Golang拼接字符串性能对比

$golang$的$string$类型是不可修改的，对于拼接字符串来说，本质上还是创建一个新的对象将数据放进去。主要有以下几种拼接方式

拼接方式介绍

1.使用$string$自带的运算符$+$

ans = ans + s

2. 使用格式化输出$fmt.Sprintf$

ans = fmt.Sprintf("%s%s", ans, s)

3. 使用$strings$的$join$函数

一般适用于将字符串数组转化为特定间隔符的字符串的情况

ans=strings.join(strs,",")

4. 使用$strings.Builder$

builder := strings.Builder{}
builder.WriteString(s)
return builder.String()

5. 使用$bytes.Buffer$

buffer := new(bytes.Buffer)
buffer.WriteString(s)
return buffer.String()

6. 使用$[]byte$，并且提前设置容量

ans := make([]byte, 0, len(s)*n)
ans = append(ans, s...)

性能对比

先写一个随机生成长度为$n$的字符串的函数

func getRandomString(n int) string {
	var tmp = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
	ans := make([]uint8, 0, n)
	for i := 0; i < n; i++ {
		ans = append(ans, tmp[rand.Intn(len(tmp))])
	}
	return string(ans)
}

接下来分别写出上述拼接方式的实现，假设每次都拼接n次字符串s后返回。

1.使用$string$自带的运算符$+$

循环$n$次，每次都令答案字符串$ans+$源字符串$s$

func plusOperatorJoin(n int, s string) string {
	var ans string
	for i := 0; i < n; i++ {
		ans = ans + s
	}
	return ans
}

2. 使用格式化输出$fmt.Sprintf$

循环$n$次，使用$fmt.Sprintf$达到拼接的目的

func sprintfJoin(n int, s string) string {
	var ans string
	for i := 0; i < n; i++ {
		ans = fmt.Sprintf("%s%s", ans, s)
	}
	return ans
}

3. 使用$strings$的$join$函数

拼接同一个字符串的话不适合用$join$函数，所以跳过这种方式

4. 使用$strings.Builder$

初始化$strings.Builder$，循环$n$次，每次调用$WriteString$方法

func stringBuilderJoin(n int, s string) string {
	builder := strings.Builder{}
	for i := 0; i < n; i++ {
		builder.WriteString(s)
	}
	return builder.String()
}

5. 使用$bytes.Buffer$

初始化$bytes.Buffer$，循环$n$次，每次调用$WriteString$方法

func bytesBufferJoin(n int, s string) string {
	buffer := new(bytes.Buffer)
	for i := 0; i < n; i++ {
		buffer.WriteString(s)
	}
	return buffer.String()
}

6. 使用$[]byte$，并且提前设置容量

定义$ans$为$byte$数组，并提前设置容量为$len(s)?n$

func bytesJoin(n int, s string) string {
	ans := make([]byte, 0, len(s)*n)
	for i := 0; i < n; i++ {
		ans = append(ans, s...)
	}
	return string(ans)
}

测试代码

先随机生成一个长度为10的字符串，然后拼接10000次。

package high_strings

import "testing"

func benchmark(b *testing.B, f func(int, string) string) {
	var str = getRandomString(10)
	for i := 0; i < b.N; i++ {
		f(10000, str)
	}
}

func BenchmarkPlusOperatorJoin(b *testing.B) {
	benchmark(b, plusOperatorJoin)
}
func BenchmarkSprintfJoin(b *testing.B) {
	benchmark(b, sprintfJoin)
}
func BenchmarkStringBuilderJoin(b *testing.B) {
	benchmark(b, stringBuilderJoin)
}
func BenchmarkBytesBufferJoin(b *testing.B) {
	benchmark(b, bytesBufferJoin)
}
func BenchmarkBytesJoin(b *testing.B) {
	benchmark(b, bytesJoin)
}

测试结果：

使用$[]byte$ > $strings.Builder$ >= $bytes.Buffer$ > $fmt.Sprintf$ > $+$运算符

源码分析

1.使用$string$自带的运算符$+$

代码在runtime\string.go里

// concatstrings implements a Go string concatenation x+y+z+...
// The operands are passed in the slice a.
// If buf != nil, the compiler has determined that the result does not
// escape the calling function, so the string data can be stored in buf
// if small enough.
func concatstrings(buf *tmpBuf, a []string) string {
	idx := 0
	l := 0
	count := 0
	for i, x := range a {
		n := len(x)
		if n == 0 {
			continue
		}
		if l+n < l {
			throw("string concatenation too long")
		}
		l += n
		count++
		idx = i
	}
	if count == 0 {
		return ""
	}

	// If there is just one string and either it is not on the stack
	// or our result does not escape the calling frame (buf != nil),
	// then we can return that string directly.
	if count == 1 && (buf != nil || !stringDataOnStack(a[idx])) {
		return a[idx]
	}
	s, b := rawstringtmp(buf, l)
	for _, x := range a {
		copy(b, x)
		b = b[len(x):]
	}
	return s
}

• 首先计算拼接后的字符串长度
• 如果只有一个字符串并且不在栈上就直接返回
• 如果$buf$不为空并且$buf$可以放下这些字符串，就把拼接后的字符串放在$buf$里，否则在堆上重新申请一块内存

func rawstringtmp(buf *tmpBuf, l int) (s string, b []byte) {
	if buf != nil && l <= len(buf) {
		b = buf[:l]
		s = slicebytetostringtmp(&b[0], len(b))
	} else {
		s, b = rawstring(l)
	}
	return
}
// rawstring allocates storage for a new string. The returned
// string and byte slice both refer to the same storage.
// The storage is not zeroed. Callers should use
// b to set the string contents and then drop b.
func rawstring(size int) (s string, b []byte) {
	p := mallocgc(uintptr(size), nil, false)
	return unsafe.String((*byte)(p), size), unsafe.Slice((*byte)(p), size)
}

• 然后遍历数组，将字符串$copy$过去

2. 使用$strings.Builder$

介绍：$strings.Builder$用于使用$Write$方法高效地生成字符串，它最大限度地减少了内存复制
拼接过程：$builder$里有一个$byte$类型的切片，每次调用$WriteString$的时候，是直接往该切片里追加字符串。因为切片底层的扩容机制是以倍数申请的，所以对比1而言，2的内存消耗要更少。
**结果返回：**在返回字符串的$String$方法里，是将$buf$数组转化为字符串直接返回的。
扩容机制： 想要缓冲区容量增加$n$个字节，扩容后容量变为$2?len+n$

// A Builder is used to efficiently build a string using Write methods.
// It minimizes memory copying. The zero value is ready to use.
// Do not copy a non-zero Builder.
type Builder struct {
	addr *Builder // of receiver, to detect copies by value
	buf  []byte
}

// String returns the accumulated string.
func (b *Builder) String() string {
	return unsafe.String(unsafe.SliceData(b.buf), len(b.buf))
}

// grow copies the buffer to a new, larger buffer so that there are at least n
// bytes of capacity beyond len(b.buf).
func (b *Builder) grow(n int) {
	buf := make([]byte, len(b.buf), 2*cap(b.buf)+n)
	copy(buf, b.buf)
	b.buf = buf
}
// WriteString appends the contents of s to b's buffer.
// It returns the length of s and a nil error.
func (b *Builder) WriteString(s string) (int, error) {
	b.copyCheck()
	b.buf = append(b.buf, s...)
	return len(s), nil
}

3. 使用$bytes.Buffer$

介绍：$bytes.Buffer$跟$strings.Builder$的底层都是$byte$数组，区别在于扩容机制和返回字符串的$String$方法。
结果返回： 因为$bytes.Buffer$实际上是一个流式的字节缓冲区，可以向尾部写入数据，也可以读取头部的数据。所以在返回字符串的$String$方法里，只返回了缓冲区里未读的部分，所以需要重新申请内存来存放返回的结果。内存会比$strings.Builder$稍慢一些。
扩容机制： 想要缓冲区容量至少增加$n$个字节，$m$是未读的长度，$c$是当前的容量。
优化点在于如果$n<=c/2?m$，也就是当前容量的一半都大于等于现有的内容（未读的字节数）加上所需要增加的字节数，就复用当前的数组，把未读的内容拷贝到头部去。

We can slide things down instead of allocating a new slice. We only need m+n <= c to slide, but we instead let capacity get twice as large so we don’t spend all our time copying.
我们可以向下滑动，而不是分配一个新的切片。我们只需要m+n<=c来滑动，但我们让容量增加了一倍，这样我们就不会把所有的时间都花在复制上。

否则的话也是$2?len+n$的扩张

// A Buffer is a variable-sized buffer of bytes with Read and Write methods.
// The zero value for Buffer is an empty buffer ready to use.
type Buffer struct {
	buf      []byte // contents are the bytes buf[off : len(buf)]
	off      int    // read at &buf[off], write at &buf[len(buf)]
	lastRead readOp // last read operation, so that Unread* can work correctly.
}
// String returns the contents of the unread portion of the buffer
// as a string. If the Buffer is a nil pointer, it returns "<nil>".
//
// To build strings more efficiently, see the strings.Builder type.
func (b *Buffer) String() string {
	if b == nil {
		// Special case, useful in debugging.
		return "<nil>"
	}
	return string(b.buf[b.off:])
}
// WriteString appends the contents of s to the buffer, growing the buffer as
// needed. The return value n is the length of s; err is always nil. If the
// buffer becomes too large, WriteString will panic with ErrTooLarge.
func (b *Buffer) WriteString(s string) (n int, err error) {
	b.lastRead = opInvalid
	m, ok := b.tryGrowByReslice(len(s))
	if !ok {
		m = b.grow(len(s))
	}
	return copy(b.buf[m:], s), nil
}

// grow grows the buffer to guarantee space for n more bytes.
// It returns the index where bytes should be written.
// If the buffer can't grow it will panic with ErrTooLarge.
func (b *Buffer) grow(n int) int {
	m := b.Len()
	// If buffer is empty, reset to recover space.
	if m == 0 && b.off != 0 {
		b.Reset()
	}
	// Try to grow by means of a reslice.
	if i, ok := b.tryGrowByReslice(n); ok {
		return i
	}
	if b.buf == nil && n <= smallBufferSize {
		b.buf = make([]byte, n, smallBufferSize)
		return 0
	}
	c := cap(b.buf)
	if n <= c/2-m {
		// We can slide things down instead of allocating a new
		// slice. We only need m+n <= c to slide, but
		// we instead let capacity get twice as large so we
		// don't spend all our time copying.
		copy(b.buf, b.buf[b.off:])
	} else if c > maxInt-c-n {
		panic(ErrTooLarge)
	} else {
		// Add b.off to account for b.buf[:b.off] being sliced off the front.
		b.buf = growSlice(b.buf[b.off:], b.off+n)
	}
	// Restore b.off and len(b.buf).
	b.off = 0
	b.buf = b.buf[:m+n]
	return m
}

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