英文:
Initialization method for slice type fails to initialize
问题
我试图为我的自定义类型实现一个初始化方法。然而，在调用该方法后，变量在main()函数中没有改变。我可能还没有完全理解切片的工作原理，下面是我的示例代码：
package main
import "fmt"
type test [][]float64
func (p *test) init(m, n int){
tmp := *p
tmp = make(test, m)
for i := 0; i < m; i++ {
tmp[i] = make([]float64, n)
}
}
func main(){
var t test
t.init(10,2)
fmt.Println(t)
}
我原本以为接收者类型的内容可以被改变，但是对于切片来说似乎并非如此。那么，我该如何正确地将初始化函数绑定到我的类型上呢？我相当确定这是我方面的一个误解...
我尝试了几种方法，比如：
var t *test = new(test)
或者
func (p *test) init(m, n int){
tmp := *p
tmp = append(tmp, make(test, m)...)
for i := 0; i < m; i++ {
tmp[i] = append(tmp[i], make([]float64, n)...)
}
}
等等，但都失败了...
我目前唯一知道的可行解决方案是一个无绑定方法，它返回一个指向新切片的指针。这暂时可以解决问题，但我想将其作为接口的先决条件。那么，我该如何绑定它呢？
英文:
I try to implement an initialization method for my own type. However after calling the method the variable is unchanged in main(). I probably haven't fully understood how slices work, here is my example code
package main
import "fmt"
type test [][]float64
func (p *test) init(m, n int){
tmp := *p
tmp = make(test, m)
for i := 0; i < m; i++ {
tmp[i] = make([]float64, n)
}
}
func main(){
var t test
t.init(10,2)
fmt.Println(t)
}
I was under the impression that the content of the receiver type * can be changed, but this doesn't seem to be the case for slices. So how do I properly bind my initialization function to my type? I am pretty sure that there is an misunderstanding on my side....
I tried several things like
var t *test = new(test)
or
func (p *test) init(m, n int){
tmp := *p
tmp = append(tmp, make(test, m)...)
for i := 0; i < m; i++ {
tmp[i] = append(tmp[i], make([]float64, n)...)
}
}
and so on but all failed..
The only working solution I currently know of is a unbound method which returns the pointer to a fresh slice. This will do for now, but I want to make this a prerequisite in an interface. So how can I bind it?
答案1
得分: 8
func (p *test) init(m, n int) {
tmp := make(test, m)
for i := 0; i < m; i++ {
tmp[i] = make([]float64, n)
}
*p = tmp
}
你离答案很接近。上面的代码可以实现你想要的功能。但是没有理由避免返回一个新的切片的函数。这是惯用的写法，感觉上很像在其他语言中编写构造函数：
func newTest(m, n int) test {
t := make(test, m)
for i := range t {
t[i] = make([]float64, n)
}
return t
}
英文:
func (p *test) init(m, n int){
tmp = make(test, m)
for i := 0; i < m; i++ {
tmp[i] = make([]float64, n)
}
*p = tmp
}
You were close. The above does what you want. But there's no reason to avoid a function that returns a fresh slice. That's idiomatic and feels much like writing a constructor in other languages:
func newTest(m, n int) test {
t = make(test, m)
for i := range t {
t[i] = make([]float64, n)
}
return t
}
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