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This article introduces the relevant knowledge of "how to understand golang Escape Analysis". In the operation of actual cases, many people will encounter such a dilemma, so let the editor lead you to learn how to deal with these situations. I hope you can read it carefully and be able to achieve something!

Background

Recently, I want to integrate protobuf variables with previously designed data objects and maintain them in memory to reduce memory request and GC performance loss.

Feature or bug,gogoproto Decoding confusion

Because gogoproto does not guarantee input and output consistency when unmarshal, the pointer variable as a result may not be consistent with the input byte slice (for example, there is no reset operation in unmarshal slice). We need to reset this pointer variable, and the reset implementation of the pb generated file is as follows.

Func (m * Data) Reset () {* m = Data {}}

I was confused when I saw Data {}. As far as I understand it, this step requires applying for memory. In this way, it is inevitable that we need to apply for memory when we throw a pb variable into memory, so this R & D requirement seems to be meaningless.

My first reaction was that this was gogoproto's problem, maybe the official goproto was not like this. However, after regeneration, it is found that the implementation of the reset method is no different. It's just that the official go proto will take the initiative to reset during unmarshal.

So, was it the wrong direction in the first place? Ah, bald.

Past dark willows and flowers in bloom lies another village

Not giving up, I began to read all kinds of documents, including various plug-ins for gogoproto, but unfortunately I couldn't find anything useful. Then I started looking at the official proto documents.

Then I found some clues.

In the daily use of protobuf, if we do not reuse the old variables, we will generally

Declare pointer variables, data: = & pb.Data {}

Decoding, proto.Unmarshal (bytes, data).

Obviously, the first step is to apply for memory. According to the source code of go proto, the reset operation of unmarshal will apply for memory again, will Google allow this kind of performance loss?

Really? I don't believe it.

Introduction to escape analysis

If you think too much, you might as well write a benchmark and try it. (be careful of some pits in microbenchmark)

Benchmarkpackage mainimport ("testing") type boy struct {name string age int} var b1 = & boy {} var b2 = & boy {} func Benchmark_1 (b * testing.B) {for I: = 0; I

< b.N; i++ { temp := &boy{} b1 = temp }}func Benchmark_2(b *testing.B) { for i := 0; i < b.N; i++ { temp := &boy{} *b1 = *temp }}func Benchmark_3(b *testing.B) { for i := 0; i < b.N; i++ { temp := &boy{} *b1 = *temp b2 = temp }} 结果如下。 goos: linuxgoarch: amd64pkg: bibleBenchmark_1-4 29142411 42.2 ns/op 32 B/op 1 allocs/opBenchmark_2-4 1000000000 0.711 ns/op 0 B/op 0 allocs/opBenchmark_3-4 28474614 39.5 ns/op 32 B/op 1 allocs/opPASSok bible 3.258s 结果是一目了然的，temp := &boy{} 确实没有重复申请内存。 编译报告 到此，只需要逐个分析就好。首先打开编译报告看一下。 go build -gcflags "-m -m" var b1 = &boy{}//go:noinlinefunc main() { temp := &boy{} // &boy literal escapes to heap: // flow: temp = &{storage for &boy literal}: // from &boy literal (spill) at ./main.go:12:10 // from temp := &boy literal (assign) at ./main.go:12:7 // flow: {heap} = temp: // from b1 = temp (assign) at ./main.go:13:5 // &boy literal escapes to heap b1 = temp} 新创建的 &boy{} 被全局变量引用，于是逃逸到堆上，成为动态变量，无法被重复利用。 var b1 = &boy{}//go:noinlinefunc main() { temp := &boy{} // &boy literal does not escape *b1 = *temp} *b1 = *temp 仅仅是赋值操作，新创建的 &boy{} 没有被引用，留在栈上，后续被重复利用。 汇编分析

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