Given the following pseudo-code:
func main() {
go runFuncOne()
}
func runFuncOne() bool {
runFuncTwo()
return true
}
func runFuncTwo() bool {
// Do some heavy work
return true
}
Would runFuncTwo only be blocking to runFuncOne (the calling goroutine) or would runFuncTwo also block main() as it is not itself running as a goroutine?
My assumption is that main() will open a thread within which runFuncOne() and runFuncTwo() will then operate. Any work performed within runFuncTwo() will then only block this instance of runFuncOne()?
runFuncTwo blocks runFuncOne only, as both are running in a separate Go routine.
Note though that main() will therefore continue and exit, causing the program to exit. To avoid this and all runFuncTwo to complete, you should use a sync.WaitGroup.
The Go Programming Language Specification
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. Instead, the function begins executing independently in a new goroutine. When the function terminates, its goroutine also terminates. If the function has any return values, they are discarded when the function completes.
func main() {
go runFuncOne()
}
main will invoke runFuncOne() as a goroutine and then exit the program immediately, terminating runFuncOne() without waiting for it to complete.
func runFuncOne() bool {
runFuncTwo()
return true
}
func runFuncTwo() bool {
// Do some heavy work
return true
}
runFuncOne() makes a function call to runFuncTwo() so it will wait for runFuncTwo() to complete.
Can also use a channel for synchronization:
package main
import (
"fmt"
"time"
)
func main() {
var ch chan int = make(chan int)
go runFuncOne(ch)
fmt.Println("Waiting..")
fmt.Println(<-ch)
}
func runFuncOne(ch chan int) {
runFuncTwo(ch)
ch <- 1
}
func runFuncTwo(ch chan int) bool {
time.Sleep(1 * time.Second)
fmt.Println("Done working..")
return true
}
https://play.golang.org/p/h1S4TSdT0w
Return type of runFuncOne won't be of consequence if you are calling it with a go routine.