On Thu, Mar 29, 2018 at 8:48 AM, hexun via golang-nuts
<golang-nuts@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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