return union type in Golang

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'.

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).