在 golang 中声明一个空的 map[string]interface{} 内存 cost/overhead 是多少？

Question

出于好奇，来自 source code

type hmap struct {
    count     int // 1 word
    flags     uint8
    B         uint8  
    noverflow uint16 
    hash0     uint32 // = 8 byte

    buckets    unsafe.Pointer // 1 word
    oldbuckets unsafe.Pointer // 1 word
    nevacuate  uintptr        // 1 word

    extra *mapextra // 1 word
}

所以至少：5 个字 + 8 个字节

但为什么 creation cost is 0？ -

package main

import (
    "fmt"
    "runtime"
)

func main() {
    var m1, m2 runtime.MemStats
    var i byte

    runtime.ReadMemStats(&m1)
    mp := map[byte]byte{}
    runtime.ReadMemStats(&m2)
    fmt.Println("Bytes allocated on creation:", m2.Alloc-m1.Alloc)
    for i = 0; i < 100; i++ {
        runtime.ReadMemStats(&m1)
        mp[i] = i
        runtime.ReadMemStats(&m2)
        fmt.Printf("Bytes allocated on assignment %d: %d\n", i, m2.Alloc-m1.Alloc)
    }
}

游乐场：https://play.golang.org/p/iyYshDzexQt

输出：

Bytes allocated on creation: 0

Answer 1

why the creation cost is 0?

Go 堆栈分配在堆上分配零字节。

mp := map[byte]byte{}

main map[byte]byte literal does not escape

package main

import (
    "fmt"
    "runtime"
)

func main() {
    var m1, m2 runtime.MemStats
    var i byte

    runtime.ReadMemStats(&m1)
    mp := map[byte]byte{}
    runtime.ReadMemStats(&m2)
    fmt.Println("Bytes allocated on creation:", m2.Alloc-m1.Alloc)
    for i = 0; i < 100; i++ {
        runtime.ReadMemStats(&m1)
        mp[i] = i
        runtime.ReadMemStats(&m2)
        fmt.Printf("Bytes allocated on assignment %d: %d\n", i, m2.Alloc-m1.Alloc)
    }
}

游乐场：https://play.golang.org/p/iyYshDzexQt

输出：

Bytes allocated on creation: 0

在 golang 中声明一个空的 map[string]interface{} 内存 cost/overhead 是多少？

What's the memory cost/overhead of declaring one empty map[string]interface{} in golang?

hashtable

go

data-structures