I have a question. The desired value is running every 5 seconds...That's how it's written If you write the code and the name is alarm.c, if you do alarm 5, the value of *argv[1] goes to 5 If we execute it, wouldn't SIGALARM be executed on signal after 5 seconds through alarm (*argv[1]) in the main function and exit (0)? Why does running keep coming out when I run it and it can't end...
include stdio.h
include stdlib.h
include signal.h
include unistd.h
void sig(int signal)
{
exit(0);
}
int main(int argc, char *argv[])
{
signal(SIGALRM, sig);
alarm(*argv[1]);
while(1){
printf("running...\n");
sleep(1);
}
return 0; }
unix
alarm(unsigned) accepts unsigned integers as call factors.
*argv[1] is treated as an integer as char. When executed with ./alarm5, the iskey code 53 for '5' instead of an integer 5.
That is, alram (*argv[1]) generates a signal after 53 seconds.
Therefore, you should call it by converting a string into an integer, as follows:
int main(int argc, char *argv[])
{
int timeout = atoi(argv[1]);
if (timeout < 0)
return -1;
signal(SIGALRM, sig);
alarm((unsigned) timeout);
As an admonition, exit() should not be used in the signal handler. Please refer to the following link.
https://en.cppreference.com/w/c/program/signal#Signal_handler
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