Consider the following Go program allowing the parallel processing of two slices of an array. Here,...

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Consider the following Go program allowing the parallel processing of two slices of an array. Here, synchronization is achieved with a WaitGroup. Rewrite the program is performing, this time, a synchronization with a boolean channel (chan bool). So you can no longer use the sync package: package main import ( > "fmt" "sync" func main() { } var wg sync.WaitGroup x = []int(3, 1, 4, 1, 5, 9, 2, 6} var y [8]int wg.Add(2) // number of goroutines to synchronize // parallel loop in 2 slices go calcul2(x[:4], y[:4], &wg) go calcul2(x[4:], y[4:], &wg) wg.wait() // waiting for the 2 goroutines fmt.Println(y) func calcul2(in (Jint, out []int, wg sync. WaitGroup) { for i, v := range in { out[i] = 2*v*v* + *v } wg.Done() // signals that the goroutine is finished. Consider the following Go program allowing the parallel processing of two slices of an array. Here, synchronization is achieved with a WaitGroup. Rewrite the program is performing, this time, a synchronization with a boolean channel (chan bool). So you can no longer use the sync package: package main import ( > "fmt" "sync" func main() { } var wg sync.WaitGroup x = []int(3, 1, 4, 1, 5, 9, 2, 6} var y [8]int wg.Add(2) // number of goroutines to synchronize // parallel loop in 2 slices go calcul2(x[:4], y[:4], &wg) go calcul2(x[4:], y[4:], &wg) wg.wait() // waiting for the 2 goroutines fmt.Println(y) func calcul2(in (Jint, out []int, wg sync. WaitGroup) { for i, v := range in { out[i] = 2*v*v* + *v } wg.Done() // signals that the goroutine is finished.

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Heres the modified version of the program using a boolean channel f... View the full answer