How to generate a random string of a fixed length in Go?
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.
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)
rand.Read(b)
return fmt.Sprintf("%x", b)[:length]
}
Check it out in the PlayGround
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"
)
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() {
rand.Seed(time.Now().UnixNano())
fmt.Println(randSeq(10))
}