Buffered vs unbuffered channels in Go

1package main
2
3import "fmt"
4
5func main() {
6	// Buffered: two sends succeed with nobody receiving yet.
7	b := make(chan string, 2)
8	b <- "first"
9	b <- "second"
10	fmt.Println("buffered len/cap:", len(b), cap(b))
11	fmt.Println(<-b, <-b)
12
13	// Unbuffered: the send needs a receiver, so it happens in a goroutine.
14	u := make(chan string)
15	go func() { u <- "handed over" }()
16	fmt.Println("unbuffered cap:", cap(u), "got:", <-u)
17}

1ch := make(chan int)
2ch <- 1 // blocks forever: main is the only goroutine, and it is stuck here
3fmt.Println(<-ch)

1package main
2
3import (
4	"fmt"
5	"sync"
6	"sync/atomic"
7	"time"
8)
9
10func main() {
11	sem := make(chan struct{}, 3) // at most 3 at a time
12	var running, peak atomic.Int32
13	var wg sync.WaitGroup
14
15	for i := 0; i < 10; i++ {
16		wg.Add(1)
17		go func() {
18			defer wg.Done()
19			sem <- struct{}{}        // take a slot (blocks if 3 are busy)
20			defer func() { <-sem }() // give it back
21
22			n := running.Add(1)
23			for {
24				p := peak.Load()
25				if n <= p || peak.CompareAndSwap(p, n) {
26					break
27				}
28			}
29			time.Sleep(5 * time.Millisecond)
30			running.Add(-1)
31		}()
32	}
33	wg.Wait()
34	fmt.Println("most running at once:", peak.Load())
35}