Pass arguments to errgroup within a loop

[GoingToGo]

Hi,

Is there a way to pass an argument to errgroup routines just like we do in classic go routines?

Thanks

classic way

for i := 1; i < 5; i++ {
  id := i
  go func(id int) {
    if id == 3 {
      return error ....
    }
    // carry on running
  }(id)
}

errgroup way

var eg errgroup.Group

for i := 1; i < 5; i++ {
  id := i
  eg.Go(func() error {
    if id == 3 { // How do I get the id here?
      return error ....
    }
    // carry on running
  })
}

[skillian (Sean Killian)]

Based on the package source code and a handful of packages I found on godoc.org that use the errgroup package, the answer seems to be that you cannot pass explicit function parameters to errgroup.(*Group).Go. You have to use a closure like you already have in your second example.

[ncw (Nick Craig-Wood)]

@GoingToGo

classic way

for i := 1; i < 5; i++ {
  id := i
  go func(id int) {
    if id == 3 {
      return error ....
    }
    // carry on running
  }(id)
}

Note that you don’t need to make an interior variable and pass in the function argument, either will do… So you can write this as

for i := 1; i < 5; i++ {
  id := i // this variable is part of the closure and it is remade each time round the look
  go func() {
    if id == 3 {
      return error ....
    }
    // carry on running
  }()
}

Which leads on the the errgroup solution (which is identical to what you’ve already written)

var eg errgroup.Group

for i := 1; i < 5; i++ {
  id := i
  eg.Go(func() error {
    if id == 3 { // This id is part of the closed over variables
      return error ....
    }
    // carry on running
  })
}

[GoingToGo]

@ncw

Note that you don’t need to make an interior variable and pass in the function argument,

I thought it was necessary. Hence the reason asked this question. Obviously learned something else now. Thanks.