initializing a struct containing a slice of structs in golang - go

Question

I have a struct that I want to initialize with a slice of structs in golang, but I'm trying to figure out if there is a more efficient version of appending every newly generated struct to the slice:
package main
import (
"fmt"
"math/rand"
)
type LuckyNumber struct {
number int
}
type Person struct {
lucky_numbers []LuckyNumber
}
func main() {
count_of_lucky_nums := 10
// START OF SECTION I WANT TO OPTIMIZE
var tmp []LuckyNumber
for i := 0; i < count_of_lucky_nums; i++ {
tmp = append(tmp, LuckyNumber{rand.Intn(100)})
}
a := Person{tmp}
// END OF SECTION I WANT TO OPTIMIZE
fmt.Println(a)
}

Answer 1

You can use make() to allocate the slice in "full-size", and then use a for range to iterate over it and fill the numbers:
tmp := make([]LuckyNumber, 10)
for i := range tmp {
tmp[i].number = rand.Intn(100)
}
a := Person{tmp}
fmt.Println(a)
Try it on the Go Playground.
Note that inside the for I did not create new "instances" of the LuckyNumber struct, because the slice already contains them; because the slice is not a slice of pointers. So inside the for loop all we need to do is just use the struct value designated by the index expression tmp[i].

Answer 2

You can use make() the way icza proposes, you can also use it this way:
tmp := make([]LuckyNumber, 0, countOfLuckyNums)
for i := 0; i < countOfLuckyNums; i++ {
tmp = append(tmp, LuckyNumber{rand.Intn(100)})
}
a := Person{tmp}
fmt.Println(a)
This way, you don't have to allocate memory for tmp several times: you just do it once, when calling make. But, contrary to the version where you would call make([]LuckyNumber, countOfLuckyNums), here, tmp only contains initialized values, not uninitialized, zeroed values. Depending on your code, it might make a difference or not.