运行 go-libp2p-http 示例时的错误

Question

问题出在go func。报错信息是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)
    }

错误是

命令行参数
.\sever.go:18:5: go中的表达式必须是函数调用

Answer 1

您忘记调用（调用）匿名函数。因此，错误： expression in go must be function call.

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

这称为立即调用函数表达式 (IIFE)。

Answer 2

如果你决定做一个匿名函数，那么

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)
}()

命名函数：

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)
}

运行 go-libp2p-http 示例时的错误

Bugs when running the example of go-libp2p-http

go

libp2p

go-libp2p