切片结构的本质是对数组的封装, 那么切片的容量是怎么增长的?
源码分析
Go1.9版本 src/runtime/slice.go 178行
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func growslice(et *_type, old slice, cap int) slice {
// ......
// 这里计算了扩容的数量
// 这里的if else判断, 整体表达的意思是:
// 1. 当原slice容量小于1024的时候, 新slice容量变成原来的2倍
// 2. 当原slice容量超过1024的时候, 新slice容量变成原来的1.25倍
newcap := old.cap
doublecap := newcap + newcap
if cap > doublecap {
newcap = cap
} else {
const threshold = 256
if old.cap < threshold {
newcap = doublecap
} else {
for 0 < newcap && newcap < cap {
newcap += (newcap + 3*threshold) / 4
}
if newcap <= 0 {
newcap = cap
}
}
}
// ........
// 以下部分重点内容是
// capmem = roundupsize(capmem) 这里的意思是做内存对齐
// 这样就导致newcap都会被重新计算
switch {
case et.size == 1:
lenmem = uintptr(old.len)
newlenmem = uintptr(cap)
capmem = roundupsize(uintptr(newcap))
overflow = uintptr(newcap) > maxAlloc
newcap = int(capmem)
case et.size == goarch.PtrSize:
lenmem = uintptr(old.len) * goarch.PtrSize
newlenmem = uintptr(cap) * goarch.PtrSize
capmem = roundupsize(uintptr(newcap) * goarch.PtrSize)
overflow = uintptr(newcap) > maxAlloc/goarch.PtrSize
newcap = int(capmem / goarch.PtrSize)
case isPowerOfTwo(et.size):
var shift uintptr
if goarch.PtrSize == 8 {
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)
overflow = uintptr(newcap) > (maxAlloc >> shift)
newcap = int(capmem >> shift)
default:
lenmem = uintptr(old.len) * et.size
newlenmem = uintptr(cap) * et.size
capmem, overflow = math.MulUintptr(et.size, uintptr(newcap))
capmem = roundupsize(capmem)
newcap = int(capmem / et.size)
}
// ......
}
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Warning
通过源码可以看出以下说法不一定准确
- 当原slice容量小于1024的时候, 新slice容量变成原来的2倍
- 当原slice容量超过1024的时候, 新slice容量变成原来的1.25倍
通过以下程序来验证:
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package main
import "fmt"
func main() {
s := make([]int, 0)
oldCap := cap(s)
for i := 0; i < 2048; i++ {
s = append(s, i)
newCap := cap(s)
if newCap != oldCap {
fmt.Printf("[%d -> %4d] cap = %-4d | after append %-4d cap = %-4d\n", 0, i-1, oldCap, i, newCap)
oldCap = newCap
}
}
}
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运行结果:
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[0 -> -1] cap = 0 | after append 0 cap = 1
[0 -> 0] cap = 1 | after append 1 cap = 2
[0 -> 1] cap = 2 | after append 2 cap = 4
[0 -> 3] cap = 4 | after append 4 cap = 8
[0 -> 7] cap = 8 | after append 8 cap = 16
[0 -> 15] cap = 16 | after append 16 cap = 32
[0 -> 31] cap = 32 | after append 32 cap = 64
[0 -> 63] cap = 64 | after append 64 cap = 128
[0 -> 127] cap = 128 | after append 128 cap = 256
[0 -> 255] cap = 256 | after append 256 cap = 512
[0 -> 511] cap = 512 | after append 512 cap = 848
[0 -> 847] cap = 848 | after append 848 cap = 1280
[0 -> 1279] cap = 1280 | after append 1280 cap = 1792
[0 -> 1791] cap = 1792 | after append 1792 cap = 2560
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Success
当老s容量小于1024的时候, 新s的容量的确是老s的2倍
Failure
当老s的容量大于等于1024的时候, 计算出来的倍数并不是1.25倍, 会比1.25大一些