These answers are all correct.
Here is another way to see the problem and address it. Not as clear to the
next reader perhaps so not the recommendation, but a good example to
contemplate. "Precisely what does this extra statement do?"
package main
import "fmt"
import "log"
func main() {
for i := 0; i < 5; i++ {
i := i
go func() {
log.Println("i=", i)
}()
}
fmt.Print("Press <Enter> to quit...")
var input string
fmt.Scanln(&input)
}
On Sun, Feb 12, 2017 at 3:12 AM, Ayan George <a...@ayan.net> wrote:
> Srivathsan Madhavan <m.srivath...@gmail.com> wrote:
>
> > Hi all,
> >
> > I still consider myself a newbie to Golang hence my question here could
> be
> > due to my fundamental misunderstanding of this language
> >
> > I have the following code:
> >
> > func main() {
> >
> > for i := 0; i < 5; i++ {
> >
> > go func() {
> >
> > log.Println("i=", i)
> >
> > }()
> >
> > }
> >
> >
> > fmt.Print("Press <Enter> to quit...")
> >
> > var input string
> >
> > fmt.Scanln(&input)
> >
> > }
> >
> >
> > The output I expect is 0, 1, 2, 3, and 4 in any jumbled order. But what
> I
> > got was 5, 5, 5, 5 and 5! Why is that so?
> >
>
> I'm learning golang too so please correct any errors here.
>
> It looks like you created a closure where i is lexically scoped to
> each of your gofunc() instances. Since they're all accessing the same
> i variable, you're seeing the race between incrementing i and spooling
> up / starting each goroutine -- that is, the value of i reaches 5
> before all of the goroutines get to print it. You can verify this by
> printing the address of i by changing your log.Println() call to
> something like:
>
> log.Printf("i = %d, &i = %p", i, &i)
>
> The addresses should be the same.
>
> If you want each goroutine to have a copy of I, you should declare and
> run gofunc() like below:
>
> gofunc(c int) {
> log.Println("c=",c)
> }(i)
>
> Which passes a copy of i to each routine. That should give you what
> you're looking for -- that is if you're not expecting i to be printed
> in order. :)
>
> -ayan
>
> --
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--
Michael T. Jones
michael.jo...@gmail.com
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