What's the difference between unbuffered channel and buffered channel? For example, arr1 := make(chan string), arr2 := make(chan string, 10), What's the difference? Espacially when using goroutine?
test1:
func main() {
c1 := make(chan int)
for i := 0; i < 2 ; i++ {
c1 <- i+1
}
for i := 0; i < 2 ; i++ {
fmt.Println(<-c1)
}
}
test2:
func main() {
c1 := make(chan int)
for i := 0; i < 2 ; i++ {
go func(i int){c1 <- i}(i)
}
for i := 0; i < 2 ; i++ {
fmt.Println(<-c1)
}
}
Why test1 will cause deadlock and test2 won't?
test3:
func main() {
c1 := make(chan int, 2)
for i := 0; i < 2 ; i++ {
c1 <- i+2
go func(i int){c1 <- i}(i)
}
for i := 0; i < 4 ; i++ {
fmt.Println(<-c1)
}
}
Why test3 outputs "2310" rather than "2301"?
test4
func main() {
c1 := make(chan int, 4)
for i := 0; i < 4 ; i++ {
go func(i int){c1 <- i}(i)
}
for i := 0; i < 4 ; i++ {
fmt.Println(<-c1)
}
}
Why test4 outputs "3012" rather than "0123"?
it's like a queue.
`arr1 := make(chan string)` you can put max one member into this chan,in order to push another one ,you should pop it first!
`arr2 := make(chan string, 10)` you can push 10 member ,before you push 11th you should pop some member
the number means max n in the chan at the same time.
#1
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for the print order.
when you use goroutine,it's hard to say which goroutine will finish first.
that's why sometime we use channel to make few goroutine in a order.
you can use more goroutine in your test4 and run it few time ,then you can see order not always the same
here a example use channel as lock
```
package main
import (
"fmt"
"time"
)
var c1 chan int
func step1() {
fmt.Println("step 1")
c1 <- 1
}
func step2() {
<-c1
fmt.Println("step 2")
}
func main() {
c1 = make(chan int)
go step2()
go step1()
time.Sleep(10)
}
```
#2
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