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| author | Denis Vlasenko <vda.linux@googlemail.com> | 2007-06-19 16:57:47 +0000 |
|---|---|---|
| committer | Denis Vlasenko <vda.linux@googlemail.com> | 2007-06-19 16:57:47 +0000 |
| commit | dcf6de552a15c1e7bddb32f1028ceb6214915425 (patch) | |
| tree | 01099725f37568f9f6a8b846cd2cba17c98cbd88 /networking | |
| parent | 5b5bcf24f5293f7f3db0aca9af98777a5aff6be2 (diff) | |
| download | busybox-dcf6de552a15c1e7bddb32f1028ceb6214915425.tar.gz | |
pscan: size optimization (Tito <farmatito@tiscali.it>)
Diffstat (limited to 'networking')
| -rw-r--r-- | networking/pscan.c | 8 |
1 files changed, 3 insertions, 5 deletions
diff --git a/networking/pscan.c b/networking/pscan.c index 9fa6993e8..fe146b8bd 100644 --- a/networking/pscan.c +++ b/networking/pscan.c @@ -35,13 +35,13 @@ int pscan_main(int argc, char **argv) { const char *opt_max_port = "1024"; /* -P: default max port */ const char *opt_min_port = "1"; /* -p: default min port */ - const char *opt_timeout = "5000"; /* -t: default timeout */ + const char *opt_timeout = "5000"; /* -t: default timeout in msec */ /* We estimate rtt and wait rtt*4 before concluding that port is * totally blocked. min rtt of 5 ms may be too low if you are * scanning an Internet host behind saturated/traffic shaped link. * Rule of thumb: with min_rtt of N msec, scanning 1000 ports * will take N seconds at absolute minimum */ - const char *opt_min_rtt = "5"; /* -T: default min rtt */ + const char *opt_min_rtt = "5"; /* -T: default min rtt in msec */ len_and_sockaddr *lsap; int s; unsigned port, max_port, nports; @@ -57,10 +57,8 @@ int pscan_main(int argc, char **argv) getopt32(argc, argv, "p:P:t:T:", &opt_min_port, &opt_max_port, &opt_timeout, &opt_min_rtt); argv += optind; max_port = xatou_range(opt_max_port, 1, 65535); - port = xatou_range(opt_min_port, 1, 65535); + port = xatou_range(opt_min_port, 1, max_port); nports = max_port - port + 1; - if ((int)nports <= 0) - bb_show_usage(); rtt_4 = timeout = xatou_range(opt_timeout, 1, INT_MAX/1000 / 4) * 1000; min_rtt = xatou_range(opt_min_rtt, 1, INT_MAX/1000 / 4) * 1000; |