Unbuffered vs Buffered channels

Unbuffered channels

• By default channels are unbuffered, for a sender there should be a corresponding receiver
• There’s no need for any other synchronization mechanism like mutexes as for each sender there’s only one receiver
• By default unbuffered channels are blocking
• You can send values into channels from one goroutine and receive those values into another goroutine
• Do not communicate by sharing memory instead share memory by communicating

package main
import (
	"fmt"
	"time"
)

func main() {
	// unbuffered channels have 0 cap by default
	oneChef := 0
	order := make(chan int, oneChef)
	customers := []int{1, 2, 3, 4, 5}

	go func() {
		for c := range customers {
			order <- c
			fmt.Printf("Placed order for customer %d\n", c)
		}
		close(order)
	}()

	for o := range order {
		fmt.Printf("Preparing order for customer %d\n", o)
		time.Sleep(1 * time.Second)
		fmt.Printf("Ready order for customer %d\n", o)
	}
}

Buffered channels

• Buffered channels accept a limited number of values without a corresponding receiver for those values
• Buffered channels allow a specified number of values to be sent without blocking until the buffer is full
• Its like creating a buffer inside channel where all data will be kept temporarily until its received by receiver
• Buffered channels have to be synchronized manually i,e main process will not fail even if receiver doesn’t exist

package main
import (
	"fmt"
	"time"
)

func main() {
	// buffered channels with 2 cap
	threeChefs := 2
	order := make(chan int, threeChefs)
	customers := []int{1, 2, 3, 4, 5}

	go func() {
		for c := range customers {
			order <- c
			fmt.Printf("Placed order for customer %d\n", c)
		}
		close(order)
	}()

	for o := range order {
		fmt.Printf("Preparing order for customer %d\n", o)
		time.Sleep(1 * time.Second)
		fmt.Printf("Ready order for customer %d\n", o)
	}
}