为了使切片追加操作更快，我们需要分配足够的容量.追加Slice有两种方式，代码如下:
func BenchmarkSliceAppend(b *testing.B) {
a := make([]int, 0, b.N)
for i := 0; i < b.N; i++ {
a = append(a, i)
}
}
func BenchmarkSliceSet(b *testing.B) {
a := make([]int, b.N)
for i := 0; i < b.N; i++ {
a[i] = i
}
}
结果是:
基准-4 200000000 7.87 ns/op 8 B/op 0 allocs/op
基准切片集-4 300000000 5.76 ns/op 8 B/op
为什么a[i] = i比a = append(a, i)快？
a[i] = i只是将值i赋值给a[i].这是not追加，只是一个简单的assignment.
现在追加:
a = append(a, i)
理论上，会发生以下情况:
这将调用内置append()函数.为此，它首先必须复制a片(片标题，后备数组不是标题的一部分)，并且它必须为将包含值i的可变参数创建临时片.
Then it has to reslice a if it has enough capacity (it has in your case) like a = a[:len(a)+1] - which involves assigning the new slice to a inside the append().
(If a would not have big enough capacity to do the append "in-place", a new array would have to be allocated, content from slice copied, and then the assign / append be executed - but it is not the case here.)
然后将i指定给a[len(a)-1].
然后从append()返回新切片，这个新切片被分配给局部变量a.
与简单的任务相比，这里发生了很多事情.即使这些步骤中的许多步骤被优化和/或内联，作为对片the local variable 101 of slice type(其为片首部)has to be updated in each cycle of the loop的元素赋值i的最小附加.
推荐阅读量:The Go Blog: Arrays, slices (and strings): The mechanics of 'append'