英文:
go concurrency prints serially
问题
我正在尝试并发打印，但无法弄清楚为什么它是串行的，以下是我放置的代码：
package main
import (
"fmt"
"sync"
)
func main() {
fmt.Println("Hello, playground")
var wg sync.WaitGroup
wg.Add(2)
go func() {
for i := 0; i < 4; i++ {
if i%2 == 0 {
fmt.Println("hi", i)
}
}
wg.Done()
}()
go func() {
for i := 0; i < 4; i++ {
if i%2 != 0 {
fmt.Println("g", i)
}
}
wg.Done()
}()
wg.Wait()
}
期望输出是：
hi0
g1
hi2
g3
但我得到的输出是：
from g 1
from g 3
hi 0
hi 2
英文:
i am trying to print concurrently but not able to figure out why its serial, have put the code below
package main
import (
"fmt"
"sync"
)
func main() {
fmt.Println("Hello, playground")
var wg sync.WaitGroup
wg.Add(2)
go func(){
for i := 0; i < 4; i++ {
if i%2 == 0 {
fmt.Println("hi", i)
}
}
wg.Done()
}()
go func() {
for i := 0; i < 4; i++ {
if i%2 != 0 {
fmt.Println("g", i)
}
}
wg.Done()
}()
wg.Wait()
}
expectation is
hi0
g1
hi2
g3
but i get
from g 1
from g 3
hi 0
hi 2
答案1
得分: 3
这样一个小函数很少能展示并发性，因为第一个goroutine可能在第二个goroutine开始之前甚至在上下文切换之前就完成了。如果在循环中添加一个小的暂停，你将观察到交错执行的结果：
package main
import (
"fmt"
"sync"
"time"
)
func main() {
var wg sync.WaitGroup
wg.Add(2)
go func() {
for i := 0; i < 4; i++ {
if i%2 == 0 {
fmt.Println("hi", i)
}
time.Sleep(10 * time.Millisecond)
}
wg.Done()
}()
go func() {
for i := 0; i < 4; i++ {
if i%2 != 0 {
fmt.Println("from g", i)
}
time.Sleep(10 * time.Millisecond)
}
wg.Done()
}()
wg.Wait()
}
英文:
Such a small function is less likely to demonstrate the concurrency, because the first goroutine may complete even before the second one starts, or before context switch happens. If you add a small pause to the loop, you will observe the interleaving:
package main
import (
"fmt"
"sync"
"time"
)
func main() {
var wg sync.WaitGroup
wg.Add(2)
go func() {
for i := 0; i < 4; i++ {
if i%2 == 0 {
fmt.Println("hi", i)
}
time.Sleep(10 * time.Millisecond)
}
wg.Done()
}()
go func() {
for i := 0; i < 4; i++ {
if i%2 != 0 {
fmt.Println("from g", i)
}
time.Sleep(10 * time.Millisecond)
}
wg.Done()
}()
wg.Wait()
}
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