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Goroutines Channels order of execution

You are calling the go routine in your code and you can't tell when the routine will end and the value will be passed to the buffered channel. As this code is asynchronous so whenever the routine will finish it will write the data to the channel and will be read on the other side. In the example above you are calling only two go routines so the behavior is certain and same output is generated somehow for most of the cases but when you will increase the go routines the output will not be same and the order will be different unless you make it synchronous.

Example

package main

import "fmt"

func sum(a []int, c chan int) {
	sum := 0
	for _, v := range a {
		sum += v
	}
	c <- sum
}

func main() {
	a := []int{17, 12, 18, 9, 24, 42, 64, 12, 68, 82, 57, 32, 9, 2, 12, 82, 52, 34, 92, 36}

	c := make(chan int)
	for i := 0; i < len(a); i = i + 5 {
		go sum(a[i:i+5], c)
	}
	output := make([]int, 5)
	for i := 0; i < 4; i++ {
		output[i] = <-c
	}
	close(c)

	fmt.Println(output)
}
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