simple DNS question

[Brad Knowles]

At 11:34 PM +0100 3/16/01, Ali Ghodsi wrote:

>  My book says that the iterative DNS lookup is the most common one, how come?
>  To me it just sounds like its much easier to do recursive since the DNS:es
>  will handle all the requests.

	Client programs will almost always be asking a caching recursive 
nameserver, but these clients do not themselves generate the largest 
number of queries.  Indeed, it is the caching recursive nameservers 
themselves that generate the largest number of queries, as they 
iterate through the answers returned by the other nameservers in the 
world, trying to answer the query that was asked of them in a 
recursive manner.

	So, in short, you are both right.

--
Brad Knowles, <brad.knowles at skynet.be>

/*     efdtt.c     Author:  Charles M. Hannum <root at ihack.net>             */
/*                                                                         */
/*     Thanks to Phil Carmody <fatphil at asdf.org> for additional tweaks.    */
/*                                                                         */
/*     Length:  434 bytes (excluding unnecessary newlines)                 */
/*                                                                         */
/*     Usage is:  cat title-key scrambled.vob | efdtt >clear.vob           */
/*     where title-key = "153 2 8 105 225" or other similar 5-byte key     */

#define m(i)(x[i]^s[i+84])<<
unsigned char x[5],y,s[2048];main(n){for(read(0,x,5);read(0,s,n=2048);write(1,s
,n))if(s[y=s[13]%8+20]/16%4==1){int i=m(1)17^256+m(0)8,k=m(2)0,j=m(4)17^m(3)9^k
*2-k%8^8,a=0,c=26;for(s[y]-=16;--c;j*=2)a=a*2^i&1,i=i/2^j&1<<24;for(j=127;++j<n
;c=c>y)c+=y=i^i/8^i>>4^i>>12,i=i>>8^y<<17,a^=a>>14,y=a^a*8^a<<6,a=a>>8^y<<9,k=s
[j],k="7Wo~'G_\216"[k&7]+2^"cr3sfw6v;*k+>/n."[k>>4]*2^k*257/8,s[j]=k^(k&k*2&34)
*6^c+~y;}}