Interesting question. I think the code is fine. From the spec
<https://golang.org/ref/spec#Go_statements>:
> The function value and parameters are evaluated as usual in the calling
goroutine, but unlike with a regular call, program execution does not wait
for the invoked function to complete.
This means `onedoer.do` is immediately evaluated, before the loop
continues. The usual issue with loops is due to closures. So, this would be
wrong and have exactly the issue you are worried about:
for _, onedoer := range d {
go func() { oneoder(&wg) }
}
But the go statement isn't equivalent to that, it doesn't close over the
function value, it evaluates it.
On Thu, Mar 4, 2021 at 5:01 PM Julien Pivotto <roidelapl...@gmail.com>
wrote:
> Hello,
>
> In the following example: https://play.golang.org/p/yz_ifHC-Hut
>
> for _, onedoer := range d {
> go onedoer.do(&wg)
> }
>
> Should I pass the function onedoer.do as a parameter of the go routine:
> https://play.golang.org/p/WHPahoayDbM ?
>
> for _, onedoer := range d {
> go func(od *doer, w *sync.WaitGroup) {
> od.do(w)
> }(onedoer, &wg)
> }
>
> I am wondering if `go func()` could return before figuring out which
> function to run, creating a race with the loop.
>
>
> Thanks in advance!
>
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