I've encountered a strange behaviour. I was playing around with buffered channels, and when using large buffers the whole program execution would block. In the following code snippet:
package main
import (
"fmt"
)
func main() {
choke := make(chan string, 150000)
go func() {
for i := 0; i < 10000000; i++ {
choke <- string(i)
fmt.Println("i=", i)
}
}()
for {
//fmt.Println(len(choke))
if len(choke) >= 150000 {
fmt.Println("Full")
}
}
}
My program blocks at ~96000 iterations and never reaches the "Full" print, unless I print out len(choke)
before evaluating it. This is probably due to the delay provided by fmt.Println
, as the issue can also be "fixed" by adding a small time.Sleep
.
Could someone explain the reason for this behaviour?
This is happening because your goroutine is never executing. The "for" loop you have outside of the goroutine is a tight loop without any blocking operations, which means it's getting all of the scheduler time. The goroutine you create is therefore never being scheduled to execute.
It works when you have the print statement, because that's a blocking operation (I/O) that causes the Go scheduler to switch to the goroutine you've created.
来源:https://stackoverflow.com/questions/48791949/getting-len-of-large-buffered-channel-blocks-for-loop