return union type in Golang-CodePudding

I want to try the union type implementation in Golang as in this answer
I try this:

package main

import (
    "fmt"
    "math/rand"
    "time"
)

type intOrString interface {
    int | string
}

func main() {
    fmt.Println(measure())

}
func measure[T intOrString]() T {
    rand.Seed(time.Now().UnixNano())
    min := 20
    max := 35
    temp := rand.Intn(max-min 1)   min

    switch {
    case temp < 20:
        return "low" //'"low"' (type string) cannot be represented by the type T
    case temp > 20:
        return T("high") //Cannot convert an expression of the type 'string' to the type 'T'
    default:
        return T(temp)

    }
}

so how I can convert an expression of the type 'string' or 'int' to the type 'T'.

CodePudding user response：

You misunderstand how generics work. For your function you must provide a type when you call the function. Like fmt.Println(measure[string]()), so in this case you expect to get a string from it. If you call it like measure[int]() then you expect an int as a result. But you cannot call it without a type parameter. Generics are for function which share the same logic for different types.

For what you want, you must use any as a result, then check it if it's a string or an int. Example:

package main

import (
    "fmt"
    "math/rand"
    "time"
)

func main() {
    res := measure()
    if v, ok := res.(int); ok {
        fmt.Printf("The temp is an int with value %v", v)
    }
    if v, ok := res.(string); ok {
        fmt.Printf("The temp is a string with value %v", v)
    }
}

func measure() any {
    rand.Seed(time.Now().UnixNano())
    min := 20
    max := 35
    temp := rand.Intn(max-min 1)   min

    switch {
    case temp < 20:
        return "low"
    case temp > 20:
        return "high"
    default:
        return temp
    }
}

Or if you want to just print it out (and don't need to know the type), you don't even need to check it, just call fmt.Printf("The temp is %v", res).