own len function with recursion in Go - recursion

Question

I'm trying to build simple function to count elements in slice (like len) It must be simple (without additional libs) and with recursion. The problem is when i try to check is slice is empty (is nul).
package main
import "fmt"
func main() {
x := []int{1, 2, 3}
fmt.Println(len2(x))
}
func len2(s []int) int {
if s == nil {
return 0
}
return 1 + len2(s[1:])
}
the result it should be in this example '3'.
It's broken in if s == nil:
panic: runtime error: slice bounds out of range

Answer 1

It panics because you have no valid termination condition.
When your len2() is called with a non-nil empty slice, it attempts to slice it like s[1:], which will be a runtime panic.
Instead of checking for nil slice, check if the slice is empty by comparing its length to 0:
func len2(s []int) int {
if len(s) == 0 {
return 0
}
return 1 + len2(s[1:])
}
Try it on the Go Playground.
If you can't use the builtin len() function (which you already did in your solution), you may use the for ... range:
func len2(s []int) int {
size := 0
for i := range s {
size = i + 1
}
return size
}
Try this on the Go Playground.
And if it must be recursive, then for example:
func len2(s []int) int {
size := 0
for range s {
size = 1 + len2(s[1:])
break
}
return size
}
Try this on the Go Playground.
But know that these are awful solutions compared to using the builtin len().

Answer 2

The following is not a solution with better performance than len but an implementation that does not use any extra libraries and depends on recursion to find length
func len2(s []int) (count int) {
defer func() {
if r := recover(); r != nil {
count = 0
}
}()
return 1 + len2(s[1:])
}
Here is sample code
package main
import "fmt"
func main() {
var x []int = nil
var x1 = []int{1, 2, 3, 4}
var x2 = []int{}
var x3 = make([]int, 10, 20)
fmt.Println(len2(x))
fmt.Println(len2(x1))
fmt.Println(len2(x2))
fmt.Println(len2(x3))
}
func len2(s []int) (count int) {
defer func() {
if r := recover(); r != nil {
count = 0
}
}()
return 1 + len2(s[1:])
}
Checkout the same in playground

Answer 3

If you can leave without recursion here is a function that does not use len() and should be faster then re-slicing recursively.
func len2(s []int) (count int) {
for i := range s {
count = i + 1
}
}

Answer 4

If you do not want to use len() func, you can use cap()
func main() {
x := []int{1, 2, 3}
fmt.Println(len2(x))
}
func len2(s []int) int {
if cap(s) == 0 {
return 0
}
return 1 + len2(s[1:])
}
Try it again
Original Answer:
In order to check if an array (slice) is empty, you should use the function len()
if len(s) == 0 {
return 0
}
Try it