I have a string mixed with characters and numerals, but i want to increment the last character which happens to be a number, here is what i have, it works, but once i reach 10 rune goes to black since 10 decimal is zero, is there a better way to do this?
package main
import (
"fmt"
)
func main() {
str := "version-1.1.0-8"
rStr := []rune(str)
last := rStr[len(rStr)-1]
rStr[len(rStr)-1] = last + 1
}
So this works for str := "version-1.1.0-8" = version-1.1.0-9
str := version-1.1.0-9 = version-1.1.0-
I understand why it is happening, but I dont know how to fix it
Your intention is to increment the number represented by the last rune, so you should do that: parse out that number, increment it as a number, and "re-encode" it into string.
You can't operate on a single rune, as once the number reaches 10, it can only be represented using 2 runes. Another issue is if the last number is 19, incrementing it needs to alter the previous rune (and not adding a new rune).
Parsing the numbers and re-encoding though is much easier than one might think.
You can take advantage of the fmt package's fmt.Sscanf() and fmt.Sprintf() functions. Parsing and re-encoding is just a single function call.
Let's wrap this functionality into a function:
const format = "version-%d.%d.%d-%d"
func incLast(s string) (string, error) {
var a, b, c, d int
if _, err := fmt.Sscanf(s, format, &a, &b, &c, &d); err != nil {
return "", err
}
d++
return fmt.Sprintf(format, a, b, c, d), nil
}
Testing it:
s := "version-1.1.0-8"
for i := 0; i < 13; i++ {
var err error
if s, err = incLast(s); err != nil {
panic(err)
}
fmt.Println(s)
}
Output (try it on the Go Playground):
version-1.1.0-9
version-1.1.0-10
version-1.1.0-11
version-1.1.0-12
version-1.1.0-13
version-1.1.0-14
version-1.1.0-15
version-1.1.0-16
version-1.1.0-17
version-1.1.0-18
version-1.1.0-19
version-1.1.0-20
version-1.1.0-21
Another option would be to just parse and re-encode the last part, and not the complete version text. This is how it would look like:
func incLast2(s string) (string, error) {
i := strings.LastIndexByte(s, '-')
if i < 0 {
return "", fmt.Errorf("invalid input")
}
d, err := strconv.Atoi(s[i+1:])
if err != nil {
return "", err
}
d++
return s[:i+1] + strconv.Itoa(d), nil
}
Testing and output is the same. Try this one on the Go Playground.
Prologue / Context
Last week my root filesystem was remounted readonly serveral times and I took a complete snapshot via ddrescue. Sadly the filesystem was damaged already and some files are missing. At the moment I try to find my ejabberd user-database which should be somewhere within the image. Testdisk found the required file (marked as deleted) but could not restore it. Since the file is pretty small and I have a backup from some month ago I thought about doing a binary search over the whole image.
So now I have a 64GB file with a damaged filesystem and would like to extract some 4kb blocks which contain a certain pattern.
Question
How can I find the data within the 64GB large file and extract the result with some context (4kb)?
Since the filesystem image resides on my server I would prefer a linux cli tool.
The Tool
Since I couldn't find a tool which meet my requirements I wrote it myself in golang. I call it bima (for binary match). It isn't pretty but it did the job:
package main
import (
"bytes"
"encoding/hex"
"fmt"
"gopkg.in/alecthomas/kingpin.v1"
"io"
"log"
"math"
"os"
)
var (
debug = kingpin.Flag("debug", "Enable debug mode.").Short('d').Bool()
bsize = kingpin.Flag("blocksize", "Blocksize").Short('b').Default("126976").Int()
debugDetail = kingpin.Flag("debugdetail", "Debug Detail").Short('v').Default("10").Int()
matchCommand = kingpin.Command("match", "Match a value")
matchCommandValue = matchCommand.Arg("value", "The value (Hex Encoded e.g.: 616263 == abc)").Required().String()
matchCommandFile = matchCommand.Arg("file", "The file").Required().String()
)
func main() {
kingpin.Version("0.1")
mode := kingpin.Parse()
if *bsize <= 0 {
log.Fatal("The blocksize has to be larger than 0")
}
if *debugDetail <= 0 {
log.Fatal("The Debug Detail has to be larger than 0")
}
if mode == "match" {
searchBytes, err := hex.DecodeString(*matchCommandValue)
if err != nil {
log.Fatal(err)
}
scanFile(searchBytes, *matchCommandFile)
}
}
func scanFile(search []byte, path string) {
searchLength := len(search)
blocksize := *bsize
f, err := os.Open(path)
if err != nil {
log.Fatal(err)
}
defer f.Close()
fi, err := f.Stat()
if err != nil {
log.Fatal(err)
}
filesize := fi.Size()
expectedRounds := int(math.Ceil(float64(filesize-int64(searchLength))/float64(blocksize)) + 1)
if expectedRounds <= 0 {
expectedRounds = 1
}
data := make([]byte, 0, blocksize+searchLength-1)
data2 := make([]byte, 0, blocksize+searchLength-1)
offset := make([]byte, searchLength-1)
//reading the len of the slice or less (but not the cap)
readCount, err := f.Read(offset)
if err == io.EOF {
fmt.Println("The files seems to be empty")
return
} else if err != nil {
log.Fatal(err)
}
data = append(data, offset...)
buffer := make([]byte, blocksize)
var blockpos int
var idx int
blockpos = 0
lastLevel := -1
roundLevel := 0
idxOffset := 0
for round := 0; ; round++ {
if *debug {
roundLevel = ((round * 100) / expectedRounds)
if (roundLevel%*debugDetail == 0) && (roundLevel > lastLevel) {
lastLevel = roundLevel
fmt.Fprintln(os.Stderr, "Starting round", round+1, "of", expectedRounds, "--", ((round * 100) / expectedRounds))
}
}
//At EOF, the count will be zero and err will be io.EOF
readCount, err = f.Read(buffer)
if err != nil {
if err == io.EOF {
if *debug {
fmt.Fprintln(os.Stderr, "Done - Found EOF")
}
break
}
fmt.Println(err)
return
}
data = append(data, buffer[:readCount]...)
data2 = data
idxOffset = 0
for {
idx = bytes.Index(data2, search)
if idx >= 0 {
fmt.Println(blockpos + idxOffset + idx)
if idx+searchLength < len(data2) {
data2 = data2[idx+searchLength:]
idxOffset += idx
} else {
break
}
} else {
break
}
}
data = data[readCount:]
blockpos += readCount
}
}
The Story
For completeness here comes what I did to solve my problem:
At first I used hexedit to find out, that all db files have the same header. Encoded in hex it looks like this: 0102030463584d0b0000004b62574c41
So I used my tool to find all occurrences within my sda.image file:
./bima match 0102030463584d0b0000004b62574c41 ./sda.image >DBfiles.txt
For the 64GB this took about 8 Minutes and I think the HDD was the limiting factor.
The result where about 1200 occurrences which I extracted from the image with dd. As I didn't know the exact size of the files I simply extracted chunks of 20.000 bytes:
for f in $(cat DBfiles.txt); do
dd if=sda.image of=$f.dunno bs=1 ibs=1 skip=$f count=20000
done
Now I had about 1200 files and had to find the right ones. In a first step I search for the passwd files (passwd.DCD and passwd.DCL). later I did the same for the roster files. As the header of the files contains the name, I simply greped for passwd:
for f in *.dunno; do
if [ "$(cat $f | head -c 200 | grep "passwd" | wc -l)" == "1" ]; then
echo "$f" | sed 's/\.$//g' >> passwd_files.list
fi
done
Because the chunks were larger than the files I had to find the end of each files manually. I did the corrections with Curses Hexedit.
During that process I could see that the head of each file contained either dcl_logk or dcd_logk. So I knew which of the files were DCL files and which were DCD files.
In the end I had each file up to ten times and had to decide which version I wanted to use. In general I took the largest file. After putting the files in the DB directory of the new ejabberd server and restarting it, all accounts are back again. :-)
I want a random string of characters only (uppercase or lowercase), no numbers, in Go. What is the fastest and simplest way to do this?
Paul's solution provides a simple, general solution.
The question asks for the "the fastest and simplest way". Let's address the fastest part too. We'll arrive at our final, fastest code in an iterative manner. Benchmarking each iteration can be found at the end of the answer.
All the solutions and the benchmarking code can be found on the Go Playground. The code on the Playground is a test file, not an executable. You have to save it into a file named XX_test.go and run it with
go test -bench . -benchmem
Foreword:
The fastest solution is not a go-to solution if you just need a random string. For that, Paul's solution is perfect. This is if performance does matter. Although the first 2 steps (Bytes and Remainder) might be an acceptable compromise: they do improve performance by like 50% (see exact numbers in the II. Benchmark section), and they don't increase complexity significantly.
Having said that, even if you don't need the fastest solution, reading through this answer might be adventurous and educational.
I. Improvements
1. Genesis (Runes)
As a reminder, the original, general solution we're improving is this:
func init() {
rand.Seed(time.Now().UnixNano())
}
var letterRunes = []rune("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ")
func RandStringRunes(n int) string {
b := make([]rune, n)
for i := range b {
b[i] = letterRunes[rand.Intn(len(letterRunes))]
}
return string(b)
}
2. Bytes
If the characters to choose from and assemble the random string contains only the uppercase and lowercase letters of the English alphabet, we can work with bytes only because the English alphabet letters map to bytes 1-to-1 in the UTF-8 encoding (which is how Go stores strings).
So instead of:
var letters = []rune("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ")
we can use:
var letters = []byte("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ")
Or even better:
const letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
Now this is already a big improvement: we could achieve it to be a const (there are string constants but there are no slice constants). As an extra gain, the expression len(letters) will also be a const! (The expression len(s) is constant if s is a string constant.)
And at what cost? Nothing at all. strings can be indexed which indexes its bytes, perfect, exactly what we want.
Our next destination looks like this:
const letterBytes = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
func RandStringBytes(n int) string {
b := make([]byte, n)
for i := range b {
b[i] = letterBytes[rand.Intn(len(letterBytes))]
}
return string(b)
}
3. Remainder
Previous solutions get a random number to designate a random letter by calling rand.Intn() which delegates to Rand.Intn() which delegates to Rand.Int31n().
This is much slower compared to rand.Int63() which produces a random number with 63 random bits.
So we could simply call rand.Int63() and use the remainder after dividing by len(letterBytes):
func RandStringBytesRmndr(n int) string {
b := make([]byte, n)
for i := range b {
b[i] = letterBytes[rand.Int63() % int64(len(letterBytes))]
}
return string(b)
}
This works and is significantly faster, the disadvantage is that the probability of all the letters will not be exactly the same (assuming rand.Int63() produces all 63-bit numbers with equal probability). Although the distortion is extremely small as the number of letters 52 is much-much smaller than 1<<63 - 1, so in practice this is perfectly fine.
To make this understand easier: let's say you want a random number in the range of 0..5. Using 3 random bits, this would produce the numbers 0..1 with double probability than from the range 2..5. Using 5 random bits, numbers in range 0..1 would occur with 6/32 probability and numbers in range 2..5 with 5/32 probability which is now closer to the desired. Increasing the number of bits makes this less significant, when reaching 63 bits, it is negligible.
4. Masking
Building on the previous solution, we can maintain the equal distribution of letters by using only as many of the lowest bits of the random number as many is required to represent the number of letters. So for example if we have 52 letters, it requires 6 bits to represent it: 52 = 110100b. So we will only use the lowest 6 bits of the number returned by rand.Int63(). And to maintain equal distribution of letters, we only "accept" the number if it falls in the range 0..len(letterBytes)-1. If the lowest bits are greater, we discard it and query a new random number.
Note that the chance of the lowest bits to be greater than or equal to len(letterBytes) is less than 0.5 in general (0.25 on average), which means that even if this would be the case, repeating this "rare" case decreases the chance of not finding a good number. After n repetition, the chance that we still don't have a good index is much less than pow(0.5, n), and this is just an upper estimation. In case of 52 letters the chance that the 6 lowest bits are not good is only (64-52)/64 = 0.19; which means for example that chances to not have a good number after 10 repetition is 1e-8.
So here is the solution:
const letterBytes = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
const (
letterIdxBits = 6 // 6 bits to represent a letter index
letterIdxMask = 1<<letterIdxBits - 1 // All 1-bits, as many as letterIdxBits
)
func RandStringBytesMask(n int) string {
b := make([]byte, n)
for i := 0; i < n; {
if idx := int(rand.Int63() & letterIdxMask); idx < len(letterBytes) {
b[i] = letterBytes[idx]
i++
}
}
return string(b)
}
5. Masking Improved
The previous solution only uses the lowest 6 bits of the 63 random bits returned by rand.Int63(). This is a waste as getting the random bits is the slowest part of our algorithm.
If we have 52 letters, that means 6 bits code a letter index. So 63 random bits can designate 63/6 = 10 different letter indices. Let's use all those 10:
const letterBytes = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
const (
letterIdxBits = 6 // 6 bits to represent a letter index
letterIdxMask = 1<<letterIdxBits - 1 // All 1-bits, as many as letterIdxBits
letterIdxMax = 63 / letterIdxBits // # of letter indices fitting in 63 bits
)
func RandStringBytesMaskImpr(n int) string {
b := make([]byte, n)
// A rand.Int63() generates 63 random bits, enough for letterIdxMax letters!
for i, cache, remain := n-1, rand.Int63(), letterIdxMax; i >= 0; {
if remain == 0 {
cache, remain = rand.Int63(), letterIdxMax
}
if idx := int(cache & letterIdxMask); idx < len(letterBytes) {
b[i] = letterBytes[idx]
i--
}
cache >>= letterIdxBits
remain--
}
return string(b)
}
6. Source
The Masking Improved is pretty good, not much we can improve on it. We could, but not worth the complexity.
Now let's find something else to improve. The source of random numbers.
There is a crypto/rand package which provides a Read(b []byte) function, so we could use that to get as many bytes with a single call as many we need. This wouldn't help in terms of performance as crypto/rand implements a cryptographically secure pseudorandom number generator so it's much slower.
So let's stick to the math/rand package. The rand.Rand uses a rand.Source as the source of random bits. rand.Source is an interface which specifies a Int63() int64 method: exactly and the only thing we needed and used in our latest solution.
So we don't really need a rand.Rand (either explicit or the global, shared one of the rand package), a rand.Source is perfectly enough for us:
var src = rand.NewSource(time.Now().UnixNano())
func RandStringBytesMaskImprSrc(n int) string {
b := make([]byte, n)
// A src.Int63() generates 63 random bits, enough for letterIdxMax characters!
for i, cache, remain := n-1, src.Int63(), letterIdxMax; i >= 0; {
if remain == 0 {
cache, remain = src.Int63(), letterIdxMax
}
if idx := int(cache & letterIdxMask); idx < len(letterBytes) {
b[i] = letterBytes[idx]
i--
}
cache >>= letterIdxBits
remain--
}
return string(b)
}
Also note that this last solution doesn't require you to initialize (seed) the global Rand of the math/rand package as that is not used (and our rand.Source is properly initialized / seeded).
One more thing to note here: package doc of math/rand states:
The default Source is safe for concurrent use by multiple goroutines.
So the default source is slower than a Source that may be obtained by rand.NewSource(), because the default source has to provide safety under concurrent access / use, while rand.NewSource() does not offer this (and thus the Source returned by it is more likely to be faster).
7. Utilizing strings.Builder
All previous solutions return a string whose content is first built in a slice ([]rune in Genesis, and []byte in subsequent solutions), and then converted to string. This final conversion has to make a copy of the slice's content, because string values are immutable, and if the conversion would not make a copy, it could not be guaranteed that the string's content is not modified via its original slice. For details, see How to convert utf8 string to []byte? and golang: []byte(string) vs []byte(*string).
Go 1.10 introduced strings.Builder. strings.Builder is a new type we can use to build contents of a string similar to bytes.Buffer. Internally it uses a []byte to build the content, and when we're done, we can obtain the final string value using its Builder.String() method. But what's cool in it is that it does this without performing the copy we just talked about above. It dares to do so because the byte slice used to build the string's content is not exposed, so it is guaranteed that no one can modify it unintentionally or maliciously to alter the produced "immutable" string.
So our next idea is to not build the random string in a slice, but with the help of a strings.Builder, so once we're done, we can obtain and return the result without having to make a copy of it. This may help in terms of speed, and it will definitely help in terms of memory usage and allocations.
func RandStringBytesMaskImprSrcSB(n int) string {
sb := strings.Builder{}
sb.Grow(n)
// A src.Int63() generates 63 random bits, enough for letterIdxMax characters!
for i, cache, remain := n-1, src.Int63(), letterIdxMax; i >= 0; {
if remain == 0 {
cache, remain = src.Int63(), letterIdxMax
}
if idx := int(cache & letterIdxMask); idx < len(letterBytes) {
sb.WriteByte(letterBytes[idx])
i--
}
cache >>= letterIdxBits
remain--
}
return sb.String()
}
Do note that after creating a new strings.Buidler, we called its Builder.Grow() method, making sure it allocates a big-enough internal slice (to avoid reallocations as we add the random letters).
8. "Mimicing" strings.Builder with package unsafe
strings.Builder builds the string in an internal []byte, the same as we did ourselves. So basically doing it via a strings.Builder has some overhead, the only thing we switched to strings.Builder for is to avoid the final copying of the slice.
strings.Builder avoids the final copy by using package unsafe:
// String returns the accumulated string.
func (b *Builder) String() string {
return *(*string)(unsafe.Pointer(&b.buf))
}
The thing is, we can also do this ourselves, too. So the idea here is to switch back to building the random string in a []byte, but when we're done, don't convert it to string to return, but do an unsafe conversion: obtain a string which points to our byte slice as the string data.
This is how it can be done:
func RandStringBytesMaskImprSrcUnsafe(n int) string {
b := make([]byte, n)
// A src.Int63() generates 63 random bits, enough for letterIdxMax characters!
for i, cache, remain := n-1, src.Int63(), letterIdxMax; i >= 0; {
if remain == 0 {
cache, remain = src.Int63(), letterIdxMax
}
if idx := int(cache & letterIdxMask); idx < len(letterBytes) {
b[i] = letterBytes[idx]
i--
}
cache >>= letterIdxBits
remain--
}
return *(*string)(unsafe.Pointer(&b))
}
(9. Using rand.Read())
Go 1.7 added a rand.Read() function and a Rand.Read() method. We should be tempted to use these to read as many bytes as we need in one step, in order to achieve better performance.
There is one small "problem" with this: how many bytes do we need? We could say: as many as the number of output letters. We would think this is an upper estimation, as a letter index uses less than 8 bits (1 byte). But at this point we are already doing worse (as getting the random bits is the "hard part"), and we're getting more than needed.
Also note that to maintain equal distribution of all letter indices, there might be some "garbage" random data that we won't be able to use, so we would end up skipping some data, and thus end up short when we go through all the byte slice. We would need to further get more random bytes, "recursively". And now we're even losing the "single call to rand package" advantage...
We could "somewhat" optimize the usage of the random data we acquire from math.Rand(). We may estimate how many bytes (bits) we'll need. 1 letter requires letterIdxBits bits, and we need n letters, so we need n * letterIdxBits / 8.0 bytes rounding up. We can calculate the probability of a random index not being usable (see above), so we could request more that will "more likely" be enough (if it turns out it's not, we repeat the process). We can process the byte slice as a "bit stream" for example, for which we have a nice 3rd party lib: github.com/icza/bitio (disclosure: I'm the author).
But Benchmark code still shows we're not winning. Why is it so?
The answer to the last question is because rand.Read() uses a loop and keeps calling Source.Int63() until it fills the passed slice. Exactly what the RandStringBytesMaskImprSrc() solution does, without the intermediate buffer, and without the added complexity. That's why RandStringBytesMaskImprSrc() remains on the throne. Yes, RandStringBytesMaskImprSrc() uses an unsynchronized rand.Source unlike rand.Read(). But the reasoning still applies; and which is proven if we use Rand.Read() instead of rand.Read() (the former is also unsynchronzed).
II. Benchmark
All right, it's time for benchmarking the different solutions.
Moment of truth:
BenchmarkRunes-4 2000000 723 ns/op 96 B/op 2 allocs/op
BenchmarkBytes-4 3000000 550 ns/op 32 B/op 2 allocs/op
BenchmarkBytesRmndr-4 3000000 438 ns/op 32 B/op 2 allocs/op
BenchmarkBytesMask-4 3000000 534 ns/op 32 B/op 2 allocs/op
BenchmarkBytesMaskImpr-4 10000000 176 ns/op 32 B/op 2 allocs/op
BenchmarkBytesMaskImprSrc-4 10000000 139 ns/op 32 B/op 2 allocs/op
BenchmarkBytesMaskImprSrcSB-4 10000000 134 ns/op 16 B/op 1 allocs/op
BenchmarkBytesMaskImprSrcUnsafe-4 10000000 115 ns/op 16 B/op 1 allocs/op
Just by switching from runes to bytes, we immediately have 24% performance gain, and memory requirement drops to one third.
Getting rid of rand.Intn() and using rand.Int63() instead gives another 20% boost.
Masking (and repeating in case of big indices) slows down a little (due to repetition calls): -22%...
But when we make use of all (or most) of the 63 random bits (10 indices from one rand.Int63() call): that speeds up big time: 3 times.
If we settle with a (non-default, new) rand.Source instead of rand.Rand, we again gain 21%.
If we utilize strings.Builder, we gain a tiny 3.5% in speed, but we also achieved 50% reduction in memory usage and allocations! That's nice!
Finally if we dare to use package unsafe instead of strings.Builder, we again gain a nice 14%.
Comparing the final to the initial solution: RandStringBytesMaskImprSrcUnsafe() is 6.3 times faster than RandStringRunes(), uses one sixth memory and half as few allocations. Mission accomplished.
You can just write code for it. This code can be a little simpler if you want to rely on the letters all being single bytes when encoded in UTF-8.
package main
import (
"fmt"
"time"
"math/rand"
)
func init() {
rand.Seed(time.Now().UnixNano())
}
var letters = []rune("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ")
func randSeq(n int) string {
b := make([]rune, n)
for i := range b {
b[i] = letters[rand.Intn(len(letters))]
}
return string(b)
}
func main() {
fmt.Println(randSeq(10))
}
Use package uniuri, which generates cryptographically secure uniform (unbiased) strings.
Disclaimer: I'm the author of the package
Simple solution for you, with least duplicate result:
import (
"fmt"
"math/rand"
"time"
)
func randomString(length int) string {
rand.Seed(time.Now().UnixNano())
b := make([]byte, length+2)
rand.Read(b)
return fmt.Sprintf("%x", b)[2 : length+2]
}
Check it out in the PlayGround
Two possible options (there might be more of course):
You can use the crypto/rand package that supports reading random byte arrays (from /dev/urandom) and is geared towards cryptographic random generation. see http://golang.org/pkg/crypto/rand/#example_Read . It might be slower than normal pseudo-random number generation though.
Take a random number and hash it using md5 or something like this.
If you want cryptographically secure random numbers, and the exact charset is flexible (say, base64 is fine), you can calculate exactly what the length of random characters you need from the desired output size.
Base 64 text is 1/3 longer than base 256. (2^8 vs 2^6; 8bits/6bits = 1.333 ratio)
import (
"crypto/rand"
"encoding/base64"
"math"
)
func randomBase64String(l int) string {
buff := make([]byte, int(math.Ceil(float64(l)/float64(1.33333333333))))
rand.Read(buff)
str := base64.RawURLEncoding.EncodeToString(buff)
return str[:l] // strip 1 extra character we get from odd length results
}
Note: you can also use RawStdEncoding if you prefer + and / characters to - and _
If you want hex, base 16 is 2x longer than base 256. (2^8 vs 2^4; 8bits/4bits = 2x ratio)
import (
"crypto/rand"
"encoding/hex"
"math"
)
func randomBase16String(l int) string {
buff := make([]byte, int(math.Ceil(float64(l)/2)))
rand.Read(buff)
str := hex.EncodeToString(buff)
return str[:l] // strip 1 extra character we get from odd length results
}
However, you could extend this to any arbitrary character set if you have a base256 to baseN encoder for your character set. You can do the same size calculation with how many bits are needed to represent your character set. The ratio calculation for any arbitrary charset is: ratio = 8 / log2(len(charset))).
Though both of these solutions are secure, simple, should be fast, and don't waste your crypto entropy pool.
Here's the playground showing it works for any size. https://play.golang.org/p/_yF_xxXer0Z
Another version, inspired from generate password in JavaScript crypto:
package main
import (
"crypto/rand"
"fmt"
)
var chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890-"
func shortID(length int) string {
ll := len(chars)
b := make([]byte, length)
rand.Read(b) // generates len(b) random bytes
for i := 0; i < length; i++ {
b[i] = chars[int(b[i])%ll]
}
return string(b)
}
func main() {
fmt.Println(shortID(18))
fmt.Println(shortID(18))
fmt.Println(shortID(18))
}
Following icza's wonderfully explained solution, here is a modification of it that uses crypto/rand instead of math/rand.
const (
letterBytes = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ" // 52 possibilities
letterIdxBits = 6 // 6 bits to represent 64 possibilities / indexes
letterIdxMask = 1<<letterIdxBits - 1 // All 1-bits, as many as letterIdxBits
)
func SecureRandomAlphaString(length int) string {
result := make([]byte, length)
bufferSize := int(float64(length)*1.3)
for i, j, randomBytes := 0, 0, []byte{}; i < length; j++ {
if j%bufferSize == 0 {
randomBytes = SecureRandomBytes(bufferSize)
}
if idx := int(randomBytes[j%length] & letterIdxMask); idx < len(letterBytes) {
result[i] = letterBytes[idx]
i++
}
}
return string(result)
}
// SecureRandomBytes returns the requested number of bytes using crypto/rand
func SecureRandomBytes(length int) []byte {
var randomBytes = make([]byte, length)
_, err := rand.Read(randomBytes)
if err != nil {
log.Fatal("Unable to generate random bytes")
}
return randomBytes
}
If you want a more generic solution, that allows you to pass in the slice of character bytes to create the string out of, you can try using this:
// SecureRandomString returns a string of the requested length,
// made from the byte characters provided (only ASCII allowed).
// Uses crypto/rand for security. Will panic if len(availableCharBytes) > 256.
func SecureRandomString(availableCharBytes string, length int) string {
// Compute bitMask
availableCharLength := len(availableCharBytes)
if availableCharLength == 0 || availableCharLength > 256 {
panic("availableCharBytes length must be greater than 0 and less than or equal to 256")
}
var bitLength byte
var bitMask byte
for bits := availableCharLength - 1; bits != 0; {
bits = bits >> 1
bitLength++
}
bitMask = 1<<bitLength - 1
// Compute bufferSize
bufferSize := length + length / 3
// Create random string
result := make([]byte, length)
for i, j, randomBytes := 0, 0, []byte{}; i < length; j++ {
if j%bufferSize == 0 {
// Random byte buffer is empty, get a new one
randomBytes = SecureRandomBytes(bufferSize)
}
// Mask bytes to get an index into the character slice
if idx := int(randomBytes[j%length] & bitMask); idx < availableCharLength {
result[i] = availableCharBytes[idx]
i++
}
}
return string(result)
}
If you want to pass in your own source of randomness, it would be trivial to modify the above to accept an io.Reader instead of using crypto/rand.
Here is my way ) Use math rand or crypto rand as you wish.
func randStr(len int) string {
buff := make([]byte, len)
rand.Read(buff)
str := base64.StdEncoding.EncodeToString(buff)
// Base 64 can be longer than len
return str[:len]
}
Here is a simple and performant solution for a cryptographically secure random string.
package main
import (
"crypto/rand"
"unsafe"
"fmt"
)
var alphabet = []byte("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ")
func main() {
fmt.Println(generate(16))
}
func generate(size int) string {
b := make([]byte, size)
rand.Read(b)
for i := 0; i < size; i++ {
b[i] = alphabet[b[i] % byte(len(alphabet))]
}
return *(*string)(unsafe.Pointer(&b))
}
Benchmark
Benchmark 95.2 ns/op 16 B/op 1 allocs/op
func Rand(n int) (str string) {
b := make([]byte, n)
rand.Read(b)
str = fmt.Sprintf("%x", b)
return
}
I usually do it like this if it takes an option to capitalize or not
func randomString(length int, upperCase bool) string {
rand.Seed(time.Now().UnixNano())
var alphabet string
if upperCase {
alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
} else {
alphabet = "abcdefghijklmnopqrstuvwxyz"
}
var sb strings.Builder
l := len(alphabet)
for i := 0; i < length; i++ {
c := alphabet[rand.Intn(l)]
sb.WriteByte(c)
}
return sb.String()
}
and like this if you don't need capital letters
func randomString(length int) string {
rand.Seed(time.Now().UnixNano())
var alphabet string = "abcdefghijklmnopqrstuvwxyz"
var sb strings.Builder
l := len(alphabet)
for i := 0; i < length; i++ {
c := alphabet[rand.Intn(l)]
sb.WriteByte(c)
}
return sb.String()
}
If you are willing to add a few characters to your pool of allowed characters, you can make the code work with anything which provides random bytes through a io.Reader. Here we are using crypto/rand.
// len(encodeURL) == 64. This allows (x <= 265) x % 64 to have an even
// distribution.
const encodeURL = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789-_"
// A helper function create and fill a slice of length n with characters from
// a-zA-Z0-9_-. It panics if there are any problems getting random bytes.
func RandAsciiBytes(n int) []byte {
output := make([]byte, n)
// We will take n bytes, one byte for each character of output.
randomness := make([]byte, n)
// read all random
_, err := rand.Read(randomness)
if err != nil {
panic(err)
}
// fill output
for pos := range output {
// get random item
random := uint8(randomness[pos])
// random % 64
randomPos := random % uint8(len(encodeURL))
// put into output
output[pos] = encodeURL[randomPos]
}
return output
}
This is a sample code which I used to generate certificate number in my app.
func GenerateCertificateNumber() string {
CertificateLength := 7
t := time.Now().String()
CertificateHash, err := bcrypt.GenerateFromPassword([]byte(t), bcrypt.DefaultCost)
if err != nil {
fmt.Println(err)
}
// Make a Regex we only want letters and numbers
reg, err := regexp.Compile("[^a-zA-Z0-9]+")
if err != nil {
log.Fatal(err)
}
processedString := reg.ReplaceAllString(string(CertificateHash), "")
fmt.Println(string(processedString))
CertificateNumber := strings.ToUpper(string(processedString[len(processedString)-CertificateLength:]))
fmt.Println(CertificateNumber)
return CertificateNumber
}
/*
korzhao
*/
package rand
import (
crand "crypto/rand"
"math/rand"
"sync"
"time"
"unsafe"
)
// Doesn't share the rand library globally, reducing lock contention
type Rand struct {
Seed int64
Pool *sync.Pool
}
var (
MRand = NewRand()
randlist = []byte("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890")
)
// init random number generator
func NewRand() *Rand {
p := &sync.Pool{New: func() interface{} {
return rand.New(rand.NewSource(getSeed()))
},
}
mrand := &Rand{
Pool: p,
}
return mrand
}
// get the seed
func getSeed() int64 {
return time.Now().UnixNano()
}
func (s *Rand) getrand() *rand.Rand {
return s.Pool.Get().(*rand.Rand)
}
func (s *Rand) putrand(r *rand.Rand) {
s.Pool.Put(r)
}
// get a random number
func (s *Rand) Intn(n int) int {
r := s.getrand()
defer s.putrand(r)
return r.Intn(n)
}
// bulk get random numbers
func (s *Rand) Read(p []byte) (int, error) {
r := s.getrand()
defer s.putrand(r)
return r.Read(p)
}
func CreateRandomString(len int) string {
b := make([]byte, len)
_, err := MRand.Read(b)
if err != nil {
return ""
}
for i := 0; i < len; i++ {
b[i] = randlist[b[i]%(62)]
}
return *(*string)(unsafe.Pointer(&b))
}
24.0 ns/op 16 B/op 1 allocs/
As a follow-up to icza's brilliant solution, below I am using rand.Reader
func RandStringBytesMaskImprRandReaderUnsafe(length uint) (string, error) {
const (
charset = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
charIdxBits = 6 // 6 bits to represent a letter index
charIdxMask = 1<<charIdxBits - 1 // All 1-bits, as many as charIdxBits
charIdxMax = 63 / charIdxBits // # of letter indices fitting in 63 bits
)
buffer := make([]byte, length)
charsetLength := len(charset)
max := big.NewInt(int64(1 << uint64(charsetLength)))
limit, err := rand.Int(rand.Reader, max)
if err != nil {
return "", err
}
for index, cache, remain := int(length-1), limit.Int64(), charIdxMax; index >= 0; {
if remain == 0 {
limit, err = rand.Int(rand.Reader, max)
if err != nil {
return "", err
}
cache, remain = limit.Int64(), charIdxMax
}
if idx := int(cache & charIdxMask); idx < charsetLength {
buffer[index] = charset[idx]
index--
}
cache >>= charIdxBits
remain--
}
return *(*string)(unsafe.Pointer(&buffer)), nil
}
func BenchmarkBytesMaskImprRandReaderUnsafe(b *testing.B) {
b.ReportAllocs()
b.ResetTimer()
const length = 16
b.RunParallel(func(pb *testing.PB) {
for pb.Next() {
RandStringBytesMaskImprRandReaderUnsafe(length)
}
})
}
package main
import (
"encoding/base64"
"fmt"
"math/rand"
"time"
)
// customEncodeURL is like `bas64.encodeURL`
// except its made up entirely of uppercase characters:
const customEncodeURL = "ABCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKLMNOPQRSTUVWXYZABCDEFGHIJKL"
// Random generates a random string.
// It is not cryptographically secure.
func Random(n int) string {
b := make([]byte, n)
rand.Seed(time.Now().UnixNano())
_, _ = rand.Read(b) // docs say that it always returns a nil error.
customEncoding := base64.NewEncoding(customEncodeURL).WithPadding(base64.NoPadding)
return customEncoding.EncodeToString(b)
}
func main() {
fmt.Println(Random(16))
}
const (
chars = "0123456789_abcdefghijkl-mnopqrstuvwxyz" //ABCDEFGHIJKLMNOPQRSTUVWXYZ
charsLen = len(chars)
mask = 1<<6 - 1
)
var rng = rand.NewSource(time.Now().UnixNano())
// RandStr 返回指定长度的随机字符串
func RandStr(ln int) string {
/* chars 38个字符
* rng.Int63() 每次产出64bit的随机数,每次我们使用6bit(2^6=64) 可以使用10次
*/
buf := make([]byte, ln)
for idx, cache, remain := ln-1, rng.Int63(), 10; idx >= 0; {
if remain == 0 {
cache, remain = rng.Int63(), 10
}
buf[idx] = chars[int(cache&mask)%charsLen]
cache >>= 6
remain--
idx--
}
return *(*string)(unsafe.Pointer(&buf))
}
BenchmarkRandStr16-8 20000000 68.1 ns/op 16 B/op 1 allocs/op
Like [a-zA-Z0-9] string:
na1dopW129T0anN28udaZ
or hexadecimal string:
8c6f78ac23b4a7b8c0182d
By long I mean 2K and more characters.
This does about 200MBps on my box. There's obvious room for improvement.
type randomDataMaker struct {
src rand.Source
}
func (r *randomDataMaker) Read(p []byte) (n int, err error) {
for i := range p {
p[i] = byte(r.src.Int63() & 0xff)
}
return len(p), nil
}
You'd just use io.CopyN to produce the string you want. Obviously you could adjust the character set on the way in or whatever.
The nice thing about this model is that it's just an io.Reader so you can use it making anything.
Test is below:
func BenchmarkRandomDataMaker(b *testing.B) {
randomSrc := randomDataMaker{rand.NewSource(1028890720402726901)}
for i := 0; i < b.N; i++ {
b.SetBytes(int64(i))
_, err := io.CopyN(ioutil.Discard, &randomSrc, int64(i))
if err != nil {
b.Fatalf("Error copying at %v: %v", i, err)
}
}
}
On one core of my 2.2GHz i7:
BenchmarkRandomDataMaker 50000 246512 ns/op 202.83 MB/s
EDIT
Since I wrote the benchmark, I figured I'd do the obvious improvement thing (call out to the random less frequently). With 1/8 the calls to rand, it runs about 4x faster, though it's a big uglier:
New version:
func (r *randomDataMaker) Read(p []byte) (n int, err error) {
todo := len(p)
offset := 0
for {
val := int64(r.src.Int63())
for i := 0; i < 8; i++ {
p[offset] = byte(val & 0xff)
todo--
if todo == 0 {
return len(p), nil
}
offset++
val >>= 8
}
}
panic("unreachable")
}
New benchmark:
BenchmarkRandomDataMaker 200000 251148 ns/op 796.34 MB/s
EDIT 2
Took out the masking in the cast to byte since it was redundant. Got a good deal faster:
BenchmarkRandomDataMaker 200000 231843 ns/op 862.64 MB/s
(this is so much easier than real work sigh)
EDIT 3
This came up in irc today, so I released a library. Also, my actual benchmark tool, while useful for relative speed, isn't sufficiently accurate in its reporting.
I created randbo that you can reuse to produce random streams wherever you may need them.
You can use the Go package uniuri to generate random strings (or view the source code to see how they're doing it). You'll want to use:
func NewLen(length int) string
NewLen returns a new random string of the provided length, consisting of standard characters.
Or, to specify the set of characters used:
func NewLenChars(length int, chars []byte) string
This is actually a little biased towards the first 8 characters in the set (since 255 is not a multiple of len(alphanum)), but this will get you most of the way there.
import (
"crypto/rand"
)
func randString(n int) string {
const alphanum = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
var bytes = make([]byte, n)
rand.Read(bytes)
for i, b := range bytes {
bytes[i] = alphanum[b % byte(len(alphanum))]
}
return string(bytes)
}
If you want to generate cryptographically secure random string, I recommend you to take a look at this page. Here is a helper function that reads n random bytes from the source of randomness of your OS and then use these bytes to base64encode it. Note that the string length would be bigger than n because of base64.
package main
import(
"crypto/rand"
"encoding/base64"
"fmt"
)
func GenerateRandomBytes(n int) ([]byte, error) {
b := make([]byte, n)
_, err := rand.Read(b)
if err != nil {
return nil, err
}
return b, nil
}
func GenerateRandomString(s int) (string, error) {
b, err := GenerateRandomBytes(s)
return base64.URLEncoding.EncodeToString(b), err
}
func main() {
token, _ := GenerateRandomString(32)
fmt.Println(token)
}
Here Evan Shaw's answer re-worked without the bias towards the first 8 characters of the string. Note that it uses lots of expensive big.Int operations so probably isn't that quick! The answer is crypto strong though.
It uses rand.Int to make an integer of exactly the right size len(alphanum) ** n, then does what is effectively a base conversion into base len(alphanum).
There is almost certainly a better algorithm for this which would involve keeping a much smaller remainder and adding random bytes to it as necessary. This would get rid of the expensive long integer arithmetic.
import (
"crypto/rand"
"fmt"
"math/big"
)
func randString(n int) string {
const alphanum = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
symbols := big.NewInt(int64(len(alphanum)))
states := big.NewInt(0)
states.Exp(symbols, big.NewInt(int64(n)), nil)
r, err := rand.Int(rand.Reader, states)
if err != nil {
panic(err)
}
var bytes = make([]byte, n)
r2 := big.NewInt(0)
symbol := big.NewInt(0)
for i := range bytes {
r2.DivMod(r, symbols, symbol)
r, r2 = r2, r
bytes[i] = alphanum[symbol.Int64()]
}
return string(bytes)
}