Channels/Goroutines throwing error

Question

I'm currently following this tutorial http://www.miek.nl/files/go/20120807-go.pdf and on chapter 7 it discusses channels/goroutines However the example code is throwing an error for me just after you run it. ``` package main import ( "fmt" "time" ) var c chan int func ready(w string, sec int) { time.Sleep(time.Duration(sec) * time.Second) fmt.Println(w, "is ready!") c <- 1 } func main() { c := make(chan int) go ready("Tea", 2) go ready("Coffee", 1) fmt.Println("Waiting...") <-c <-c } ``` Here is the output when you execute the code ``` daniel:go> go run goroutines.go Waiting... Coffee is ready! Tea is ready! throw: all goroutines are asleep - deadlock! goroutine 1 [chan receive]: main.main() /home/daniel/Dropbox/code/go/goroutines.go:21 +0xee goroutine 2 [syscall]: created by runtime.main /build/buildd/golang-1/src/pkg/runtime/proc.c:221 goroutine 3 [chan send (nil chan)]: main.ready(0x80bb0d4, 0x3, 0x2, 0x0) /home/daniel/Dropbox/code/go/goroutines.go:13 +0xe5 created by main.main /ho…

Accepted Answer

Yes, just a typo:

```
package main

import (
    "fmt"
    "time"
)

var c chan int

func ready(w string, sec int) {
    time.Sleep(time.Duration(sec) * time.Second)
    fmt.Println(w, "is ready!")
    c <- 1
}

func main() {
    c = make(chan int) // previously c := make(chan int)
    go ready("Tea", 2)
    go ready("Coffee", 1)
    fmt.Println("Waiting...")
    <-c
    <-c
}
```

`main()` didn't use the global `c` since it declared a new one.

Notice that you don't have to make the channel inside `main()`, make it while declaring it:

```
var c = make(chan int)
```

Channels/Goroutines throwing error

Solution

Problem