Do sending more than 8 element to a channel can result in deadlock?

package main

import (
	"fmt"
	"time"
)

func printCount(c chan int) {
	num := 0
	for num >= 0 {
		num = <-c
		fmt.Print(num, " ")

	}
}

func main() {

	a := []int{8, 6, 7, 5, 3, 0, 9, -1, 3, 4}
	// If i use a := []int{8, 6, 7, 5, 3, 0, 9, -1} then it works fine

	c := make(chan int)

	go printCount(c)

	for _, v := range a {

		c <- v

	}

	time.Sleep(time.Second * 1)
	fmt.Println("End of main")
}

I try to send element of the slice to the channel and print it using golang.
Now upto 8 element the program works fine.
As i am adding 9th element in slice then deadlock conditions occures dont know why. I used to solve that by using the waitgroup but still not working.

Sending a value through an (unbuffered) channel is always blocking, until it is received somewhere else.

I think your intention here is to send all values and then only print them if they are 0 or greater? You could alter your printCount like this:

func printCount(c chan int) {
    for num := range c {
        if num >= 0 {
            fmt.Print(num, " ")
        }
    }
}

https://goplay.space/#m5eT9AYDH-Q

Hi just change c := make(chan int) to c := make(chan int, 100)
where 100 is highest capacity of your channel:

package main
import (
	"fmt"
	"time"
)
func printCount(c chan int) {
	num := 0
	for num >= 0 {
		num = <-c
		fmt.Print(num, " ")
	}
}
func main() {
	a := []int{8, 6, 7, 5, 3, 0, 9, -1, 3, 4}
	// If i use a := []int{8, 6, 7, 5, 3, 0, 9, -1} then it works fine
	c := make(chan int, 40)
	go printCount(c)
	for _, v := range a {
		c <- v
	}
	time.Sleep(time.Second * 1)
	fmt.Println("End of main")
}

result:

8 6 7 5 3 0 9 -1 End of main

The printCount goroutine exits if a negative value is processed.
More precisely, the value -1 is stopping it.

Here the printCount goroutine is the only goroutine that can read the channel values. When this goroutine exits no more writing are possible in the channel because the channel is unbuffered.

NB: The channel is unbuffered and thus cannot contains any value. A unbuffered channel is a golang idiom to concurrently safe pass values from a goroutine to another synchronously.

Solution

You should read the channel until no more element need to be processed.

No more need to use time.Sleep(time.Second * 1), the printCount is now blocking the main goroutine until finished !

package main

import (
    "fmt"
    "time"
)

func printCount(c chan int) {
    // exits if the channel is empty (or unbuffered) and closed.
    for num := range c {
        if num >= 0 {
            fmt.Print(num, " ")
        }
    }
}

func main() {

    a := []int{8, 6, 7, 5, 3, 0, 9, -1, 3, 4}

    c := make(chan int)

    go func(){
        // close the channel after sending the last value
        defer close(c)

        for _, v := range a {

            c <- v

        }
    }()


    printCount(c)

    fmt.Println("End of main")
}