Hi guys, why this simple program goes in deadlock? func main()

Question

Hi guys, why this simple program goes in deadlock? func main()

{
messages := make(chan string)

messages <- "msg1"
messages <- "msg2"

fmt.Println(<-messages)
fmt.Println(<-messages)
}

0

29.01.2022

3 ответов

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Ernesto A. Santana 🇨🇺

use this to run your example func main() { messages := make(chan string, 2) //buffered channel messages <- "msg1" messages <- "msg2" fmt.Println(<-messages) fmt.Println(<-messages) }

0

29.01.2022

*(&Fulvio) Автор вопроса

Ernesto A. Santana 🇨🇺
use this to run your example func main() { mess...

Yup i knew with buffered work, i was curious about why with unbuffered did not work

0

29.01.2022

Ernesto A. Santana 🇨🇺 · Accepted Answer

Ernesto A. Santana 🇨🇺

In unbuffered channel there is no capacity to hold any value before it's received. In this type of channels both a sending and receiving goroutine needs to be ready at the same instant before any send or receive operation can complete

0

29.01.2022

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Hi guys, why this simple program goes in deadlock? func main()

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