Do Continuous Goroutine Loops Starve Other Goroutines in Go?

Goroutine Cooperative Scheduling and Execution Order

Goroutines in Go are cooperatively scheduled, meaning that they yield execution to other goroutines explicitly rather than being preempted by the operating system. This raises the question: if a goroutine does not explicitly yield, will it prevent other goroutines from running?

Influence of Continuous Loops

According to documentation, goroutines that run continuously without yielding can "starve" other goroutines on the same thread. This is because a goroutine that perpetually executes a loop prevents the scheduler from allocating time slices to other goroutines.

Function Calls as Yielding Points

However, it's important to note that function calls within a goroutine can act as implicit yield points. Specifically, calls to functions that involve system resources, such as network input, sleeping, channel operations, or blocking primitives, trigger the scheduler.

Example Scenario

Consider a simple Go program that launches four goroutines, each executing a loop and printing a sum:

func sum(x int) {
  sum := 0
  for i := 0; i < x; i++ {
    sum += i
  }
  fmt.Println(sum)
}

If these goroutines are launched concurrently with:

go sum(100)
go sum(200)
go sum(300)
go sum(400)

the goroutines will not necessarily execute one by one. The fmt.Println() call made at the end of each goroutine will trigger the scheduler, allowing other goroutines to run.

Additional Notes

Recent improvements in Go's runtime have introduced additional ways for goroutines to yield execution. For example, a proposal exists to insert scheduler calls into loops without any function calls.

In summary, while continuous loops without yielding can potentially starve other goroutines in a single-threaded environment, function calls and recent runtime enhancements provide mechanisms for goroutine scheduling even in the absence of explicit yielding.

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