8 bits of the return code and 8 bits of the number of the killing signal are mixed into a single value on the return from wait(2)
& co..
#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <unistd.h>
#include <signal.h>
int main() {
int status;
pid_t child = fork();
if (child <= 0)
exit(42);
waitpid(child, &status, 0);
if (WIFEXITED(status))
printf("first child exited with %u\n", WEXITSTATUS(status));
/* prints: "first child exited with 42" */
child = fork();
if (child <= 0)
kill(getpid(), SIGSEGV);
waitpid(child, &status, 0);
if (WIFSIGNALED(status))
printf("second child died with %u\n", WTERMSIG(status));
/* prints: "second child died with 11" */
}
How are you determining the exit status? Traditionally, the shell only stores an 8-bit return code, but sets the high bit if the process was abnormally terminated.
$ sh -c 'exit 42'; echo $?
42
$ sh -c 'kill -SEGV $$'; echo $?
Segmentation fault
139
$ expr 139 - 128
11
If you're seeing anything other than this, then the program probably has a SIGSEGV
signal handler which then calls exit
normally, so it isn't actually getting killed by the signal. (Programs can chose to handle any signals aside from SIGKILL
and SIGSTOP
.)