I really want to when call handlerSIGSEGV end, it will continue back to
golang ,then golang will recover it.. Is there anyway?
在 2018年3月30日星期五 UTC+8下午2:20:23，Ian Lance Taylor写道：
>
> On Thu, Mar 29, 2018 at 6:34 PM, hexun via golang-nuts
> <golan...@googlegroups.com <javascript:>> wrote:
> > func mustSendSIGSEGV(){
> > defer func() {
> > if err := recover(); err != nil {
> > fmt.Print("enter recover 111")
> > // logger.PrintPanicStack()
> > }
> > }()
> > var r *Test
> > r.Num = 0
> > }
> > i rewrite mustSendSIGSEGV like this。it is same as before.. how to modify
> > the default Go SIGSEGV handler?
>
> When you install your own SIGSEGV handler, you are replacing the Go
> handler.
>
> You can't really modify the Go handler. You can use the os/signal
> package to catch a SIGSEGV signal sent to the process, but it won't
> catch a SIGSEGV caused by the process.
>
> What are you really trying to do?
>
> Ian
>
>
>
> > 在 2018年3月30日星期五 UTC+8上午12:42:42，Ian Lance Taylor写道：
> >>
> >> On Thu, Mar 29, 2018 at 8:48 AM, hexun via golang-nuts
> >> <golan...@googlegroups.com> wrote:
> >> > From "The Go Programming Language" I see this:
> >> >
> >> > If the non-Go code installs any signal handlers, it must use the
> >> > SA_ONSTACK
> >> > flag with sigaction. Failing to do so is likely to cause the program
> to
> >> > crash if the signal is received.
> >> >
> >> >
> >> >
> >> >
> >> > Can any one tell me why syscall.Kill(pid, syscall.SIGSEGV) only print
> >> > "handlerSIGSEGV Sent by 0" once ,but mustSendSIGSEGV will print
> >> > "handlerSIGSEGV Sent by 0" Unlimited times。 I want golang SIGSEGV
> pass
> >> > to c
> >> > ,only handle once ,not many times.can anyone help me?
> >> > package main
> >> > /*
> >> > #include <stdio.h>
> >> > #include <signal.h>
> >> > #include <string.h>
> >> >
> >> > struct sigaction old_action;
> >> >
> >> >
> >> > void handlerSIGSEGV(int signum, siginfo_t *info, void *context) {
> >> > printf("handlerSIGSEGV Sent by %d\n", info->si_pid);
> >> > }
> >> >
> >> >
> >> >
> >> > void testSIGSEGV() {
> >> > struct sigaction action;
> >> > sigaction(SIGSEGV, NULL, &action);
> >> > memset(&action, 0, sizeof action);
> >> > sigfillset(&action.sa_mask);
> >> > action.sa_sigaction = handlerSIGSEGV;
> >> > action.sa_flags = SA_NOCLDSTOP | SA_SIGINFO | SA_ONSTACK;
> >> > sigaction(SIGSEGV, &action, &old_action);
> >> > }
> >> > */
> >> > import "C"
> >> >
> >> > import (
> >> > "os"
> >> > "syscall"
> >> > "time"
> >> > "fmt"
> >> > )
> >> > type Test struct {
> >> > Num int
> >> > }
> >> >
> >> > func mustSendSIGSEGV(){
> >> > var r *Test
> >> > r.Num = 0
> >> > }
> >> >
> >> > func main() {
> >> > // C.test()
> >> > C.testSIGSEGV()
> >> > pid := os.Getpid()
> >> > syscall.Kill(pid, syscall.SIGSEGV)
> >> > // mustSendSIGSEGV()
> >> > for {
> >> > // syscall.Kill(pid, syscall.SIGUSR1)
> >> > fmt.Print("33")
> >> > time.Sleep(time.Second)
> >> > }
> >> > }
> >>
> >> Your SIGSEGV handler just returns, which means that the program
> >> resumes execution at the point of failure. In your mustSendSIGSEGV
> >> function you haven't done anything to fix the code, so it just gets
> >> another SIGSEGV.
> >>
> >> Note that the default Go SIGSEGV handler does not simply return. It
> >> panics.
> >>
> >> Ian
> >
> > --
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