Bugs when running the example of go-libp2p-http

Question

The problem lies in go func. The error message is expression in go must be function call

 listener, _ := gostream.Listen(host1, p2phttp.DefaultP2PProtocol)
    defer listener.Close()
    go func() {
        http.HandleFunc("/hello", func(w http.ResponseWriter, r *http.Request) {
            w.Write([]byte("Hi!"))
        })
        server := &http.Server{}
        server.Serve(listener)
    }

The error is

command-line-arguments
.\sever.go:18:5: expression in go must be function call

Answer 1

You forget to invoke (call) the anonymous function. Hence, the error: expression in go must be function call.

go func() { fmt.Prinln("Im an IIFE") }()  // note the parenthesis

This is called immediate invoked function expression (IIFE).

Answer 2

If you decide to make an anonymous function, then

listener, _ := gostream.Listen(host1, p2phttp.DefaultP2PProtocol)
defer listener.Close()
go func() {
    http.HandleFunc("/hello", func(w http.ResponseWriter, r *http.Request) {
        w.Write([]byte("Hi!"))
    })
    server := &http.Server{}
    server.Serve(listener)
}()

Named function:

listener, _ := gostream.Listen(host1, p2phttp.DefaultP2PProtocol)
defer listener.Close()
go Greet()

func Greet() {
    http.HandleFunc("/hello", func(w http.ResponseWriter, r *http.Request) {
        w.Write([]byte("Hi!"))
    })
    server := &http.Server{}
    server.Serve(listener)
}