`var a chan int`和`a：= make（chan int）`有什么区别？

Question

Today I'm learning channels and goroutine of go. And I met some phenomenon which confuses me.

My go file looks like this:

package main

import (
    "fmt"
)

func testRoutine(number int, channel chan int) {
    channel <- number
}


func main() {
//    var a chan int
    a := make(chan int)
    b := make(chan int)
    go testRoutine(1, a)
    go testRoutine(2, b)

    c, d := <-a, <-b
    fmt.Printf("%d %d
", c, d)
}

it works well when I used syntax a := make(chan int).

But when I changed a := make(chan int) to var a chan int, I got panic report:

fatal error: all goroutines are asleep - deadlock!

goroutine 1 [chan receive (nil chan)]:
main.main()
    /Users/marioluisgarcia/Local/practice/go/cache/var_make_diff.go:19 +0xc7

goroutine 18 [chan send (nil chan)]:
main.testRoutine(0x1, 0x0)
    /Users/marioluisgarcia/Local/practice/go/cache/var_make_diff.go:8 +0x3f
created by main.main
    /Users/marioluisgarcia/Local/practice/go/cache/var_make_diff.go:16 +0x7c

goroutine 19 [chan send]:
main.testRoutine(0x2, 0xc42008a060)
    /Users/marioluisgarcia/Local/practice/go/cache/var_make_diff.go:8 +0x3f
created by main.main
    /Users/marioluisgarcia/Local/practice/go/cache/var_make_diff.go:17 +0xa7

So, is there any difference between var a chan int and a := make(chan int), and why this panic phenomenon was triggered?

Answer 1

Recall that var name type creates a variable named name of type type set to the default value for that type. This means that var a chan int creates a channel such that a == nil.

var a chan int = make(chan int) and a := make(chan int) are, however, the same.