I have a struct:
type Name struct {
hexID string
age uint8
}
What is the easiest way to check that hexID field is a valid hex string? And if not - rise an error.
For example:
var n Name
n.hexID = "Hello World >)" // not a valid hex
n.hexID = "aaa12Eb9990101010101112cC" // valid hex
Or maybe there are somewhere struct tag exists?
Iterate over the characters of the string, and check if each is a valid hex digit.
func isValidHex(s string) bool {
for _, r := range s {
if !(r >= '0' && r <= '9' || r >= 'a' && r <= 'f' || r >= 'A' && r <= 'F') {
return false
}
}
return true
}
Testing it:
fmt.Println(isValidHex("Hello World >)"))
fmt.Println(isValidHex("aaa12Eb9990101010101112cC"))
Output (try it on the Go Playground):
false
true
Note: you'd be tempted to use hex.DecodeString() and check the returned error: if it's nil, it's valid. But do note that this function expects that the string has even length (as it produces bytes from the hex digits, and 2 hex digits forms a byte). Not to mention that if you don't need the result (as a byte slice), this is slower and creates unnecessary garbage (for the gc to collect).
Another solution could be using big.Int.SetString():
func isValidHex(s string) bool {
_, ok := new(big.Int).SetString(s, 16)
return ok
}
This outputs the same, try it on the Go Playground. But this again is slower and uses memory allocations (generates garbage).
What about this one
regexp.MatchString("[^0-9A-Fa-f]", n.hexID)
True if string contains HEX illegal characters
Comment: I'm completely confused now which one to use :( – armaka
Different inplementations have different performance. For example,
func isHexRock(s string) bool {
for _, b := range []byte(s) {
if !(b >= '0' && b <= '9' || b >= 'a' && b <= 'f' || b >= 'A' && b <= 'F') {
return false
}
}
return true
}
func isHexIcza(s string) bool {
for _, r := range s {
if !(r >= '0' && r <= '9' || r >= 'a' && r <= 'f' || r >= 'A' && r <= 'F') {
return false
}
}
return true
}
var rxNotHex = regexp.MustCompile("[^0-9A-Fa-f]")
func isHexOjacoGlobal(s string) bool {
return !rxNotHex.MatchString(s)
}
func isHexOjacoLocal(s string) bool {
notHex, err := regexp.MatchString("[^0-9A-Fa-f]", s)
if err != nil {
panic(err)
}
return !notHex
}
Some benchmark results:
BenchmarkRock-4 36386997 30.92 ns/op 0 B/op 0 allocs/op
BenchmarkIcza-4 21100798 52.86 ns/op 0 B/op 0 allocs/op
BenchmarkOjacoGlobal-4 5958829 209.9 ns/op 0 B/op 0 allocs/op
BenchmarkOjacoLocal-4 227672 4648 ns/op 1626 B/op 22 allocs/op
I think ParseInt method can also tackle this -
_, err := strconv.ParseInt("deadbeef", 16, 64)
if err != nil {
fmt.Println("not a hex string")
} else {
fmt.Println("it is a hex string")
}
Is there a better way to insert "|" into a string
given a binary string representation of decimal 200 = 11001000
this function returns a string = 11|001|000
While this function works, it seems very kludgy!! Why is it so
hard in GO to do a simple character insertion???
func (i Binary) FString() string {
a := strconv.FormatUint(i.Get(), 2)
y := make([]string, len(a), len(a)*2)
data := []rune(a)
r := []rune{}
for i := len(data) - 1; i >= 0; i-- {
r = append(r, data[i])
}
for j := len(a) - 1; j >= 0; j-- {
y = append(y, string(r[j]))
if ((j)%3) == 0 && j > 0 {
y = append(y, "|")
}
}
return strings.Join(y, "")
}
Depends on what you call better. I'd use regular expressions.
In this case, the complexity arises from inserting separators from the right. If we padded the string so that its length was a multiple of 3, we could insert the separator from the left. And we could easily use a regular expression to insert | before every three characters. Then, we can just strip off the leading | + padding.
func (i Binary) FString() string {
a := strconv.FormatUint(i.Get(), 2)
pad_req := len(a) % 3
padding := strings.Repeat("0", (3 - pad_req))
a = padding + a
re := regexp.MustCompile("([01]{3})")
a = re.ReplaceAllString(a, "|$1")
start := len(padding) + 1
if len(padding) == 3 {
// If we padded with "000", we want to remove the `|` before *and* after it
start = 5
}
a = a[start:]
return a
}
Snippet on the Go Playground
If performance is not critical and you just want a compact version, you may copy the input digits to output, and insert a | symbol whenever a group of 2 has been written to the output.
Groups are counted from right-to-left, so when copying the digits from left-to-right, the first group might be smaller. So the counter of digits inside a group may not necessarily start from 0 in case of the first group, but from len(input)%3.
Here is an example of it:
func Format(s string) string {
b, count := &bytes.Buffer{}, len(s)%3
for i, r := range s {
if i > 0 && count == i%3 {
b.WriteRune('|')
}
b.WriteRune(r)
}
return b.String()
}
Testing it:
for i := uint64(0); i < 10; i++ {
fmt.Println(Format(strconv.FormatUint(i, 2)))
}
fmt.Println(Format(strconv.FormatInt(1234, 2)))
Output (try it on the Go Playground):
0
1
10
11
100
101
110
111
1|000
1|001
10|011|010|010
If you have to do this many times and performance does matter, then check out my answer to the question: How to fmt.Printf an integer with thousands comma
Based on that a fast solution can be:
func Format(s string) string {
out := make([]byte, len(s)+(len(s)-1)/3)
for i, j, k := len(s)-1, len(out)-1, 0; ; i, j = i-1, j-1 {
out[j] = s[i]
if i == 0 {
return string(out)
}
if k++; k == 3 {
j, k = j-1, 0
out[j] = '|'
}
}
}
Output is the same of course. Try it on the Go Playground.
This is a partitioning problem. You can use this function:
func partition(s, separator string, pLen int) string {
if pLen < 1 || len(s) == 0 || len(separator) == 0 {
return s
}
buffer := []rune(s)
L := len(buffer)
pCount := L / pLen
result := []string{}
index := 0
for ; index < pCount; index++ {
_from := L - (index+1)*pLen
_to := L - index*pLen
result = append(result, string(buffer[_from:_to]))
}
if L%pLen != 0 {
result = append(result, string(buffer[0:L-index*pLen]))
}
for h, t := 0, len(result)-1; h < t; h, t = h+1, t-1 {
result[t], result[h] = result[h], result[t]
}
return strings.Join(result, separator)
}
And s := partition("11001000", "|", 3) will give you 11|001|000.
Here is a little test:
func TestSmokeTest(t *testing.T) {
input := "11001000"
s := partition(input, "|", 3)
if s != "11|001|000" {
t.Fail()
}
s = partition(input, "|", 2)
if s != "11|00|10|00" {
t.Fail()
}
input = "0111001000"
s = partition(input, "|", 3)
if s != "0|111|001|000" {
t.Fail()
}
s = partition(input, "|", 2)
if s != "01|11|00|10|00" {
t.Fail()
}
}
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 4 years ago.
Improve this question
I want to make a function that calculates the length of the common segment (starting from the beginning) in two strings. For example:
foo:="Makan"
bar:="Makon"
The result should be 3.
foo:="Indah"
bar:="Ihkasyandehlo"
The result should be 1.
It's not clear what you are asking because you limited your test cases to ASCII characters.
I've added a Unicode test case and I've included answers for bytes, runes, or both.
play.golang.org:
package main
import (
"fmt"
"unicode/utf8"
)
func commonBytes(s, t string) (bytes int) {
if len(s) > len(t) {
s, t = t, s
}
i := 0
for ; i < len(s); i++ {
if s[i] != t[i] {
break
}
}
return i
}
func commonRunes(s, t string) (runes int) {
if len(s) > len(t) {
s, t = t, s
}
i := 0
for ; i < len(s); i++ {
if s[i] != t[i] {
break
}
}
return utf8.RuneCountInString(s[:i])
}
func commonBytesRunes(s, t string) (bytes, runes int) {
if len(s) > len(t) {
s, t = t, s
}
i := 0
for ; i < len(s); i++ {
if s[i] != t[i] {
break
}
}
return i, utf8.RuneCountInString(s[:i])
}
func main() {
Tests := []struct {
word1, word2 string
}{
{"Makan", "Makon"},
{"Indah", "Ihkasyandehlo"},
{"日本語", "日本語"},
}
for _, test := range Tests {
fmt.Println("Words: ", test.word1, test.word2)
fmt.Println("Bytes: ", commonBytes(test.word1, test.word2))
fmt.Println("Runes: ", commonRunes(test.word1, test.word2))
fmt.Print("Bytes & Runes: ")
fmt.Println(commonBytesRunes(test.word1, test.word2))
}
}
Output:
Words: Makan Makon
Bytes: 3
Runes: 3
Bytes & Runes: 3 3
Words: Indah Ihkasyandehlo
Bytes: 1
Runes: 1
Bytes & Runes: 1 1
Words: 日本語 日本語
Bytes: 9
Runes: 3
Bytes & Runes: 9 3
Note that if you were working with Unicode characters, the result could be quite different.
Try for instance using utf8.DecodeRuneInString().
See this example:
package main
import "fmt"
import "unicode/utf8"
func index(s1, s2 string) int {
res := 0
for i, w := 0, 0; i < len(s2); i += w {
if i >= len(s1) {
return res
}
runeValue1, width := utf8.DecodeRuneInString(s1[i:])
runeValue2, width := utf8.DecodeRuneInString(s2[i:])
if runeValue1 != runeValue2 {
return res
}
if runeValue1 == utf8.RuneError || runeValue2 == utf8.RuneError {
return res
}
w = width
res = i + w
}
return res
}
func main() {
foo := "日本本a語"
bar := "日本本b語"
fmt.Println(index(foo, bar))
foo = "日本語"
bar = "日otest"
fmt.Println(index(foo, bar))
foo = "\xF0"
bar = "\xFF"
fmt.Println(index(foo, bar))
}
Here, the result would be:
9 (3 common runes of width '3')
3 (1 rune of width '3')
0 (invalid rune, meaning utf8.RuneError)
You mean like this. Please note, this will not handle UTF 8, only ascii.
package main
import (
"fmt"
)
func equal(s1, s2 string) int {
eq := 0
if len(s1) > len(s2) {
s1, s2 = s2, s1
}
for key, _ := range s1 {
if s1[key] == s2[key] {
eq++
} else {
break
}
}
return eq
}
func main() {
fmt.Println(equal("buzzfizz", "buzz"))
fmt.Println(equal("Makan", "Makon"))
fmt.Println(equal("Indah", "Ihkasyandehlo"))
}
Does anyone know how to split a string in Golang by length?
For example to split "helloworld" after every 3 characters, so it should ideally return an array of "hel" "low" "orl" "d"?
Alternatively a possible solution would be to also append a newline after every 3 characters..
All ideas are greatly appreciated!
Make sure to convert your string into a slice of rune: see "Slice string into letters".
for automatically converts string to rune so there is no additional code needed in this case to convert the string to rune first.
for i, r := range s {
fmt.Printf("i%d r %c\n", i, r)
// every 3 i, do something
}
r[n:n+3] will work best with a being a slice of rune.
The index will increase by one every rune, while it might increase by more than one for every byte in a slice of string: "世界": i would be 0 and 3: a character (rune) can be formed of multiple bytes.
For instance, consider s := "世a界世bcd界efg世": 12 runes. (see play.golang.org)
If you try to parse it byte by byte, you will miss (in a naive split every 3 chars implementation) some of the "index modulo 3" (equals to 2, 5, 8 and 11), because the index will increase past those values:
for i, r := range s {
res = res + string(r)
fmt.Printf("i %d r %c\n", i, r)
if i > 0 && (i+1)%3 == 0 {
fmt.Printf("=>(%d) '%v'\n", i, res)
res = ""
}
}
The output:
i 0 r 世
i 3 r a <== miss i==2
i 4 r 界
i 7 r 世 <== miss i==5
i 10 r b <== miss i==8
i 11 r c ===============> would print '世a界世bc', not exactly '3 chars'!
i 12 r d
i 13 r 界
i 16 r e <== miss i==14
i 17 r f ===============> would print 'd界ef'
i 18 r g
i 19 r 世 <== miss the rest of the string
But if you were to iterate on runes (a := []rune(s)), you would get what you expect, as the index would increase one rune at a time, making it easy to aggregate exactly 3 characters:
for i, r := range a {
res = res + string(r)
fmt.Printf("i%d r %c\n", i, r)
if i > 0 && (i+1)%3 == 0 {
fmt.Printf("=>(%d) '%v'\n", i, res)
res = ""
}
}
Output:
i 0 r 世
i 1 r a
i 2 r 界 ===============> would print '世a界'
i 3 r 世
i 4 r b
i 5 r c ===============> would print '世bc'
i 6 r d
i 7 r 界
i 8 r e ===============> would print 'd界e'
i 9 r f
i10 r g
i11 r 世 ===============> would print 'fg世'
Here is another variant playground.
It is by far more efficient in terms of both speed and memory than other answers. If you want to run benchmarks here they are benchmarks. In general it is 5 times faster than the previous version that was a fastest answer anyway.
func Chunks(s string, chunkSize int) []string {
if len(s) == 0 {
return nil
}
if chunkSize >= len(s) {
return []string{s}
}
var chunks []string = make([]string, 0, (len(s)-1)/chunkSize+1)
currentLen := 0
currentStart := 0
for i := range s {
if currentLen == chunkSize {
chunks = append(chunks, s[currentStart:i])
currentLen = 0
currentStart = i
}
currentLen++
}
chunks = append(chunks, s[currentStart:])
return chunks
}
Please note that the index points to a first byte of a rune on iterating over a string. The rune takes from 1 to 4 bytes. Slicing also treats the string as a byte array.
PREVIOUS SLOWER ALGORITHM
The code is here playground. The conversion from bytes to runes and then to bytes again takes a lot of time actually. So better use the fast algorithm at the top of the answer.
func ChunksSlower(s string, chunkSize int) []string {
if chunkSize >= len(s) {
return []string{s}
}
var chunks []string
chunk := make([]rune, chunkSize)
len := 0
for _, r := range s {
chunk[len] = r
len++
if len == chunkSize {
chunks = append(chunks, string(chunk))
len = 0
}
}
if len > 0 {
chunks = append(chunks, string(chunk[:len]))
}
return chunks
}
Please note that these two algorithms treat invalid UTF-8 characters in a different way. First one processes them as is when second one replaces them by utf8.RuneError symbol ('\uFFFD') that has following hexadecimal representation in UTF-8: efbfbd.
Also needed a function to do this recently, see example usage here
func SplitSubN(s string, n int) []string {
sub := ""
subs := []string{}
runes := bytes.Runes([]byte(s))
l := len(runes)
for i, r := range runes {
sub = sub + string(r)
if (i + 1) % n == 0 {
subs = append(subs, sub)
sub = ""
} else if (i + 1) == l {
subs = append(subs, sub)
}
}
return subs
}
Here is another example (you can try it here):
package main
import (
"fmt"
"strings"
)
func ChunkString(s string, chunkSize int) []string {
var chunks []string
runes := []rune(s)
if len(runes) == 0 {
return []string{s}
}
for i := 0; i < len(runes); i += chunkSize {
nn := i + chunkSize
if nn > len(runes) {
nn = len(runes)
}
chunks = append(chunks, string(runes[i:nn]))
}
return chunks
}
func main() {
fmt.Println(ChunkString("helloworld", 3))
fmt.Println(strings.Join(ChunkString("helloworld", 3), "\n"))
}
An easy solution using regex
re := regexp.MustCompile((\S{3}))
x := re.FindAllStringSubmatch("HelloWorld", -1)
fmt.Println(x)
https://play.golang.org/p/mfmaQlSRkHe
I tried 3 version to implement the function, the function named "splitByWidthMake" is fastest.
These functions ignore the unicode but only the ascii code.
package main
import (
"fmt"
"strings"
"time"
"math"
)
func splitByWidthMake(str string, size int) []string {
strLength := len(str)
splitedLength := int(math.Ceil(float64(strLength) / float64(size)))
splited := make([]string, splitedLength)
var start, stop int
for i := 0; i < splitedLength; i += 1 {
start = i * size
stop = start + size
if stop > strLength {
stop = strLength
}
splited[i] = str[start : stop]
}
return splited
}
func splitByWidth(str string, size int) []string {
strLength := len(str)
var splited []string
var stop int
for i := 0; i < strLength; i += size {
stop = i + size
if stop > strLength {
stop = strLength
}
splited = append(splited, str[i:stop])
}
return splited
}
func splitRecursive(str string, size int) []string {
if len(str) <= size {
return []string{str}
}
return append([]string{string(str[0:size])}, splitRecursive(str[size:], size)...)
}
func main() {
/*
testStrings := []string{
"hello world",
"",
"1",
}
*/
testStrings := make([]string, 10)
for i := range testStrings {
testStrings[i] = strings.Repeat("#", int(math.Pow(2, float64(i))))
}
//fmt.Println(testStrings)
t1 := time.Now()
for i := range testStrings {
_ = splitByWidthMake(testStrings[i], 2)
//fmt.Println(t)
}
elapsed := time.Since(t1)
fmt.Println("for loop version elapsed: ", elapsed)
t1 = time.Now()
for i := range testStrings {
_ = splitByWidth(testStrings[i], 2)
}
elapsed = time.Since(t1)
fmt.Println("for loop without make version elapsed: ", elapsed)
t1 = time.Now()
for i := range testStrings {
_ = splitRecursive(testStrings[i], 2)
}
elapsed = time.Since(t1)
fmt.Println("recursive version elapsed: ", elapsed)
}
Not the most efficient, will work for most use-cases.
Go playground: https://play.golang.org/p/0JSqv3OMdCR
// splitBy splits a string s by int n.
func splitBy(s string, n int) []string {
var ss []string
for i := 1; i < len(s); i++ {
if i%n == 0 {
ss = append(ss, s[:i])
s = s[i:]
i = 1
}
}
ss = append(ss, s)
return ss
}
// test
s := "helloworld"
ss := splitBy(s, 3)
fmt.Println(ss)
# output
$ go run main.go
[hel low orl d]
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