James Carlson wrote:
<snip>
> 847: not your code ... but what does '& 0xffff' accomplish here?
This is the following piece of code (current version of netcat.c):
784 /* Randomly swap ports. */
785 if (rflag) {
786 int y;
787 char *c;
788
789 srandom(time((time_t *)0));
790
791 for (x = 0; x <= (hi - lo); x++) {
792 y = (random() & 0xFFFF) % (hi - lo);
793 c = portlist[x];
794 portlist[x] = portlist[y];
795 portlist[y] = c;
796 }
797 }
random(3C) returns long which is 4 bytes in /usr/bin/nc (nc is delivered
only as 32-bit prog). This means (random() & 0xFFFF) will be just the 2
less significant bytes of the 4 returned by random().
The purpose of the expression on line 792 is preparation for permuting
the numbers in (lo,hi) interval. It seems that the goal of the binary
operation is to cap the random number returned from random() to 16 bits
(so it fits in the range of usable TCP/UDP ports) and then further limit
it with the range 'hi - lo'.
Although without the binary operation this will remain functional
(thanks to the modulo operation), IMHO the idea is logical and should be
preserved. There is probably a better way how to do it, though. (cast to
in_port_t ?)
v.
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