共计 4725 个字符,预计需要花费 12 分钟才能阅读完成。
在进行本地 file
文件内容读取,或进行 HTTP
网络接口通信的时候,咱们常常应用 io.ReadAll
来读取近程接口返回的 resp.Body
,但接口返回数据量有大有小,io.ReadAll
是怎么实现全副数据的读取的?
带着此疑难,让咱们走近 io.ReadAll
源码一探到底:
1. Demo 读取文件内容
package main
import (
"fmt"
"io"
"os"
)
func main() {
// 读取文件内容
fileInfo, err := os.Open("./abc.go")
if err != nil {panic(err)
}
contentBytes, err := io.ReadAll(fileInfo)
if err != nil {panic(err)
}
fmt.Println(string(contentBytes))
}
此时读取的 IO stream
大小并不知道,io.ReadAll
应用什么策略读取全副数据呢?滑动窗口?线性 / 指数递增读取?Talk is cheap. Show me the code
.
2. io.ReadAll Code
go1.16/src/io/io.go#L626
// ReadAll reads from r until an error or EOF and returns the data it read.
// A successful call returns err == nil, not err == EOF. Because ReadAll is
// defined to read from src until EOF, it does not treat an EOF from Read
// as an error to be reported.
func ReadAll(r Reader) ([]byte, error) {b := make([]byte, 0, 512)
for {if len(b) == cap(b) {// Add more capacity (let append pick how much).
b = append(b, 0)[:len(b)]
}
//println(cap(b))
n, err := r.Read(b[len(b):cap(b)])
b = b[:len(b)+n]
if err != nil {
if err == EOF {err = nil}
return b, err
}
}
}
源码解析:
从下面源码能够看到,应用 make
先默认申请 cap = 512
的 []byte
,而后进入 for
循环迭代,直到数据全副读取实现。for
循环中,首先通过 len(b) == cap(b)
判断 b
的容量是否满了,如果曾经满了,应用 append(b, 0)
追加一个元素,此时会产生什么呢?
咱们晓得,一个 slice
容量不够了须要扩容,但扩容机制是怎么的呢?持续 Show me the code
.
3. slice 扩容机制
go1.16/src/runtime/slice.go#L125
// growslice handles slice growth during append.
// It is passed the slice element type, the old slice, and the desired new minimum capacity,
// and it returns a new slice with at least that capacity, with the old data
// copied into it.
// The new slice's length is set to the old slice's length,
// NOT to the new requested capacity.
// This is for codegen convenience. The old slice's length is used immediately
// to calculate where to write new values during an append.
// TODO: When the old backend is gone, reconsider this decision.
// The SSA backend might prefer the new length or to return only ptr/cap and save stack space.
func growslice(et *_type, old slice, cap int) slice {
...
newcap := old.cap
doublecap := newcap + newcap
//println("newcap:", newcap)
//println("cap:", cap)
if cap > doublecap {newcap = cap} else {
if old.cap < 1024 {newcap = doublecap} else {
// Check 0 < newcap to detect overflow
// and prevent an infinite loop.
for 0 < newcap && newcap < cap {newcap += newcap / 4}
// Set newcap to the requested cap when
// the newcap calculation overflowed.
if newcap <= 0 {newcap = cap}
}
}
...
}
源码解析:
从下面源码能够看到,slice
扩容算法为:
1). 当须要的容量 (cap
) 超过原切片容量的两倍 (doublecap
) 时,会应用须要的容量作为新容量 (newcap
);
2). 当原切片容量 < 1024
时,新切片的容量(newcap
) 会间接翻倍 (doublecap
);
3). 当原切片容量 >= 1024
时,会按原切片容量重复地减少 1/4
,直到新容量(newcap
) 超过所须要的容量;
举例说明:
在下面 io.ReadAll
源码中,初始 slice cap = 512
,前面扩容将会:
512
1024(doublecap)
1280(1024 + 1024/4)
1600(1280 + 1280/4)
2000(1600 + 1600/4)
...
理论扩容 cap
是这样的吗?让咱们验证一下:
before newcap: 1024
-after newcap: 1024
before newcap: 1280
-after newcap: 1280
before newcap: 1600
-after newcap: 1792
before newcap: 2240
-after newcap: 2304
奇怪?发现 after newcap
并没有依照下面料想的值扩容,认真挖代码,发现除了依照下面 slice cap
扩容外,还对内存调配进行了“对齐”:
go1.16/src/runtime/slice.go#L198
println("before newcap:", newcap)
var overflow bool
var lenmem, newlenmem, capmem uintptr
// Specialize for common values of et.size.
// For 1 we don't need any division/multiplication.
// For sys.PtrSize, compiler will optimize division/multiplication into a shift by a constant.
// For powers of 2, use a variable shift.
switch {
...
case isPowerOfTwo(et.size):
var shift uintptr
if sys.PtrSize == 8 {
// Mask shift for better code generation.
shift = uintptr(sys.Ctz64(uint64(et.size))) & 63
} else {shift = uintptr(sys.Ctz32(uint32(et.size))) & 31
}
lenmem = uintptr(old.len) << shift
newlenmem = uintptr(cap) << shift
capmem = roundupsize(uintptr(newcap) << shift) // 进入到内存块 (memory block) 调配
overflow = uintptr(newcap) > (maxAlloc >> shift)
newcap = int(capmem >> shift)
...
}
println("after newcap:", newcap)
进入到内存块 (memory block
) 调配:go1.16/src/runtime/msize.go#L13
// Returns size of the memory block that mallocgc will allocate if you ask for the size.
func roundupsize(size uintptr) uintptr {
if size < _MaxSmallSize {
if size <= smallSizeMax-8 {return uintptr(class_to_size[size_to_class8[divRoundUp(size, smallSizeDiv)]])
} else {return uintptr(class_to_size[size_to_class128[divRoundUp(size-smallSizeMax, largeSizeDiv)]])
}
}
if size+_PageSize < size {return size}
return alignUp(size, _PageSize)
}
获取 spanClass
对应的 size
:go1.16/src/runtime/sizeclasses.go#L84
const (_NumSizeClasses = 68)
var class_to_size = [_NumSizeClasses]uint16{0, 8, 16, 24, 32, 48, 64, 80, 96, 112, 128,
144, 160, 176, 192, 208, 224, 240, 256, 288, 320, 352, 384, 416, 448, 480, 512, 576, 640,
704, 768, 896, 1024, 1152, 1280, 1408, 1536, 1792, 2048, 2304, 2688, 3072, 3200, 3456,
4096, 4864, 5376, 6144, 6528, 6784, 6912, 8192, 9472, 9728, 10240, 10880, 12288, 13568,
14336, 16384, 18432, 19072, 20480, 21760, 24576, 27264, 28672, 32768}
从下面 68
类 spanClass
能够看到,咱们想要调配 1600
被对齐到了 1792
,2240
被对齐到了 2304
,合乎上面的验证后果:
before newcap: 1024
-after newcap: 1024
before newcap: 1280
-after newcap: 1280
before newcap: 1600
-after newcap: 1792
before newcap: 2240
-after newcap: 2304
4. 小结
从下面的源码剖析能够看到,io.ReadAll
通过应用 slice append
主动扩容 + 内存对齐机制,应用减少的容量来实现对 io stream
的全副读取。slice append
扩容算法为:
1). 当须要的容量 (cap
) 超过原切片容量的两倍 (doublecap
) 时,会应用须要的容量作为新容量 (newcap
);
2). 当原切片容量 < 1024
时,新切片的容量(newcap
) 会间接翻倍 (doublecap
);
3). 当原切片容量 >= 1024
时,会按原切片容量重复地减少 1/4
,直到新容量(newcap
) 超过所须要的容量;
前面将会有更多系列文章,解读内存调配、GC
机制、GPM
调度、面试系列、K8s
系列、etcd
系列等,如有谬误恳请斧正。最初,祝大家端午节高兴~