CODEWITHSUNDEEP
go
Bort
Bort

Reputation: 2491

Initialization method for slice type fails to initialize

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?

Upvotes: 2

Views: 223

Answers (1)

Sonia
Sonia

Reputation: 28355

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
}

Upvotes: 8

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