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<>1. 直接使用运算符
func BenchmarkAddStringWithOperator(b *testing.B) { hello := "hello" world :=
"world" for i := 0; i < b.N; i++ { _ = hello + "," + world } }
golang 里面的字符串都是不可变的，每次运算都会产生一个新的字符串，所以会产生很多临时的无用的字符串，不仅没有用，还会给 gc
带来额外的负担，所以性能比较差
<>2. fmt.Sprintf()
func BenchmarkAddStringWithSprintf(b *testing.B) { hello := "hello" world :=
"world" for i := 0; i < b.N; i++ { _ = fmt.Sprintf("%s,%s", hello, world) } }
内部使用 []byte 实现，不像直接运算符这种会产生很多临时的字符串，但是内部的逻辑比较复杂，有很多额外的判断，还用到了
interface，所以性能也不是很好
<>3. strings.Join()
func BenchmarkAddStringWithJoin(b *testing.B) { hello := "hello" world :=
"world" for i := 0; i < b.N; i++ { _ = strings.Join([]string{hello, world},
",") } }
join会先根据字符串数组的内容，计算出一个拼接之后的长度，然后申请对应大小的内存，一个一个字符串填入，在已有一个数组的情况下，这种效率会很高，但是本来没有，去构造这个数据的代价也不小
<>4. buffer.WriteString()
func BenchmarkAddStringWithBuffer(b *testing.B) { hello := "hello" world :=
"world" for i := 0; i < 1000; i++ { var buffer bytes.Buffer
buffer.WriteString(hello) buffer.WriteString(",") buffer.WriteString(world) _ =
buffer.String() } }
这个比较理想，可以当成可变字符使用，对内存的增长也有优化，如果能预估字符串的长度，还可以用 buffer.Grow() 接口来设置 capacity
