How to efficiently concatenate strings in go

Question

In Go, a string is a primitive type, which means it is read-only, and every manipulation of it will create a new string.

So if I want to concatenate strings many times without knowing the length of the resulting string, what's the best way to do it?

The naive way would be:

var s string
for i := 0; i < 1000; i++ {
    s += getShortStringFromSomewhere()
}
return s

but that does not seem very efficient.

Answer 1

New Way:

From Go 1.10 there is a strings.Builder type, please take a look at this answer for more detail.

Old Way:

Use the bytes package. It has a Buffer type which implements io.Writer.

package main

import (
    "bytes"
    "fmt"
)

func main() {
    var buffer bytes.Buffer

    for i := 0; i < 1000; i++ {
        buffer.WriteString("a")
    }

    fmt.Println(buffer.String())
}

This does it in O(n) time.

Answer 2

In Go 1.10+ there is strings.Builder, here.

A Builder is used to efficiently build a string using Write methods. It minimizes memory copying. The zero value is ready to use.

---

Example

It's almost the same with bytes.Buffer.

package main

import (
    "strings"
    "fmt"
)

func main() {
    // ZERO-VALUE:
    //
    // It's ready to use from the get-go.
    // You don't need to initialize it.
    var sb strings.Builder

    for i := 0; i < 1000; i++ {
        sb.WriteString("a")
    }

    fmt.Println(sb.String())
}

Click to see this on the playground.

---

Supported Interfaces

strings.Builder's methods are being implemented with the existing interfaces in mind so that you can switch to the new Builder type easily in your code.

Method Signature	Interface	Description
Grow(int)	bytes.Buffer	Grows the buffer's capacity by the specified amount. See bytes.Buffer#Grow for more information.
Len() int	bytes.Buffer	Returns the number of bytes in the buffer. See bytes.Buffer#Len for more information.
Reset()	bytes.Buffer	Resets the buffer to be empty. See bytes.Buffer#Reset for more information.
String() string	fmt.Stringer	Returns the contents of the buffer as a string. See fmt.Stringer for more information.
Write([]byte) (int, error)	io.Writer	Writes the given bytes to the buffer. See io.Writer for more information.
WriteByte(byte) error	io.ByteWriter	Writes the given byte to the buffer. See io.ByteWriter for more information.
WriteRune(rune) (int, error)	bufio.Writer or bytes.Buffer	Writes the given rune to the buffer. See bufio.Writer#WriteRune or bytes.Buffer#WriteRune for more information.
WriteString(string) (int, error)	io.stringWriter	Writes the given string to the buffer. See io.stringWriter for more information.

---

Differences from bytes.Buffer

• It can only grow or reset.
• It has a copyCheck mechanism built-in that prevents accidentally copying it. In bytes.Buffer, one can access the underlying bytes like this: (*Buffer).Bytes(). strings.Builder prevents this problem. Sometimes, this is not a problem, though, and is desired instead. For example: For the peeking behavior when the bytes are passed to an io.Reader etc.
• bytes.Buffer.Reset() rewinds and reuses the underlying buffer whereas the strings.Builder.Reset() does not, it detaches the buffer.

---

Note

• Do not copy a strings.Builder value as it caches the underlying data.
• If you want to share a strings.Builder value, use a pointer to it.

---

Check out its source code for more details, here.

Answer 3

If you know the total length of the string that you're going to preallocate then the most efficient way to concatenate strings may be using the builtin function copy. If you don't know the total length before hand, do not use copy, and read the other answers instead.

In my tests, that approach is ~3x faster than using bytes.Buffer and much much faster (~12,000x) than using the operator +. Also, it uses less memory.

I've created a test case to prove this and here are the results:

BenchmarkConcat  1000000    64497 ns/op   502018 B/op   0 allocs/op
BenchmarkBuffer  100000000  15.5  ns/op   2 B/op        0 allocs/op
BenchmarkCopy    500000000  5.39  ns/op   0 B/op        0 allocs/op

---

Below is code for testing:

package main

import (
    "bytes"
    "strings"
    "testing"
)

func BenchmarkConcat(b *testing.B) {
    var str string
    for n := 0; n < b.N; n++ {
        str += "x"
    }
    b.StopTimer()

    if s := strings.Repeat("x", b.N); str != s {
        b.Errorf("unexpected result; got=%s, want=%s", str, s)
    }
}

func BenchmarkBuffer(b *testing.B) {
    var buffer bytes.Buffer
    for n := 0; n < b.N; n++ {
        buffer.WriteString("x")
    }
    b.StopTimer()

    if s := strings.Repeat("x", b.N); buffer.String() != s {
        b.Errorf("unexpected result; got=%s, want=%s", buffer.String(), s)
    }
}

func BenchmarkCopy(b *testing.B) {
    bs := make([]byte, b.N)
    bl := 0

    b.ResetTimer()
    for n := 0; n < b.N; n++ {
        bl += copy(bs[bl:], "x")
    }
    b.StopTimer()

    if s := strings.Repeat("x", b.N); string(bs) != s {
        b.Errorf("unexpected result; got=%s, want=%s", string(bs), s)
    }
}

// Go 1.10
func BenchmarkStringBuilder(b *testing.B) {
    var strBuilder strings.Builder

    b.ResetTimer()
    for n := 0; n < b.N; n++ {
        strBuilder.WriteString("x")
    }
    b.StopTimer()

    if s := strings.Repeat("x", b.N); strBuilder.String() != s {
        b.Errorf("unexpected result; got=%s, want=%s", strBuilder.String(), s)
    }
}

Answer 4

If you have a string slice that you want to efficiently convert to a string then you can use this approach. Otherwise, take a look at the other answers.

There is a library function in the strings package called Join: http://golang.org/pkg/strings/#Join

A look at the code of Join shows a similar approach to Append function Kinopiko wrote: https://golang.org/src/strings/strings.go#L420

Usage:

import (
    "fmt";
    "strings";
)

func main() {
    s := []string{"this", "is", "a", "joined", "string\n"};
    fmt.Printf(strings.Join(s, " "));
}

$ ./test.bin
this is a joined string

Answer 5

I just benchmarked the top answer posted above in my own code (a recursive tree walk) and the simple concat operator is actually faster than the BufferString.

func (r *record) String() string {
    buffer := bytes.NewBufferString("");
    fmt.Fprint(buffer,"(",r.name,"[")
    for i := 0; i < len(r.subs); i++ {
        fmt.Fprint(buffer,"\t",r.subs[i])
    }
    fmt.Fprint(buffer,"]",r.size,")\n")
    return buffer.String()
}

This took 0.81 seconds, whereas the following code:

func (r *record) String() string {
    s := "(\"" + r.name + "\" ["
    for i := 0; i < len(r.subs); i++ {
        s += r.subs[i].String()
    }
    s += "] " + strconv.FormatInt(r.size,10) + ")\n"
    return s
} 

only took 0.61 seconds. This is probably due to the overhead of creating the new BufferString.

Update: I also benchmarked the join function and it ran in 0.54 seconds.

func (r *record) String() string {
    var parts []string
    parts = append(parts, "(\"", r.name, "\" [" )
    for i := 0; i < len(r.subs); i++ {
        parts = append(parts, r.subs[i].String())
    }
    parts = append(parts, strconv.FormatInt(r.size,10), ")\n")
    return strings.Join(parts,"")
}

Answer 6

package main

import (
  "fmt"
)

func main() {
    var str1 = "string1"
    var str2 = "string2"
    out := fmt.Sprintf("%s %s ",str1, str2)
    fmt.Println(out)
}

Answer 7

This is the fastest solution that does not require you to know or calculate the overall buffer size first:

var data []byte
for i := 0; i < 1000; i++ {
    data = append(data, getShortStringFromSomewhere()...)
}
return string(data)

By my benchmark, it's 20% slower than the copy solution (8.1ns per append rather than 6.72ns) but still 55% faster than using bytes.Buffer.

Answer 8

You could create a big slice of bytes and copy the bytes of the short strings into it using string slices. There is a function given in "Effective Go":

func Append(slice, data[]byte) []byte {
    l := len(slice);
    if l + len(data) > cap(slice) { // reallocate
        // Allocate double what's needed, for future growth.
        newSlice := make([]byte, (l+len(data))*2);
        // Copy data (could use bytes.Copy()).
        for i, c := range slice {
            newSlice[i] = c
        }
        slice = newSlice;
    }
    slice = slice[0:l+len(data)];
    for i, c := range data {
        slice[l+i] = c
    }
    return slice;
}

Then when the operations are finished, use string ( ) on the big slice of bytes to convert it into a string again.

Answer 9

Note added in 2018

From Go 1.10 there is a strings.Builder type, please take a look at this answer for more detail.

Pre-201x answer

The benchmark code of @cd1 and other answers are wrong. b.N is not supposed to be set in benchmark function. It's set by the go test tool dynamically to determine if the execution time of the test is stable.

A benchmark function should run the same test b.N times and the test inside the loop should be the same for each iteration. So I fix it by adding an inner loop. I also add benchmarks for some other solutions:

package main

import (
    "bytes"
    "strings"
    "testing"
)

const (
    sss = "xfoasneobfasieongasbg"
    cnt = 10000
)

var (
    bbb      = []byte(sss)
    expected = strings.Repeat(sss, cnt)
)

func BenchmarkCopyPreAllocate(b *testing.B) {
    var result string
    for n := 0; n < b.N; n++ {
        bs := make([]byte, cnt*len(sss))
        bl := 0
        for i := 0; i < cnt; i++ {
            bl += copy(bs[bl:], sss)
        }
        result = string(bs)
    }
    b.StopTimer()
    if result != expected {
        b.Errorf("unexpected result; got=%s, want=%s", string(result), expected)
    }
}

func BenchmarkAppendPreAllocate(b *testing.B) {
    var result string
    for n := 0; n < b.N; n++ {
        data := make([]byte, 0, cnt*len(sss))
        for i := 0; i < cnt; i++ {
            data = append(data, sss...)
        }
        result = string(data)
    }
    b.StopTimer()
    if result != expected {
        b.Errorf("unexpected result; got=%s, want=%s", string(result), expected)
    }
}

func BenchmarkBufferPreAllocate(b *testing.B) {
    var result string
    for n := 0; n < b.N; n++ {
        buf := bytes.NewBuffer(make([]byte, 0, cnt*len(sss)))
        for i := 0; i < cnt; i++ {
            buf.WriteString(sss)
        }
        result = buf.String()
    }
    b.StopTimer()
    if result != expected {
        b.Errorf("unexpected result; got=%s, want=%s", string(result), expected)
    }
}

func BenchmarkCopy(b *testing.B) {
    var result string
    for n := 0; n < b.N; n++ {
        data := make([]byte, 0, 64) // same size as bootstrap array of bytes.Buffer
        for i := 0; i < cnt; i++ {
            off := len(data)
            if off+len(sss) > cap(data) {
                temp := make([]byte, 2*cap(data)+len(sss))
                copy(temp, data)
                data = temp
            }
            data = data[0 : off+len(sss)]
            copy(data[off:], sss)
        }
        result = string(data)
    }
    b.StopTimer()
    if result != expected {
        b.Errorf("unexpected result; got=%s, want=%s", string(result), expected)
    }
}

func BenchmarkAppend(b *testing.B) {
    var result string
    for n := 0; n < b.N; n++ {
        data := make([]byte, 0, 64)
        for i := 0; i < cnt; i++ {
            data = append(data, sss...)
        }
        result = string(data)
    }
    b.StopTimer()
    if result != expected {
        b.Errorf("unexpected result; got=%s, want=%s", string(result), expected)
    }
}

func BenchmarkBufferWrite(b *testing.B) {
    var result string
    for n := 0; n < b.N; n++ {
        var buf bytes.Buffer
        for i := 0; i < cnt; i++ {
            buf.Write(bbb)
        }
        result = buf.String()
    }
    b.StopTimer()
    if result != expected {
        b.Errorf("unexpected result; got=%s, want=%s", string(result), expected)
    }
}

func BenchmarkBufferWriteString(b *testing.B) {
    var result string
    for n := 0; n < b.N; n++ {
        var buf bytes.Buffer
        for i := 0; i < cnt; i++ {
            buf.WriteString(sss)
        }
        result = buf.String()
    }
    b.StopTimer()
    if result != expected {
        b.Errorf("unexpected result; got=%s, want=%s", string(result), expected)
    }
}

func BenchmarkConcat(b *testing.B) {
    var result string
    for n := 0; n < b.N; n++ {
        var str string
        for i := 0; i < cnt; i++ {
            str += sss
        }
        result = str
    }
    b.StopTimer()
    if result != expected {
        b.Errorf("unexpected result; got=%s, want=%s", string(result), expected)
    }
}

Environment is OS X 10.11.6, 2.2 GHz Intel Core i7

Test results:

BenchmarkCopyPreAllocate-8         20000             84208 ns/op          425984 B/op          2 allocs/op
BenchmarkAppendPreAllocate-8       10000            102859 ns/op          425984 B/op          2 allocs/op
BenchmarkBufferPreAllocate-8       10000            166407 ns/op          426096 B/op          3 allocs/op
BenchmarkCopy-8                    10000            160923 ns/op          933152 B/op         13 allocs/op
BenchmarkAppend-8                  10000            175508 ns/op         1332096 B/op         24 allocs/op
BenchmarkBufferWrite-8             10000            239886 ns/op          933266 B/op         14 allocs/op
BenchmarkBufferWriteString-8       10000            236432 ns/op          933266 B/op         14 allocs/op
BenchmarkConcat-8                     10         105603419 ns/op        1086685168 B/op    10000 allocs/op

Conclusion:

1. CopyPreAllocate is the fastest way; AppendPreAllocate is pretty close to No.1, but it's easier to write the code.
2. Concat has really bad performance both for speed and memory usage. Don't use it.
3. Buffer#Write and Buffer#WriteString are basically the same in speed, contrary to what @Dani-Br said in the comment. Considering string is indeed []byte in Go, it makes sense.
4. bytes.Buffer basically use the same solution as Copy with extra book keeping and other stuff.
5. Copy and Append use a bootstrap size of 64, the same as bytes.Buffer
6. Append use more memory and allocs, I think it's related to the grow algorithm it use. It's not growing memory as fast as bytes.Buffer

Suggestion:

1. For simple task such as what OP wants, I would use Append or AppendPreAllocate. It's fast enough and easy to use.
2. If need to read and write the buffer at the same time, use bytes.Buffer of course. That's what it's designed for.

Answer 10

My original suggestion was

s12 := fmt.Sprint(s1,s2)

But above answer using bytes.Buffer - WriteString() is the most efficient way.

My initial suggestion uses reflection and a type switch. See (p *pp) doPrint and (p *pp) printArg
There is no universal Stringer() interface for basic types, as I had naively thought.

At least though, Sprint() internally uses a bytes.Buffer. Thus

`s12 := fmt.Sprint(s1,s2,s3,s4,...,s1000)`

is acceptable in terms of memory allocations.

=> Sprint() concatenation can be used for quick debug output.
=> Otherwise use bytes.Buffer ... WriteString

Answer 11

Expanding on cd1's answer: You might use append() instead of copy(). append() makes ever bigger advance provisions, costing a little more memory, but saving time. I added two more benchmarks at the top of yours. Run locally with

go test -bench=. -benchtime=100ms

On my thinkpad T400s it yields:

BenchmarkAppendEmpty    50000000         5.0 ns/op
BenchmarkAppendPrealloc 50000000         3.5 ns/op
BenchmarkCopy           20000000        10.2 ns/op

Answer 12

This is actual version of benchmark provided by @cd1 (Go 1.8, linux x86_64) with the fixes of bugs mentioned by @icza and @PickBoy.

Bytes.Buffer is only 7 times faster than direct string concatenation via + operator.

package performance_test

import (
    "bytes"
    "fmt"
    "testing"
)

const (
    concatSteps = 100
)

func BenchmarkConcat(b *testing.B) {
    for n := 0; n < b.N; n++ {
        var str string
        for i := 0; i < concatSteps; i++ {
            str += "x"
        }
    }
}

func BenchmarkBuffer(b *testing.B) {
    for n := 0; n < b.N; n++ {
        var buffer bytes.Buffer
        for i := 0; i < concatSteps; i++ {
            buffer.WriteString("x")
        }
    }
}

Timings:

BenchmarkConcat-4                             300000          6869 ns/op
BenchmarkBuffer-4                            1000000          1186 ns/op

Answer 13

goutils.JoinBetween

 func JoinBetween(in []string, separator string, startIndex, endIndex int) string {
    if in == nil {
        return ""
    }

    noOfItems := endIndex - startIndex

    if noOfItems <= 0 {
        return EMPTY
    }

    var builder strings.Builder

    for i := startIndex; i < endIndex; i++ {
        if i > startIndex {
            builder.WriteString(separator)
        }
        builder.WriteString(in[i])
    }
    return builder.String()
}

Answer 14

I do it using the following :-

package main

import (
    "fmt"
    "strings"
)

func main (){
    concatenation:= strings.Join([]string{"a","b","c"},"") //where second parameter is a separator. 
    fmt.Println(concatenation) //abc
}

Answer 15

package main

import (
"fmt"
)

func main() {
    var str1 = "string1"
    var str2 = "string2"
    result := make([]byte, 0)
    result = append(result, []byte(str1)...)
    result = append(result, []byte(str2)...)
    result = append(result, []byte(str1)...)
    result = append(result, []byte(str2)...)

    fmt.Println(string(result))
}

Answer 16

Simple and easy to digest solution. Details in the comments. Copy overwrites the elements of slice. We are slicing single-single element and overwriting it.

package main

import (
    "fmt"
)

var N int = 100000

func main() {
    slice1 := make([]rune, N, N)
    //Efficient with fast performance, Need pre-allocated memory
    //We can add a check if we reached the limit then increase capacity
    //using append, but would be fined for data copying to new array. Also append happens after the length of current slice.
    for i := 0; i < N; i++ {
        copy(slice1[i:i+1], []rune{'N'})
    }
    fmt.Println(slice1)

    //Simple but fast solution, Every time the slice capacity is reached we get a fine of effort that goes
    //in copying data to new array
    slice2 := []rune{}
    for i := 0; i <= N; i++ {
        slice2 = append(slice2, 'N')
    }
    fmt.Println(slice2)

}

Answer 17

benchmark result with memory allocation statistics. check benchmark code at github.

use strings.Builder to optimize performance.

go test -bench . -benchmem
goos: darwin
goarch: amd64
pkg: github.com/hechen0/goexp/exps
BenchmarkConcat-8                1000000             60213 ns/op          503992 B/op          1 allocs/op
BenchmarkBuffer-8               100000000               11.3 ns/op             2 B/op          0 allocs/op
BenchmarkCopy-8                 300000000                4.76 ns/op            0 B/op          0 allocs/op
BenchmarkStringBuilder-8        1000000000               4.14 ns/op            6 B/op          0 allocs/op
PASS
ok      github.com/hechen0/goexp/exps   70.071s

Answer 18

strings.Join() from the "strings" package

If you have a type mismatch(like if you are trying to join an int and a string), you do RANDOMTYPE (thing you want to change)

EX:

package main

import (
    "fmt"
    "strings"
)

var intEX = 0
var stringEX = "hello all you "
var stringEX2 = "people in here"


func main() {
    s := []string{stringEX, stringEX2}
    fmt.Println(strings.Join(s, ""))
}

Output :

hello all you people in here

Answer 19

s := fmt.Sprintf("%s%s", []byte(s1), []byte(s2))