asking is good! first, it helps people who are too shy or insecure to ask. second, it is part of the adventure of learning.
it has often been my experience that the people with the most and best questions end up with the best understanding and spirit. On Sun, Feb 12, 2017 at 9:36 PM, Vatsan <m.srivath...@gmail.com> wrote: > Thanks to all who answered; I got explanation I needed. > > @"Michael Jones": yes, your solution also works - "i := i" overshadows the > i defined as part of the for-loop and that was also explained in one of the > Go-blogs. > > Anyways, as I have already acknowledged earlier, this was an RTFM moment > for me. Had I read the relevant blogs / articles earlier, I wouldn't have > asked this at all. > > rgds, > Vatsan > > Google+ <https://google.com/+SrivathsanMadhavan> > Maverick Software Engineer > <https://plus.google.com/b/102005838874122947401/102005838874122947401/posts> > > On Sun, Feb 12, 2017 at 8:20 PM, Michael Jones <michael.jo...@gmail.com> > wrote: > >> 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 >>> >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "golang-nuts" group. >>> To unsubscribe from this group and stop receiving emails from it, send >>> an email to golang-nuts+unsubscr...@googlegroups.com. >>> For more options, visit https://groups.google.com/d/optout. >>> >> >> >> >> -- >> Michael T. Jones >> michael.jo...@gmail.com >> > > -- Michael T. Jones michael.jo...@gmail.com -- You received this message because you are subscribed to the Google Groups "golang-nuts" group. To unsubscribe from this group and stop receiving emails from it, send an email to golang-nuts+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
