On Wednesday, April 19, 2017 at 3:37:29 AM UTC+8, Keith Randall wrote:
>
> This is a weird corner case in string concatenation optimization.
>
> runtime.concatstrings (what the + in this code gets rewritten to) has an
> optimization where if all the strings but one that it is concatenating are
> 0 length, then it returns the remaining one without doing any allocation.
> Because of that optimization, runtime.concatstrings might return its
> argument. Thus, we can't do the zero-copy conversion of []byte to string
> for arguments to runtime.concatstrings.
> Except when we know that at least one of the input strings has non-zero
> length. Then we can. That's what is happening in f2.
>
Keith, thanks. But I haven't fully get your explanation.
Do you mean that compiler can't confirm whether or not the optimization
which might return its argument can be used for f1 at compile time,
but compiler can confirm it can't be used for f2 at compile time?
>
> On Monday, April 17, 2017 at 11:06:26 PM UTC-7, T L wrote:
>>
>>
>> package main
>>
>> import "fmt"
>> import "testing"
>>
>> var s string
>> var a = make([]byte, 1000)
>>
>> func f0() {
>> x := string(a)
>> s = x + x + x +
>> x + x + x +
>> x + x + x +
>> x
>> }
>>
>> func f1() {
>> s = string(a) + string(a) + string(a) +
>> string(a) + string(a) + string(a) +
>> string(a) + string(a) + string(a) +
>> string(a)
>> }
>>
>> func f2() {
>> s = (" " +
>> string(a) + string(a) + string(a) +
>> string(a) + string(a) + string(a) +
>> string(a) + string(a) + string(a) +
>> string(a) )[1:]
>> }
>>
>> func main() {
>> fmt.Println(testing.AllocsPerRun(1, f0)) // 2
>> fmt.Println(testing.AllocsPerRun(1, f1)) // 11
>> fmt.Println(testing.AllocsPerRun(1, f2)) // 1
>> }
>>
>> why doesn't gc make optimization for f1?
>>
>
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