这种“模式”背后的动机是什么？ [英] What is the motivation behind this "pattern"?

问题描述

  bigBox：=& BigBox {} $ b当我看到以下代码时，我有点困惑： $ b bigBox.BubbleGumsCount = 4 //正确... 
 bigBox.SmallBox.AnyMagicItem = true //也正确
  

为什么或者什么时候，我想要做 bigBox：=& BigBox {} 而不是 bigBox：= BigBox {} ？在某种程度上它更有效率吗？代码示例取自 > here 。

样品2：

  package main 
 
 importfmt
 
 type Ints struct {
x int 
y int 
} 
 
 func build_struct（）Ints {
 return Ints {0,0} 
} 
 $ b func build_pstruct（）* Ints {
 return& Ints {0,0 } 
 
 
 func main（）{
 fmt.Println（build_struct（））
 fmt.Println（build_pstruct（））
} 
  

样品编号3：（为什么我会在这个例子中使用& BigBox，而不是直接使用BigBox作为结构？）

  func main（）{
 bigBox：=&BigBox {} 
 bigBox.BubbleGumsCount = 4 
 fmt.Println（bigBox.BubbleGumsCount）
} 
  

有没有理由调用build_pstruct而不是build_struct变体？这不就是为什么我们有GC吗？

解决方案

我想出了这种代码的一个动机：避免struct复制意外。

---

如果您使用结构变量来保存新创建的结构：

  bigBox：= BigBox {} 
  

  myBox：= bigBox //你只需要一个bigBox的refence 。 
 myBox.BubbleGumsCount = 4 
  

或者像这样

  changeBoxColorToRed（bigBox）
  

code> changeBoxColorToRed 是

  //它将整个结构的副本作为参数。 
 func changeBoxColorToRed（box bigBox）{
 // !!!!这个功能是越野车。它不会按预期工作！ 
 //请看最后的修正。 
 box.Color = red 
} 
  

---

但是如果你使用一个结构体指针：

pre $ bigbox：=& BigBox {}

code>

不会复制

  myBox：= bigBox 
  

和

  changeBoxColorToRed（bigBox）
  

将无法编译，你有机会重新考虑 changeBoxColorToRed 的设计。修复很明显：

  func changeBoxColorToRed（box * bigBox）{
 box.Color = red 
 } 
  

新版本的 changeBoxColorToRed 不会复制整个结构并正常工作。

I'm a bit confused when I see code such as:

bigBox := &BigBox{}
bigBox.BubbleGumsCount = 4          // correct...
bigBox.SmallBox.AnyMagicItem = true // also correct

Why, or when, would I want to do bigBox := &BigBox{} instead of bigBox := BigBox{} ? Is it more efficient in some way?

Code sample was taken from here.

Sample no.2:

package main

import "fmt"

type Ints struct {
  x int
  y int
}

func build_struct() Ints {
  return Ints{0,0}
}

func build_pstruct() *Ints {
  return &Ints{0,0}
}

func main() {
  fmt.Println(build_struct())
  fmt.Println(build_pstruct())
}

Sample no. 3: ( why would I go with &BigBox in this example, and not with BigBox as a struct directly ? )

func main() {
  bigBox := &BigBox{}
  bigBox.BubbleGumsCount = 4 
  fmt.Println(bigBox.BubbleGumsCount)
}

Is there ever a reason to call build_pstruct instead of the the build_struct variant? Isn't that why we have the GC?

解决方案

I figured out one motivation for this kind of code: avoidance of "struct copying by accident".

---

If you use a struct variable to hold the newly created struct:

bigBox := BigBox{}

you may copy the struct by accident like this

myBox := bigBox // Where you just want a refence of bigBox.
myBox.BubbleGumsCount = 4

or like this

changeBoxColorToRed(bigBox)

where changeBoxColorToRed is

// It makes a copy of entire struct as parameter. 
func changeBoxColorToRed(box bigBox){
    // !!!! This function is buggy. It won't work as expected !!!
    // Please see the fix at the end.
    box.Color=red
}

---

But if you use a struct pointer:

bigBox := &BigBox{}

there will be no copying in

myBox := bigBox

and

changeBoxColorToRed(bigBox)

will fail to compile, giving you a chance to rethink the design of changeBoxColorToRed. The fix is obvious:

func changeBoxColorToRed(box *bigBox){
    box.Color=red
}

The new version of changeBoxColorToRed does not copy the entire struct and works correctly.

这篇关于这种“模式”背后的动机是什么？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！