Go并发编程小测验答案解析

1,368 阅读2分钟

题目 colobu.com/2019/04/28/…

Mutex

package main

import (
	"fmt"
	"sync"
)

var mu sync.Mutex
var chain string

func main() {
	chain = "main"
	A()
	fmt.Println(chain)
}

func A() {
	mu.Lock()
	defer mu.Unlock()
	chain = chain + " --> A"
	B()
}

func B() {
	chain = chain + " --> B"
	C()
}

func C() {
	mu.Lock()
	defer mu.Unlock()
	chain = chain + " --> C"
}

fatal error: all goroutines are asleep - deadlock! 引发panic mu未解锁又上锁引发死锁

RWMutex

package main

import (
"fmt"
"sync"
"time"
)

var mu sync.RWMutex
var count int

func main() {
go A()
time.Sleep(2 * time.Second)
mu.Lock()
defer mu.Unlock()
count++
fmt.Println(count)
}

func A() {
mu.RLock()
defer mu.RUnlock()
B()
}

func B() {
time.Sleep(5 * time.Second)
C()
}

func C() {
mu.RLock()
defer mu.RUnlock()
}

fatal error: all goroutines are asleep - deadlock! 引发panic mu未解锁又上锁引发死锁

因为go为了防止写锁饥饿,当有写锁等待时,后来的读锁获取不了,会等待写锁完成后再获取读锁。

Waitgroup

package main

import (
	"sync"
	"time"
)

func main() {
	var wg sync.WaitGroup
	wg.Add(1)
	go func() {
		time.Sleep(time.Millisecond)
		wg.Done()
		wg.Add(1)
	}()
	go func() {
		time.Sleep(time.Second)
		wg.Done()
	}()
	wg.Wait()
}

panic: sync: WaitGroup is reused before previous Wait has returned

双检查实现单例

package doublecheck

import (
	"sync"
)

type Once struct {
	m    sync.Mutex
	done uint32
}

func (o *Once) Do(f func()) {
	if o.done == 1 {
		return
	}

	o.m.Lock()
	defer o.m.Unlock()
	if o.done == 0 {
		o.done = 1
		f()
	}
}

可以编译,有并发问题,f函数可能会被执行多次

o.done在多CPU运行下会有并发问题,可使用atomic包解决问题

mutex

package main

import (
	"fmt"
	"sync"
)

type MyMutex struct {
	count int
	sync.Mutex
}

func main() {
	var mu MyMutex

	mu.Lock()
	var mu2 = mu
	mu.count++
	mu.Unlock()

	mu2.Lock()
	mu2.count++
	mu2.Unlock()

	fmt.Println(mu.count, mu2.count)
}

fatal error: all goroutines are asleep - deadlock! 死锁

sync.pool

package main

import (
	"bytes"
	"fmt"
	"runtime"
	"sync"
	"time"
)

var pool = sync.Pool{New: func() interface{} { return new(bytes.Buffer) }}

func main() {
	go func() {
		for {
			processRequest(1 << 28) // 256MiB
		}
	}()
	for i := 0; i < 1000; i++ {
		go func() {
			for {
				processRequest(1 << 10) // 1KiB
			}
		}()
	}

	var stats runtime.MemStats
	for i := 0; ; i++ {
		runtime.ReadMemStats(&stats)
		fmt.Printf("Cycle %d: %dB\n", i, stats.Alloc)
		time.Sleep(time.Second)
		runtime.GC()
	}
}

func processRequest(size int) {
	b := pool.Get().(*bytes.Buffer)
	time.Sleep(500 * time.Millisecond)
	b.Grow(size)
	pool.Put(b)
	time.Sleep(1 * time.Millisecond)
}

可以编译,运行时内存可能暴涨

channel

package main

import (
	"fmt"
	"runtime"
	"time"
)

func main() {
	var ch chan int
	go func() {
		ch = make(chan int, 1)
		ch <- 1
	}()

	go func(ch chan int) {
		time.Sleep(time.Second)
		<-ch
	}(ch)

	c := time.Tick(1 * time.Second)
	for range c {
		fmt.Printf("#goroutines: %d\n", runtime.NumGoroutine())
	}
}

一段时间后总是输出 #goroutines: 2

channel

package main

import "fmt"

func main() {
	var ch chan int
	var count int

	go func() {
		ch <- 1
	}()

	go func() {
		count++
		close(ch)
	}()

	<-ch

	fmt.Println(count)
}

panic: close of nil channel

map

package main

import (
	"fmt"
	"sync"
)

func main() {
	var m sync.Map
	m.LoadOrStore("a", 1)
	m.Delete("a")

	fmt.Println(m.Len())
}

不能编译 没有len()方法

happens before

package main

var c = make(chan int)
var a int

func f() {
	a = 1
	<-c
}
func main() {
	go f()
	c <- 0
	print(a)
}

输出 1

自定义Map

package main

import "sync"

type Map struct {
	m map[int]int
	sync.Mutex
}

func (m *Map) Get(key int) (int, bool) {
	m.Lock()
	defer m.Unlock()

	i, ok := m.m[key]
	return i, ok
}

func (m *Map) Put(key, value int) {
	m.Lock()
	defer m.Unlock()
	m.m[key] = value
}

func (m *Map) Len() int {
	return len(m.m)
}

func main() {
	var wg sync.WaitGroup
	wg.Add(2)

	m := Map{m: make(map[int]int)}
	go func() {
		for i := 0; i < 10000000; i++ {
			m.Put(i, i)
		}

		wg.Done()
	}()

	go func() {
		for i := 0; i < 10000000; i++ {
			m.Len()
		}

		wg.Done()
	}()

	wg.Wait()
}

可运行,有并发问题

slice

package main

import (
	"fmt"
	"sync"
)

func main() {
	var wg sync.WaitGroup
	wg.Add(2)

	var ints = make([]int, 0, 1000)

	go func() {
		for i := 0; i < 1000; i++ {
			ints = append(ints, i)
		}
		wg.Done()
	}()

	go func() {
		for i := 0; i < 1000; i++ {
			ints = append(ints, i)
		}
		wg.Done()
	}()

	wg.Wait()
	fmt.Println(len(ints))
}

输出可能不是2000