如何double(i)在切片迭代中同時創(chuàng)建通道切片和運行函數(shù)：package mainimport ( "fmt" "time")func double(i int) int { result := 2 * i fmt.Println(result) time.Sleep(500000000) return result}func notParallel(arr []int) (outArr []int) { for _, i := range arr { outArr = append(outArr, double(i)) } return}// how to do the same as notParallel func in parallel way.// For each element of array double func should evaluate concuruntly// without waiting each next element to evalfunc parallel(arr []int) (outArr []int) { var chans []chan int for i := 0; i < len(arr); i++ { chans[i] = make(chan int) // i = 0 : panic: runtime error: index out of range } for counter, number := range arr { go func() { chans[counter] <- double(number) }() } return}func main() { arr := []int{7, 8, 9} fmt.Printf("%d\n", notParallel(arr)) fmt.Printf("%d\n", parallel(arr))}操場由于函數(shù)double(i)休眠 500 毫秒，函數(shù)notParallel(arr []int)對 3 個元素的工作時間為 1500 毫秒，arr []int但函數(shù)parallel(arr []int)將工作約 500 毫秒。在我的實現(xiàn)中有錯誤...panic: runtime error: index out of range... 在線的 ...chans[i] = make(chan int) // i = 0
1 回答
侃侃爾雅
TA貢獻1801條經(jīng)驗 獲得超16個贊
在這種情況下，您不需要使用 chan。
package main
import (
"fmt"
"sync"
"time"
)
func double(i int) int {
result := 2 * i
fmt.Println(result)
time.Sleep(500000000)
return result
}
func notParallel(arr []int) (outArr []int) {
for _, i := range arr {
outArr = append(outArr, double(i))
}
return
}
// how to do the same as notParallel func in parallel way.
// For each element of array double func should evaluate concuruntly
// without waiting each next element to eval
func parallel(arr []int) (outArr []int) {
outArr = make([]int, len(arr))
var wg sync.WaitGroup
for counter, number := range arr {
wg.Add(1)
go func(counter int, number int) {
outArr[counter] = double(number)
wg.Done()
}(counter, number)
}
wg.Wait()
return
}
func main() {
arr := []int{7, 8, 9}
fmt.Printf("%d\n", notParallel(arr))
fmt.Printf("%d\n", parallel(arr))
}
因為并行必須等待 goroutine(s) 的所有完成。
而且我注意到您的代碼不工作，因為你是指counter，number在同樣的功能范圍。
- 1 回答
- 0 關注
- 196 瀏覽
添加回答
舉報
0/150
提交
取消