问题
Here is a very small "kill" related program .The program is executing but am unable to figure out the code.Can any one please make me understand the code below.
int main(int argc ,char **argv)
{
if(argc < 2)
{
printf("usage : ./kill PID");
return -1;
}
kill(atoi(argv[1]),SIGKILL);
return 0;
}
回答1:
Basically, it just checks to see if an argument has been provided. if(argc < 2) means are there less than two arguments to the program. Note that the program name itself is an argument too, which is why it's argc < 2 and not argc < 1. Once this has been determined the builtin kill function is called. The first argument to this method is an integer, which is why the second argument (the PID, represented as a string) is parsed to an integer with atoi. The second argument to kill is the signal, in this case SIGKILL. Other signals, such as SIGHUP could also have been used, but since this program "kills", it uses SIGKILL.
回答2:
it takes the first argument to the program (argv[1]), converts it to an integer (atoi - ascii to integer) and sends the SIGKILl (9) signal to the process with that process id.
if(argc < 2) simply checks if there were enough parameters provided, and return -1 exits the program with an exit code != 0 to signal unsuccessful termination.
来源：https://stackoverflow.com/questions/4134848/unable-to-understand-the-kill-program-of-linux