In your second example, the address of wg doesn't change with each
iteration so you can use:
for _, onedoer := range d {
go func(od *doer) {
od.do(&wg)
}(onedoer)
}
https://play.golang.org/p/A1dAUuyvg9T
On Thursday, March 4, 2021 at 8:50:19 AM UTC-8 Julien Pivotto wrote:
> Thank you for your quick answer, it is of a great help.
>
> have a nice day.
>
> On Thursday, March 4, 2021 at 5:11:40 PM UTC+1 [email protected]
> wrote:
>
>> 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 <[email protected]>
>> 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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>>
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