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1. 前言

如果你正在使用imroc/req io.Copyioutil.ReadAll,或者尝试对大量对象的内存分配和释放的场景进行优化,这篇文章可能对你有帮助

2. 起因

在我们的一个程序中,使用库imroc/req请求后端的HTTP服务, 从HTTP响应读取结果

有如下调用关系 Resq.Bytes() -> Resq.ToBytes() -> ioutil.ReadAll(r io.Reader) -> Buff.ReadFrom(r io.Reader)

使用pprof收集内存累计分配情况可以发现大量的内存分配由Buffer.grow(n int) 触发

image_1cp8krn3i1kpi1td317khkigqg09.png-132kB

3. 原因分析

ReadFrom的源码如下

// MinRead is the minimum slice size passed to a Read call by
// Buffer.ReadFrom. As long as the Buffer has at least MinRead bytes beyond
// what is required to hold the contents of r, ReadFrom will not grow the
// underlying buffer.
const MinRead = 512
// ReadFrom reads data from r until EOF and appends it to the buffer, growing
// the buffer as needed. The return value n is the number of bytes read. Any
// error except io.EOF encountered during the read is also returned. If the
// buffer becomes too large, ReadFrom will panic with ErrTooLarge.
func (b *Buffer) ReadFrom(r io.Reader) (n int64, err error) {
	b.lastRead = opInvalid
	for {
		i := b.grow(MinRead)
		m, e := r.Read(b.buf[i:cap(b.buf)])
		if m < 0 {
			panic(errNegativeRead)
		}

		b.buf = b.buf[:i+m]
		n += int64(m)
		if e == io.EOF {
			return n, nil // e is EOF, so return nil explicitly
		}
		if e != nil {
			return n, e
		}
	}
}

由于ioutil.ReadAll不知道最终需要多大的空间来存储结果数据,它采取的做法是,初始分配一个较小的Buff(大小为MinRead), 一边从输入中读取数据放入Buff, 一边看是否能够存下所有数据,如果不能则尝试扩大这个Buff(调用Buffer.grow())

// 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
	}
	// Check if we can make use of bootstrap array.
	if b.buf == nil && n <= len(b.bootstrap) {
		b.buf = b.bootstrap[:n]
		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 {
		// Not enough space anywhere, we need to allocate.
		buf := makeSlice(2*c + n) // **注意这里**
		copy(buf, b.buf[b.off:])  
		b.buf = buf
	}
	// Restore b.off and len(b.buf).
	b.off = 0
	b.buf = b.buf[:m+n]
	return m
}

如果需要读取的结果数据较大,则可能反复 调用makeSlice(n int)创建字节数组。这将会产生大量无用的临时对象,且增加GC的难度。

4.优化方案

  • 如果我们大致可以知道HTTP响应的大小,那么可以提前给Buff分配一个合适且较大的字节数组,避免反复调用makeSlice(n int)
  • 使用sync.Pool 重复使用临时对象,减少重复创建的开销

为了验证方案,这里做出对比试验

1)未优化代码

mem_allocate1.go

2)优化代码

mem_allocate2.go

执行10,000次的操作的结果对比

指标未优化方案优化方案
mem.TotalAlloc(为堆对象总计分配的字节数)186,170,65642,164,192
mem.Mallocs(为创建堆对象总计的内存申请次数)732,161681,532
mem.Frees(为销毁堆对象总计的内存释放次数)717,737661,557

由上面图表知,临时对象的申请和释放次数都有明显下降,优化方案是有效的(事实上随着执行次数的增加,指标的差距还会进一步拉大)。

其它类似场景的优化

在某些情况下,你可能会用到io.Copy

io.Copy(dst Writer, src Reader)

io.Copy 从src拷贝数据到dst,中间会借助字节数组作为buf
你可以使用io.CopyBuffer 替代 io.Copy

io.CopyBuffer(dst Writer, src Reader, buf []byte)

并使用sync.Pool缓存buf, 以减少临时对象的申请和释放。


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