Problem
In go programming language, how to create an array of length 5, with all elements has same value, eg, 42.
Preference order
readability, conciseness, performance.
Problem
In go programming language, how to create an array of length 5, with all elements has same value, eg, 42.
Preference order
readability, conciseness, performance.
For example,
package main
import (
"fmt"
)
func main() {
s := make([]int, 5)
for i := range s {
s[i] = 42
}
fmt.Println(len(s), s)
}
Playground: https://play.golang.org/p/GjTXruMsJ5h
Output:
5 [42 42 42 42 42]
Some benchmarks:
package main
import (
"fmt"
"testing"
)
func BenchmarkStack(b *testing.B) {
for N := 0; N < b.N; N++ {
s := make([]int, 5)
for i := range s {
s[i] = 42
}
}
}
func BenchmarkHeap(b *testing.B) {
var s []int
for N := 0; N < b.N; N++ {
s = make([]int, 5)
for i := range s {
s[i] = 42
}
}
}
func BenchmarkHygull(b *testing.B) {
for N := 0; N < b.N; N++ {
var s []int
for i := 0; i < 5; i++ {
s = append(s, 42)
}
}
}
Output:
$ go test slice42_test.go -bench=. -benchmem
BenchmarkStack-8 1000000000 2.05 ns/op 0 B/op 0 allocs/op
BenchmarkHeap-8 100000000 26.9 ns/op 48 B/op 1 allocs/op
BenchmarkHygull-8 10000000 123 ns/op 120 B/op 4 allocs/op
$
BenchmarkHygull
demonstrates how inefficient @hygull's solution is.
A "one-liner":
package main
import (
"fmt"
)
func newInts(n, v int) []int {
s := make([]int, n)
for i := range s {
s[i] = v
}
return s
}
func main() {
s := newInts(5, 42)
fmt.Println(len(s), s)
}
Playground: https://play.golang.org/p/t8J-AjYQ72l
Output:
5 [42 42 42 42 42]