Why does this select always run the default case when the first case actually is executed?

Question

I'm trying to get a better understanding of golang channels. While reading this article I'm toying with non-blocking sends and have come up with the following code:

package main
import (
    "fmt"
    "time"
)

func main() {
    stuff := make(chan int)
    go func(){
        for i := 0; i < 5; i ++{
            select {
            case stuff <- i:
                fmt.Printf("Sent %v\n", i)
            default:
                fmt.Printf("Default on %v\n", i)
            }
        }
        println("Closing")
        close(stuff)
    }()
    time.Sleep(time.Second)
    fmt.Println(<-stuff)
    fmt.Println(<-stuff)
    fmt.Println(<-stuff)
    fmt.Println(<-stuff)
    fmt.Println(<-stuff)
}

This will print:

Default on 0
Default on 1
Default on 2
Default on 3
Default on 4
Closing
0
0
0
0
0

While I do understand that only 0s will get printed I do not really understand why the first send does still trigger the default branch of the select?

What is the logic behind the behavior of a select in this case?

Example at the Go Playground

Answer 1

You never send any values to stuff, you execute all the default cases before you get to any of the receive operations in the fmt.Println statements. The default case is taken immediately if there is no other operation than can proceed, which means that your loop will execute and return as quickly as possible.

You want to block the loop, so you don't need the default case. You don't need the close at the end either, because you're not relying on the closed channel unblocking a receive or breaking from a range clause.

stuff := make(chan int)
go func() {
    for i := 0; i < 5; i++ {
        select {
        case stuff <- i:
            fmt.Printf("Sent %v\n", i)
        }
    }
    println("Closing")
}()
time.Sleep(time.Second)
fmt.Println(<-stuff)
fmt.Println(<-stuff)
fmt.Println(<-stuff)
fmt.Println(<-stuff)
fmt.Println(<-stuff)

https://play.golang.org/p/k2rmRDP38f

Notice also that the last "Sent" and the "Closing" line aren't printed, because you have no other synchronization waiting for the goroutine to finish, however that doesn't effect the outcome of this example.