在Go中，字符串是一种基本类型，这意味着它是只读的，对它的每次操作都会创建一个新字符串。
如果我想多次连接字符串而不知道结果字符串的长度，最好的方法是什么?
最天真的做法是:
var s string
for i := 0; i < 1000; i++ {
s += getShortStringFromSomewhere()
}
return s
但这似乎不是很有效。
当前回答
我只是在我自己的代码(递归树遍历)中对上面发布的顶部答案进行了基准测试，简单的concat操作符实际上比BufferString更快。
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()
}
这花了0.81秒，而下面的代码:
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
}
只花了0.61秒。这可能是由于创建新BufferString的开销。
更新:我还对连接函数进行了基准测试，它在0.54秒内运行。
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,"")
}
其他回答
我最初的建议是
s12 := fmt.Sprint(s1,s2)
但以上答案使用字节。Buffer - WriteString()是最有效的方法。
我最初的建议是使用反射和类型开关。参见(p *pp) doPrint和(p *pp) printArg 我曾经天真地认为，基本类型没有通用的Stringer()接口。
至少Sprint()内部使用bytes.Buffer。因此
`s12 := fmt.Sprint(s1,s2,s3,s4,...,s1000)`
在内存分配方面是可接受的。
Sprint()连接可用于快速调试输出。 =>否则使用bytes。缓冲……WriteString
2018年新增说明
从Go 1.10开始，有一个字符串。建造者类型，请看看这个答案的更多细节。
pre - 201 x的答案
@cd1的基准代码和其他答案是错误的。b.N不应该在基准函数中设置。它由go测试工具动态设置，以确定测试的执行时间是否稳定。
基准测试函数应该运行相同的测试b.N次，循环中的测试应该在每次迭代中都是相同的。所以我通过添加一个内循环来解决这个问题。我还添加了一些其他解决方案的基准:
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)
}
}
环境是OS X 10.11.6, 2.2 GHz英特尔酷睿i7
测试结果:
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
结论:
CopyPreAllocate is the fastest way; AppendPreAllocate is pretty close to No.1, but it's easier to write the code. Concat has really bad performance both for speed and memory usage. Don't use it. 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. bytes.Buffer basically use the same solution as Copy with extra book keeping and other stuff. Copy and Append use a bootstrap size of 64, the same as bytes.Buffer 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
建议:
对于OP需要的简单任务，我将使用Append或AppendPreAllocate。它足够快而且容易使用。 如果需要同时读取和写入缓冲区，则使用字节。当然是缓冲区。这就是它的设计目的。
我只是在我自己的代码(递归树遍历)中对上面发布的顶部答案进行了基准测试，简单的concat操作符实际上比BufferString更快。
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()
}
这花了0.81秒，而下面的代码:
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
}
只花了0.61秒。这可能是由于创建新BufferString的开销。
更新:我还对连接函数进行了基准测试，它在0.54秒内运行。
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,"")
}
如果你有一个字符串切片，你想要有效地转换成一个字符串，那么你可以使用这种方法。否则，看看其他答案。
在strings包中有一个名为Join的库函数: http://golang.org/pkg/strings/#Join
看看Join的代码，可以看到Kinopiko写的类似于Append函数的方法:https://golang.org/src/strings/strings.go#L420
用法:
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
package main
import (
"fmt"
)
func main() {
var str1 = "string1"
var str2 = "string2"
out := fmt.Sprintf("%s %s ",str1, str2)
fmt.Println(out)
}
