This is a prime number code, is able to run normally, but when I changed his 14 lines of code by the break for the return, the code to run complains, but although error, is also able to print the correct results first and then prompt error (see picture), could you tell me why to use the return
Complains, the feeling in this code break and should have the same effect to return, please teach bosses
Package the main
The import "FMT"
Func putNum (intChan chan int) {
For I:=1; i<=80; I++ {
IntChan<-i
}
Close (intChan)
}
Func primeNum (intChan chan int, primeChan chan int, exitChan chan bool) {
For {
Var flag bool=true
V, ok:=& lt; - intChan
if ! Ok {
Break
}
For I:=2; iIf I==0 v % {
Flag=false
Break
}
}
The if flag {
PrimeChan}
}
ExitChan<- true
}
Func main () {
IntChan:=make (chan int, 800)
PrimeChan:=make (chan int, 600)
ExitChan:=make (chan bool, 4)
Go putNum (intChan)
For I:=1; i<=4; I++ {
Go primeNum (intChan primeChan, exitChan)
}
Go func () {
For I:=1; i<=4; I++ {
<- exitChan
}
Close (primeChan)
Close (exitChan)
} ()
For {
V, ok:=& lt; - primeChan
if ! Ok {
Break
}
FMT. Println (v)
}
}

CodePudding user response:

This error is call lock,
Reason is because change needs 14 lines break to return after the last number when putNum read after closing intChan, 14 lines primeNum function return directly, not to value the exitChan, result in the main, from exitChan receive four Boolean value operation blocks, cause a deadlock,
You can change the
Func primeNum (intChan chan int, primeChan chan int, exitChan chan bool) {
.
ExitChan<- true
}

Func primeNum (intChan chan int, primeChan chan int, exitChan chan bool) {
Defer func () {
ExitChan<- true
} ()
.
}
Such modifications break for return should be no problem,

CodePudding user response:

reference 1st floor gaorx1019 response:

this error is call lock,
Reason is because change needs 14 lines break to return after the last number when putNum read after closing intChan, 14 lines primeNum function return directly, not to value the exitChan, result in the main, from exitChan receive four Boolean value operation blocks, cause a deadlock,
You can change the
Func primeNum (intChan chan int, primeChan chan int, exitChan chan bool) {
.
ExitChan<- true
}

Func primeNum (intChan chan int, primeChan chan int, exitChan chan bool) {
Defer func () {
ExitChan<- true
} ()
.
}
Such modifications break as return should have no questions,

See, didn't realize that before the return is the entire putNum function exits, it seems the return and break or can not be used casually, thank bosses!

CodePudding user response:

Uh huh,,, this belongs to the standardization degree is a code,

CodePudding user response:

Break only out of this cycle, if it is a multiple nested loop, will pay attention to, such as
For {
The select {
Case & lt; Ch:
Break
}
}

Break out of the select only above, can't jump out the for, if you want to jump out the for, can

Label:
For {
The select {
Case & lt; Ch:
Break the Label
}
}
Label: can only be defined in front of the for, if you want to define behind, can use the goto statement
For {
The select {
Case & lt; Ch:
Break
}
}

Break out of the select only above, can't jump out the for, if you want to jump out the for, can

Label:
For {
The select {
Case & lt; Ch:
Goto Label
}
}
Label:

CodePudding user response:

Break only out of this cycle, if it is a multiple nested loop, will pay attention to, such as
For {
The select {
Case & lt; Ch:
Break
}
}

Break out of the select only above, can't jump out the for, if you want to jump out the for, can

Label:
For {
The select {
Case & lt; Ch:
Break the Label
}
}
Label: can only be defined in front of the for, if you want to define behind, can use the goto statement

For {
The select {
Case & lt; Ch:
Goto Label
}
}
Label: