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Diodes are ring buffers manipulated via atomics.

Diodes are optimized for high throughput scenarios where losing data is acceptable. Unlike a channel, a diode will overwrite data on writes in lieu of blocking. A diode does its best to not "push back" on the producer. In other words, invoking Set() on a diode never blocks.

Installation

go get code.cloudfoundry.org/go-diodes

Example: Basic Use

d := diodes.NewOneToOne(1024, diodes.AlertFunc(func(missed int) {
	log.Printf("Dropped %d messages", missed)
}))

// writer
go func() {
	for i := 0; i < 2048; i++ {
		// Warning: Do not use i. By taking the address,
		// you would not get each value
		j := i
		d.Set(diodes.GenericDataType(&j))
	}
}()

// reader
poller := diodes.NewPoller(d)
for {
	i := poller.Next()
	fmt.Println(*(*int)(i))
}

Example: Creating a Concrete Shell

Diodes accept and return diodes.GenericDataType. It is recommended to not use these generic pointers directly. Rather, it is a much better experience to wrap the diode in a concrete shell that accepts the types your program works with and does the type casting for you. Here is an example of how to create a concrete shell for []byte:

type OneToOne struct {
	d *diodes.Poller
}

func NewOneToOne(size int, alerter diodes.Alerter) *OneToOne {
	return &OneToOne{
		d: diodes.NewPoller(diodes.NewOneToOne(size, alerter)),
	}
}

func (d *OneToOne) Set(data []byte) {
	d.d.Set(diodes.GenericDataType(&data))
}

func (d *OneToOne) TryNext() ([]byte, bool) {
	data, ok := d.d.TryNext()
	if !ok {
		return nil, ok
	}

	return *(*[]byte)(data), true
}

func (d *OneToOne) Next() []byte {
	data := d.d.Next()
	return *(*[]byte)(data)
}

Creating a concrete shell gives you the following advantages:

• The compiler will tell you if you use a diode to read or write data of the wrong type.
• The type casting syntax in go is not common and should be hidden.
• It prevents the generic pointer type from escaping in to client code.

Dropping Data

The diode takes an Alerter as an argument to alert the user code to when the read noticed it missed data. It is important to note that the go-routine consuming from the diode is used to signal the alert.

When the diode notices it has fallen behind, it will move the read index to the new write index and therefore drop more than a single message.

There are two things to consider when choosing a diode:

1. Storage layer
2. Access layer

Storage Layer

OneToOne

The OneToOne diode is meant to be used by one producing (invoking Set()) go-routine and a (different) consuming (invoking TryNext()) go-routine. It is not thread safe for multiple readers or writers.

ManyToOne

The ManyToOne diode is optimized for many producing (invoking Set()) go-routines and a single consuming (invoking TryNext()) go-routine. It is not thread safe for multiple readers.

It is recommended to have a larger diode buffer size if the number of producers is high. This is to avoid the diode from having to mitigate write collisions (it will call its alert function if this occurs).

Access Layer

Poller

The Poller uses polling via time.Sleep(...) when Next() is invoked. While polling might seem sub-optimal, it allows the producer to be completely decoupled from the consumer. If you require very minimal push back on the producer, then the Poller is a better choice. However, if you require several diodes (e.g. one per connected client), then having several go-routines polling (sleeping) may be hard on the scheduler.

Waiter

The Waiter uses a conditional mutex to manage when the reader is alerted of new data. While this method is great for the scheduler, it does have extra overhead for the producer. Therefore, it is better suited for situations where you have several diodes and can afford slightly slower producers.

Benchmarks

There are benchmarks that compare the various storage and access layers to channels. To run them:

go test -bench=. -run=NoTest

Known Issues

If a diode was to be written to 18446744073709551615+1 times it would overflow a uint64. This will cause problems if the size of the diode is not a power of two (2^x). If you write into a diode at the rate of one message every nanosecond, without restarting your process, it would take you 584.54 years to encounter this issue.