Re: [Help-bash] What is the effect of `kill -s INT -$$` within the INT trap of the same script?

[Peng Yu]

> Bash tries to prevent recursive trap handler invocations. ksh93 and dash,
> for instance, are happy to run the trap handler until the process runs out
> of stack space and the kernel kills it.

OK. The C library allows handling multiple INT. I typed ctrl-C twice
and the handle was called twice as shown below.

$ cat main.c
#include <stdio.h>
#include <signal.h>
#include <unistd.h>

void sig_handler(int signo) {
  if (signo == SIGINT)
    printf("received SIGINT\n");
  sleep(3);
  printf("end SIGINT\n");
}

int main() {
  printf("My process ID : %d\n", getpid());
  if (signal(SIGINT, sig_handler) == SIG_ERR)
    printf("\ncan't catch SIGINT\n");
  sleep(60);
  return 0;
}
$ ./main.exe
My process ID : 60334
^Creceived SIGINT
^Cend SIGINT
received SIGINT
end SIGINT

Bash only accepts the signal once. For example, if I try to kill a
process started the following script the trap only is called once. How
does Bash know to only handle a signal of a given signal type once?

$ cat  main.sh
#!/usr/bin/env bash
# vim: set noexpandtab tabstop=2:

trap 'echo in INT' INT
echo $$
sleep 10
echo END
$ ./main.sh # Run kill -s INT 58812 in another terminal multiple times
58812
in INT
END

-- 
Regards,
Peng