but crc64 hash does.
To do a runtime check:
h64, ok := h.(hash.Hash64)
if ok {
fmt.Printf("64-bit: %d\n", h64.Sum64())
}
Working example: https://play.golang.org/p/uLUfw0gMZka
(See Peter's comment for the simpler version of this.)
Interpreting a series of bytes as a big.Int is the same as interpreting a series of decimal digits as an arbitrarily large number. For example, to convert the digits 1234 into a "number", you'd do this:
Start with 0
Multiply by 10 = 0
Add 1 = 1
Multiply by 10 = 10
Add 2 = 12
Multiply by 10 = 120
Add 3 = 123
Multiply by 10 = 1230
Add 4 = 1234
The same applies to bytes. The "digits" are just base-256 rather than base-10:
val := big.NewInt(0)
for i := 0; i < h.Size(); i++ {
val.Lsh(val, 8)
val.Add(val, big.NewInt(int64(d[i])))
}
(Lsh is a left-shift. Left shifting by 8 bits is the same as multiplying by 256.)
Playground
Is there a built-in function to convert a uint to a slice of binary integers {0,1} ?
>> convert_to_binary(2)
[1, 0]
I am not aware of such a function, however you can use strconv.FormatUint for that purpose.
Example (on play):
func Bits(i uint64) []byte {
bits := []byte{}
for _, b := range strconv.FormatUint(i, 2) {
bits = append(bits, byte(b - rune('0')))
}
return bits
}
FormatUint will return the string representation of the given uint to a base, in this case 2, so we're encoding it in binary. So the returned string for i=2 looks like this: "10". In bytes this is [49 48] as 1 is 49 and 0 is 48 in ASCII and Unicode. So we just need to iterate over the string, subtracting 48 from each rune (unicode character) and converting it to a byte.
Here is another method:
package main
import (
"bytes"
"fmt"
"math/bits"
)
func unsigned(x uint) []byte {
b := make([]byte, bits.UintSize)
for i := range b {
if bits.LeadingZeros(x) == 0 {
b[i] = 1
}
x = bits.RotateLeft(x, 1)
}
return b
}
func trimUnsigned(x uint) []byte {
return bytes.TrimLeft(unsigned(x), string(0))
}
func main() {
b := trimUnsigned(2)
fmt.Println(b) // [1 0]
}
https://golang.org/pkg/math/bits#LeadingZeros
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
https://github.com/golang/go/blob/go1.16.5/src/encoding/gob/encode.go#L113-L117
You can try musgo_int. All you need to do is to cast your variable:
package main
import (
"github.com/ymz-ncnk/musgo_int"
)
func main() {
var myInt int = 1234
// from int to []byte
buf := make([]byte, musgo_int.Int(myInt).SizeMUS())
musgo_int.Int(myInt).MarshalMUS(buf)
// from []byte to int
_, err := (*musgo_int.Int)(&myInt).UnmarshalMUS(buf)
if err != nil {
panic(err)
}
}
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)
}
This is the most straight forward (and shortest (and safest) (and maybe most performant)) way:
buf.Bytes() is of type bytes slice.
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())
see also https://stackoverflow.com/a/74819602/589493
What's wrong with converting it to a string?
[]byte(fmt.Sprintf("%d", myint))
I need to read [100]byte to transfer a bunch of string data.
Because not all of the strings are precisely 100 characters long, the remaining part of the byte array is padded with 0s.
If I convert [100]byte to string by: string(byteArray[:]), the tailing 0s are displayed as ^#^#s.
In C, the string will terminate upon 0, so what's the best way to convert this byte array to string in Go?
Methods that read data into byte slices return the number of bytes read. You should save that number and then use it to create your string. If n is the number of bytes read, your code would look like this:
s := string(byteArray[:n])
To convert the full string, this can be used:
s := string(byteArray[:len(byteArray)])
This is equivalent to:
s := string(byteArray[:])
If for some reason you don't know n, you could use the bytes package to find it, assuming your input doesn't have a null character embedded in it.
n := bytes.Index(byteArray[:], []byte{0})
Or as icza pointed out, you can use the code below:
n := bytes.IndexByte(byteArray[:], 0)
Use:
s := string(byteArray[:])
Simplistic solution:
str := fmt.Sprintf("%s", byteArray)
I'm not sure how performant this is though.
For example,
package main
import "fmt"
func CToGoString(c []byte) string {
n := -1
for i, b := range c {
if b == 0 {
break
}
n = i
}
return string(c[:n+1])
}
func main() {
c := [100]byte{'a', 'b', 'c'}
fmt.Println("C: ", len(c), c[:4])
g := CToGoString(c[:])
fmt.Println("Go:", len(g), g)
}
Output:
C: 100 [97 98 99 0]
Go: 3 abc
The following code is looking for '\0', and under the assumptions of the question the array can be considered sorted since all non-'\0' precede all '\0'. This assumption won't hold if the array can contain '\0' within the data.
Find the location of the first zero-byte using a binary search, then slice.
You can find the zero-byte like this:
package main
import "fmt"
func FirstZero(b []byte) int {
min, max := 0, len(b)
for {
if min + 1 == max { return max }
mid := (min + max) / 2
if b[mid] == '\000' {
max = mid
} else {
min = mid
}
}
return len(b)
}
func main() {
b := []byte{1, 2, 3, 0, 0, 0}
fmt.Println(FirstZero(b))
}
It may be faster just to naively scan the byte array looking for the zero-byte, especially if most of your strings are short.
When you do not know the exact length of non-nil bytes in the array, you can trim it first:
string(bytes.Trim(arr, "\x00"))
Use this:
bytes.NewBuffer(byteArray).String()
Only use for performance tuning.
package main
import (
"fmt"
"reflect"
"unsafe"
)
func BytesToString(b []byte) string {
return *(*string)(unsafe.Pointer(&b))
}
func StringToBytes(s string) []byte {
return *(*[]byte)(unsafe.Pointer(&s))
}
func main() {
b := []byte{'b', 'y', 't', 'e'}
s := BytesToString(b)
fmt.Println(s)
b = StringToBytes(s)
fmt.Println(string(b))
}
Though not extremely performant, the only readable solution is:
// Split by separator and pick the first one.
// This has all the characters till null, excluding null itself.
retByteArray := bytes.Split(byteArray[:], []byte{0}) [0]
// OR
// If you want a true C-like string, including the null character
retByteArray := bytes.SplitAfter(byteArray[:], []byte{0}) [0]
A full example to have a C-style byte array:
package main
import (
"bytes"
"fmt"
)
func main() {
var byteArray = [6]byte{97,98,0,100,0,99}
cStyleString := bytes.SplitAfter(byteArray[:], []byte{0}) [0]
fmt.Println(cStyleString)
}
A full example to have a Go style string excluding the nulls:
package main
import (
"bytes"
"fmt"
)
func main() {
var byteArray = [6]byte{97, 98, 0, 100, 0, 99}
goStyleString := string(bytes.Split(byteArray[:], []byte{0}) [0])
fmt.Println(goStyleString)
}
This allocates a slice of slice of bytes. So keep an eye on performance if it is used heavily or repeatedly.
Use slices instead of arrays for reading. For example, io.Reader accepts a slice, not an array.
Use slicing instead of zero padding.
Example:
buf := make([]byte, 100)
n, err := myReader.Read(buf)
if n == 0 && err != nil {
log.Fatal(err)
}
consume(buf[:n]) // consume() will see an exact (not padded) slice of read data
Here is an option that removes the null bytes:
package main
import "golang.org/x/sys/windows"
func main() {
b := []byte{'M', 'a', 'r', 'c', 'h', 0}
s := windows.ByteSliceToString(b)
println(s == "March")
}
https://pkg.go.dev/golang.org/x/sys/unix#ByteSliceToString
https://pkg.go.dev/golang.org/x/sys/windows#ByteSliceToString
I am looking to convert a string array to a byte array in GO so I can write it down to a disk. What is an optimal solution to encode and decode a string array ([]string) to a byte array ([]byte)?
I was thinking of iterating the string array twice, first one to get the actual size needed for the byte array and then a second one to write the length and actual string ([]byte(str)) for each element.
The solution must be able to convert it the other-way; from a []byte to a []string.
Lets ignore the fact that this is Go for a second. The first thing you need is a serialization format to marshal the []string into.
There are many option here. You could build your own or use a library. I am going to assume you don't want to build your own and jump to serialization formats go supports.
In all examples, data is the []string and fp is the file you are reading/writing to. Errors are being ignored, check the returns of functions to handle errors.
Gob
Gob is a go only binary format. It should be relatively space efficient as the number of strings increases.
enc := gob.NewEncoder(fp)
enc.Encode(data)
Reading is also simple
var data []string
dec := gob.NewDecoder(fp)
dec.Decode(&data)
Gob is simple and to the point. However, the format is only readable with other Go code.
Json
Next is json. Json is a format used just about everywhere. This format is just as easy to use.
enc := json.NewEncoder(fp)
enc.Encode(data)
And for reading:
var data []string
dec := json.NewDecoder(fp)
dec.Decode(&data)
XML
XML is another common format. However, it has pretty high overhead and not as easy to use. While you could just do the same you did for gob and json, proper xml requires a root tag. In this case, we are using the root tag "Strings" and each string is wrapped in an "S" tag.
type Strings struct {
S []string
}
enc := xml.NewEncoder(fp)
enc.Encode(Strings{data})
var x Strings
dec := xml.NewDecoder(fp)
dec.Decode(&x)
data := x.S
CSV
CSV is different from the others. You have two options, use one record with n rows or n records with 1 row. The following example uses n records. It would be boring if I used one record. It would look too much like the others. CSV can ONLY hold strings.
enc := csv.NewWriter(fp)
for _, v := range data {
enc.Write([]string{v})
}
enc.Flush()
To read:
var err error
var data string
dec := csv.NewReader(fp)
for err == nil { // reading ends when an error is reached (perhaps io.EOF)
var s []string
s, err = dec.Read()
if len(s) > 0 {
data = append(data, s[0])
}
}
Which format you use is a matter of preference. There are many other possible encodings that I have not mentioned. For example, there is an external library called bencode. I don't personally like bencode, but it works. It is the same encoding used by bittorrent metadata files.
If you want to make your own encoding, encoding/binary is a good place to start. That would allow you to make the most compact file possible, but I hardly thing it is worth the effort.
The gob package will do this for you http://godoc.org/encoding/gob
Example to play with http://play.golang.org/p/e0FEZm-qiS
same source code is below.
package main
import (
"bytes"
"encoding/gob"
"fmt"
)
func main() {
// store to byte array
strs := []string{"foo", "bar"}
buf := &bytes.Buffer{}
gob.NewEncoder(buf).Encode(strs)
bs := buf.Bytes()
fmt.Printf("%q", bs)
// Decode it back
strs2 := []string{}
gob.NewDecoder(buf).Decode(&strs2)
fmt.Printf("%v", strs2)
}
to convert []string to []byte
var str = []string{"str1","str2"}
var x = []byte{}
for i:=0; i<len(str); i++{
b := []byte(str[i])
for j:=0; j<len(b); j++{
x = append(x,b[j])
}
}
to convert []byte to string
str := ""
var x = []byte{'c','a','t'}
for i := 0; i < len(x); i++ {
str += string(x[i])
}
To illustrate the problem, convert []string to []byte and then convert []byte back to []string, here's a simple solution:
package main
import (
"encoding/binary"
"fmt"
)
const maxInt32 = 1<<(32-1) - 1
func writeLen(b []byte, l int) []byte {
if 0 > l || l > maxInt32 {
panic("writeLen: invalid length")
}
var lb [4]byte
binary.BigEndian.PutUint32(lb[:], uint32(l))
return append(b, lb[:]...)
}
func readLen(b []byte) ([]byte, int) {
if len(b) < 4 {
panic("readLen: invalid length")
}
l := binary.BigEndian.Uint32(b)
if l > maxInt32 {
panic("readLen: invalid length")
}
return b[4:], int(l)
}
func Decode(b []byte) []string {
b, ls := readLen(b)
s := make([]string, ls)
for i := range s {
b, ls = readLen(b)
s[i] = string(b[:ls])
b = b[ls:]
}
return s
}
func Encode(s []string) []byte {
var b []byte
b = writeLen(b, len(s))
for _, ss := range s {
b = writeLen(b, len(ss))
b = append(b, ss...)
}
return b
}
func codecEqual(s []string) bool {
return fmt.Sprint(s) == fmt.Sprint(Decode(Encode(s)))
}
func main() {
var s []string
fmt.Println("equal", codecEqual(s))
s = []string{"", "a", "bc"}
e := Encode(s)
d := Decode(e)
fmt.Println("s", len(s), s)
fmt.Println("e", len(e), e)
fmt.Println("d", len(d), d)
fmt.Println("equal", codecEqual(s))
}
Output:
equal true
s 3 [ a bc]
e 19 [0 0 0 3 0 0 0 0 0 0 0 1 97 0 0 0 2 98 99]
d 3 [ a bc]
equal true
I would suggest to use PutUvarint and Uvarint for storing/retrieving len(s) and using []byte(str) to pass str to some io.Writer. With a string length known from Uvarint, one can buf := make([]byte, n) and pass the buf to some io.Reader.
Prepend the whole thing with length of the string array and repeat the above for all of its items. Reading the whole thing back is again reading first the outer length and repeating n-times the item read.
You can do something like this:
var lines = []string
var ctx = []byte{}
for _, s := range lines {
ctx = append(ctx, []byte(s)...)
}
It can be done easily using strings package. First you need to convert the slice of string to a string.
func Join(elems []string, sep string) string
You need to pass the slice of strings and the separator you need to separate the elements in the string. (examples: space or comma)
Then you can easily convert the string to a slice of bytes by type conversion.
package main
import (
"fmt"
"strings"
)
func main() {
//Slice of Strings
sliceStr := []string{"a","b","c","d"}
fmt.Println(sliceStr) //prints [a b c d]
//Converting slice of String to String
str := strings.Join(sliceStr,"")
fmt.Println(str) // prints abcd
//Converting String to slice of Bytes
sliceByte := []byte(str) //prints [97 98 99 100]
fmt.Println(sliceByte)
//Converting slice of bytes a String
str2 := string(sliceByte)
fmt.Println(str2) // prints abcd
//Converting string to a slice of Strings
sliceStr2 := strings.Split(str2,"")
fmt.Println(sliceStr2) //prints [a b c d]
}