Why can’t I ignore SIGSEGV signal?

Your code is ignoring SIGSEGV instead of catching it. Recall that the instruction that triggered the signal is restarted after handling the signal. In your case, handling the signal didn’t change anything so the next time round the offending instruction is tried, it fails the same way.

If you intend to catch the signal change this


to this

signal(SIGSEGV, sighandler);

You should probably also use sigaction() instead of signal(). See relevant man pages.

In your case the offending instruction is the one which tries to dereference the NULL pointer.

printf("%d", *p);

What follows is entirely dependent on your platform.

You can use gdb to establish what particular assembly instruction triggers the signal. If your platform is anything like mine, you’ll find the instruction is

movl    (%rax), %esi

with rax register holding value 0, i.e. NULL. One (non-portable!) way to fix this in your signal handler is to use the third argument signal your handler gets, i.e. the user context. Here is an example:

#include <signal.h>
#include <stdio.h>

#define __USE_GNU
#include <ucontext.h>

int *p = NULL;
int n = 100;

void sighandler(int signo, siginfo_t *si, ucontext_t* context)
  printf("Handler executed for signal %d\n", signo);
  context->uc_mcontext.gregs[REG_RAX] = &n;

int main(int argc,char ** argv)
  signal(SIGSEGV, sighandler);
  printf("%d\n", *p); // ... movl (%rax), %esi ...
  return 0;

This program displays:

Handler executed for signal 11

It first causes the handler to be executed by attempting to dereference a NULL address. Then the handler fixes the issue by setting rax to the address of variable n. Once the handler returns the system retries the offending instruction and this time succeeds. printf() receives 100 as its second argument.

I strongly recommend against using such non-portable solutions in your programs, though.

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