Golang如何快速删除map所有元素

a := make(map[string]int)

a["a"] = 1
a["b"] = 2

// clear all
a = make(map[string]int)

package main

func main() {
        m := make(map[byte]int)

        m[1] = 1
        m[2] = 2

        for k := range m {
	        delete(m, k)
        }
}

go tool compile -S map_clear.go

go tool compile -l -N -S map_clear.go

func mapclear(t *maptype, h *hmap) {
	...
	
	// 把oldbuckets置nil，如果有oldbuckets就让GC处理
	h.oldbuckets = nil

	// 初始化溢出数、元素个数
	h.nevacuate = 0
	h.noverflow = 0
	h.count = 0	

	// 重新申请一个新的extra，旧的交给GC回收
	if h.extra != nil {
		*h.extra = mapextra{}
	}

	// 清空bucket
	_, nextOverflow := makeBucketArray(t, h.B, h.buckets)
	...
}

func makeBucketArray(t *maptype, b uint8, dirtyalloc unsafe.Pointer) (buckets unsafe.Pointer, nextOverflow *bmap) {
	base := bucketShift(b)
	nbuckets := base
	...
	
	// 没有分配过内存，则申请一个新的
	if dirtyalloc == nil {
		buckets = newarray(t.bucket, int(nbuckets))
	} else {
		// 直接释放整个buckets
		buckets = dirtyalloc
		size := t.bucket.size * nbuckets
		if t.bucket.kind&kindNoPointers == 0 {
			memclrHasPointers(buckets, size)
		} else {
			memclrNoHeapPointers(buckets, size)
		}
	}
	...
}