CODEWITHSUNDEEP
go
Ming
Ming

Reputation: 1602

go routine, function executed out of desired order

Hi I'm learning goroutine, and I have a question I have a code where I have a publisher and a listener, I need the subscribe call to happen first then the publish call.

useCase := New(tt.fields.storage)

            tt.fields.wg.Add(1)
            go func() {
                ch, _, err := useCase.Subscribe(tt.args.ctx, tt.args.message.TopicName)
                message, ok :=  <- ch
                if !ok {
                    close(ch)
                    tt.fields.wg.Done()
                }

                require.NoError(t, err)
                assert.Equal(t, tt.want, message)
            }()

            err := useCase.Publish(tt.args.ctx, tt.args.message)
            tt.fields.wg.Wait()

and as I send messages inside these two functions I need Go Func, how would I get subscribe to always execute first, and then publish.

Upvotes: 0

Views: 43

Answers (1)

Burak Serdar
Burak Serdar

Reputation: 51587

You can use a channel:

subscribed:=make(chan struct{})
go func() {
   ch, _, err := useCase.Subscribe(tt.args.ctx, tt.args.message.TopicName)
   if err!=nil {
    ...
   }
   close(subscribed)
   message, ok :=  <- ch
   ...
}()
<-subscribed
err := useCase.Publish(tt.args.ctx, tt.args.message)

Upvotes: 3

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