about function argument type infer?

[xie cui]

package main

import (
        "fmt"
        "reflect"
)
import "golang.org/x/exp/constraints"
// Double returns a new slice that contains all the elements of s, doubled.
func Double[E constraints.Integer](s []E) []E {
        r := make([]E, len(s))
        for i, v := range s {
                r[i] = v
        }
        return r
}

func main() {
        // MySlice is a slice of ints.
        type MySlice []int

        // The type of V1 will be []int, not MySlice.
        // Here we are using function argument type inference,
        // but not constraint type inference.
        var V1 = Double(MySlice{1})
        fmt.Println(reflect.TypeOf(V1).String())
}

why use function argument type infer, and why function argument type infer int to []int?

[Axel Wagner]

On Sun, Mar 20, 2022 at 12:20 PM xie cui <cuiw...@gmail.com> wrote:

why use function argument type infer, and why function argument type infer int to []int?

Because the signature says `[]E`. So, the inference tries to determine `E`, infers it to be `int` and then substitutes that into the return type to make it `[]int`. `MySlice` is assignable to `[]int`, as they have the same underlying type and the latter is not a defined type. So, you can pass `MySlice` to `Double[int]` just fine.

You have to actually mention a separate type for the compiler to emit one, otherwise it will just do what it told you:

https://go.dev/play/p/l1ochwkavNN

 

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