* John Morrice:
> Thought I'd be kind and illustrate your bug vs my solution with code. I
> use WaitGroups in both examples to keep things clearer.
>
> Your bug:
> https://play.golang.org/p/yoNPmbXnlW
>
> I.e.
>
> Ringo wrote Yellow Submarine
> Ringo wrote I am the Walrus
> Ringo wrote Eleanor Rigby
> Ringo wrote Come Together
This is misleading because there is too little work involved.
With a larger work queue, I get a distribution which is pretty even:
sum: 41385055007
sum: 40815802419
sum: 41898216012
sum: 41672900827
sum: 41802902931
sum: 40815422395
sum: 42163589241
sum: 41552235636
sum: 41989489360
sum: 42218461137
sum: 41673625671
sum: 42011799364
total: 499999500000
(499999500000)
You are right that the specification permits that a receive operation
drains an arbitrary number of values from the channel into a
goroutine-local buffer (and draining more than one value at a time may
be beneficial for performance erasons). But I don't think the current
implementation does this.
I haven't bothered to convert this example to the
one-task-per-goroutine style because creating one goroutine per task
is too wasteful for such small tasks.
package main
import (
"fmt"
)
func summation(source chan int, result chan int) {
sum := 0
for i := range source {
sum += i
}
result <- sum
}
func main() {
integerCount := 1000000
integers := make(chan int, integerCount)
for i := 0; i < integerCount; i++ {
integers <- i
}
close(integers)
threads := 12
result := make(chan int)
for i := 0; i < threads; i++ {
go summation(integers, result)
}
total := 0
for i := 0; i < threads; i++ {
sum := <- result
fmt.Printf("sum: %d\n", sum)
total += sum
}
fmt.Printf("total: %d\n", total)
fmt.Printf(" (%d)\n",
(integerCount - 1) * integerCount / 2)
}
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