Comments on: signals https://www.emblogic.com/blog/07/signals-3/ Embedded System and ARM Training Mon, 07 Oct 2013 04:00:03 +0000 hourly 1 http://wordpress.org/?v=3.9.1 By: hemant.kumar https://www.emblogic.com/blog/07/signals-3/#comment-1024 Sun, 22 Jul 2012 08:29:51 +0000 http://emblogic.org/blog/?p=4000#comment-1024 (void) signal(SIGINT,SIG_DFL);
(void)signal(SIGINT,SIG_IGN);

try this..

here your signal(SIGINT) will be ignored ..
becuse you have declared it after the SIG_DFL

]]> By: yashvardhan https://www.emblogic.com/blog/07/signals-3/#comment-1007 Fri, 20 Jul 2012 18:21:36 +0000 http://emblogic.org/blog/?p=4000#comment-1007 Hi Akash,
No, there is nothing wrong with the code. It is the expected behaviour.
See when you will press ctrl + c for the first time, it will call your signal handler, and it will print the first statement and then the execution moves to next statement and there you are re-registering the behavior with the kernel for SIGINT (SIG_IGN)and then it will move to the next statement where you have registered it again. So your current handler is SIG_DFL which is registered with the kernel. So when you press ctrl+c for the second time, current handler is SIG_DFL, so the process will be terminated NOT IGNORED.

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