Go: what determines the iteration order for map keys?

Question

The Go Programming Language Specification says:

3. The iteration order over maps is not specified. [...]

That\'s to be expected

Answer 1

To extend @user811773 answer. A semi-random range clause iteration does not mean that the chance of returning each element in a map is also equal. See https://medium.com/i0exception/map-iteration-in-go-275abb76f721 and https://play.golang.org/p/GpSd8i7XZoG.

package main

import "fmt"

type intSet map[int]struct{}

func (s intSet) put(v int) { s[v] = struct{}{} }
func (s intSet) get() (int, bool) {
    for k := range s {
        return k, true
    }
    return 0, false
}
func main() {
    s := make(intSet)
    for i := 0; i < 4; i++ {
        s.put(i)
    }
    counts := make(map[int]int)
    for i := 0; i < 1024*1024; i++ {
        v, ok := s.get()
        if !ok {return}
        counts[v]++
    }
    for k, v := range counts {
        fmt.Printf("Value: %v, Count: %v\n", k, v)
    }
}
/*
Value: 1, Count: 130752
Value: 2, Count: 130833
Value: 0, Count: 655840
Value: 3, Count: 131151
*/