go - 是否可以只记录收到的每个信号而不改变行为？

Question

我明白如何捕捉每一个信号

idleConnsClosed := make(chan bool)
SignalChannel := make(chan os.Signal, 1)

// Notify to SignalChannel any signal received
signal.Notify(SignalChannel)

go func() {
    for {
        sig := <-SignalChannel
        log.Notice("Signal %#v received", sig)
        switch sig {
        case syscall.SIGTERM:
            // ok sigterm, I know how to handle with it
            log.Info("ShutDown HTTP server (SIGTERM)")
            if err := server.Shutdown(context.Background()); err != nil {
                // Error from closing listeners, or context timeout:
                log.Error("HTTP server Shutdown: %v", err)
            }
        case syscall.SIGHUP:
            //reinit configurations and do some stuff, I know how to handle this                 
            continue
        case syscall.SIGPIPE:
            //Don't know what to do, just wanted to log it
            continue
        case syscall.SIGABRT:
            //exit with core dump...how to do that ?
            continue
        default:
            // unhandled signal?, ok how to not change it's behavior?
            log.Warning("Unhandled Signal %s received!", sig)
        }
        close(idleConnsClosed)
    }
}()

一般来说我只是想

1. 处理一些可以理解的信号 - X
2. 记录我收到的每个信号 - Y
3. 不改变 Y - X 信号的行为

Answer 1

Notify 禁用一组给定异步信号的默认行为，而是通过一个或多个已注册 channel 传递它们。我认为这意味着您无法在不改变信号行为的情况下拦截信号。

docs描述默认行为:

By default, a synchronous signal is converted into a run-time panic. A SIGHUP, SIGINT, or SIGTERM signal causes the program to exit. A SIGQUIT, SIGILL, SIGTRAP, SIGABRT, SIGSTKFLT, SIGEMT, or SIGSYS signal causes the program to exit with a stack dump

应该可以捕获信号、处理它，然后模拟适合该信号的默认行为。