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+--- Thread: Conversion of IP-s in bits (/Thread-Conversion-of-IP-s-in-bits)
Conversion of IP-s in bits - hoxy - 04-03-2012
Conversion of IP-s in bits
Greetings.
In this tutorial will tell you that you to convert an ip, in bits.
As the dim IP Verssion 4, we use is 32 bitsh, as consists of 4 octets with 8 bits, 8 x 4 = 32 bits.
Let's continue with the tutorial, we get 1 IP, 192.168.24.8 for example, will now start with converting the bits.
First octet is below 192 now denote by
128, 64, 32, 16, 8, 4, 2, 1
--------------------------------
FIRST octet (192)
1.Tash 128 to 192, goes to her and mark a 1 1 (192-128 = 64) ie residue 64
2.Tash 64 runs in her first 64 (remainder of 192-168) and remark the one, now 64-64 = 0, ie remaining 0
3.Tash 32, the rest 0, no runs, thus: to mark a 0
4.Tash 16, the rest 0, no runs, thus: to mark a 0
5.Tash 8, the rest 0, no runs, thus: to mark a 0
6.Tash 4, the rest 0, no runs, thus: to mark a 0
7.Tash 2, the rest 0, no runs, thus: to mark a 0
8.Tash 1, the rest 0, no runs, thus: to mark a 0
From this conclude that the first octet is converted into 192 bits gives us this:
11000000, ie the first octet is 8 bits (after ip has 4 octets of 8 bits 32 bits total)
--------------------------------
Second octet is below 168 now denote by
128, 64, 32, 16, 8, 4, 2, 1
Octet is DYT (168)
1.Tash 128 to 168, goes to her and mark a 1 1 (168-128 = 40) ie residue 40
2.Tash 64 does not go to 40 (remainder of 168-128) and remark the 0
3.Tash 32 runs, her first in 40 (remainder of 168-128) and remark the 1 (40-32 = 8) residue ie 8
4.Tash 16 does not go to 8 (40-32 = 8) and remark the 0
5.Tash 8 runs, 1 times for 8 (residue from 0-32 = 8) and remark the 1 (8-8 = 0) ie remainder 0
6.Tash 4, the rest 0, no runs, thus: to mark a 0
7.Tash 2, the rest 0, no runs, thus: to mark a 0
8.Tash 1, the rest 0, no runs, thus: to mark a 0
From this conclude that the second octet is converted into 168 bits gives us this:
10101000
--------------------------------
The third octet is now 24 below mark with
128, 64, 32, 16, 8, 4, 2, 1
THREE octet (24)
1.Tash 128, at 24, does not go, thus: to mark a 0
2.Tash 64, at 24, does not go, thus: to mark a 0
3.Tash 32, at 24, does not go, thus: to mark a 0
4.Tash 16, at 24, goes 1her Internet, so that mark a 1, (24-16 = 8) residue ie 8
5.Tash 8, at 8 (remainder of 24-16 = 8), runs 1her Internet, so that mark a 1, (8-8 = 0) ie remainder 0
6.Tash 4, the rest 0, no runs, thus: to mark a 0
7.Tash 2, the rest 0, no runs, thus: to mark a 0
8.Tash 1, the rest 0, no runs, thus: to mark a 0
From this conclude that the third octet is converted into 24 bits gives us this:
00011000
--------------------------------
The fourth octet is 8 now denote with down
128, 64, 32, 16, 8, 4, 2, 1
FOURTH octet (8)
1.Tash 128, 8, no runs, thus: to mark a 0
2.Tash 64, 8, no runs, thus: to mark a 0
3.Tash 32, 8, no runs, thus: to mark a 0
4.Tash 16, 8, no runs, thus: to mark a 0
5.Tash 8, 8, goes 1her Internet, so that mark a 1, (8-9 = 0) ie remainder 0
6.Tash 4, the rest 0, no runs, thus: to mark a 0
7.Tash 2, the rest 0, no runs, thus: to mark a 0
8.Tash 1, the rest 0, no runs, thus: to mark a 0
From this conclude that the fourth octet is converted into 8 bits gives us this:
00001000
--------------------------------
Ie we converted the 4 octet one by one:
192:11000000
168:10101000
24:00011000
8:00001000
Tanii conclude that the ip 192.168.24.8 bits of konertojm we obtain this value:
11000000 10101000 00011000 00001000
So: 192.168.24.8 = 11000000 10101000 00011000 00001000
I hope you kuptur this tutorial, since that is really very important since all communication is done in this way, meaning each communication is the NAZ.
Ie when there is then a signal, and when no signal is 0
originag tutorial in albanian
Konvertimi i IP-s ne bita
Përshëndetje .
Në kët tutorial do tju tregoj se ti ta konvertojme 1 ip , ne bita .
Siq e dim IP Verssion 4 , qe perdorim ne eshte 32 bitsh , pasi perbehet nga 4 oktete me nga 8 bita , 8 x 4 = 32 bita .
Le të vazhdojmë me tutorialin , e marrim 1 IP , përshembull 192.168.24.8 , tash do te fillojme me konvertimin ne bita .
Okteti i parë 192 tash shënojmë me posht
128 , 64 , 32 , 16 , 8 , 4 , 2 , 1
--------------------------------
OKTETI I PARE ( 192 )
1.Tash 128 në 192 , shkon 1 her dhe shenojm 1 (192-128=64) dmth mbetja 64
2.Tash 64 shkon 1 her ne 64 (Mbetja nga 192-168) dhe shenojme 1 , tash 64-64= 0 , dmth mbetja 0
3.Tash 32 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
4.Tash 16 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
5.Tash 8 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
6.Tash 4 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
7.Tash 2 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
8.Tash 1 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
Nga kjo perfundojm se kur okteti i pare 192 shendrrohet ne bita na jep ket :
11000000 , dmth okteti i pare i ka 8 bita ( pasi ip i ka 4 oktete nga 8 bita gjithsej 32 bita)
--------------------------------
Okteti i dytë 168 tash shënojmë me posht
128 , 64 , 32 , 16 , 8 , 4 , 2 , 1
OKTETI I DYT (168)
1.Tash 128 në 168 , shkon 1 her dhe shenojm 1 (168-128=40) dmth mbetja 40
2.Tash 64 nuk shkon ne 40 (Mbetja nga 168-128) dhe shenojme 0
3.Tash 32 shkon , 1 her ne 40 (Mbetja nga 168-128) dhe shenojme 1 (40-32=8) dmth mbetja 8
4.Tash 16 nuk shkon ne 8 (40-32=8) dhe shenojme 0
5.Tash 8 shkon , 1 her ne 8 (Mbetja nga 0-32=8) dhe shenojme 1 (8-8=0) dmth mbetja 0
6.Tash 4 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
7.Tash 2 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
8.Tash 1 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
Nga kjo perfundojm se kur okteti i dyte 168 shendrrohet ne bita na jep ket :
10101000
--------------------------------
Okteti i trete 24 tash shënojmë me posht
128 , 64 , 32 , 16 , 8 , 4 , 2 , 1
OKTETI I TRETE (24)
1.Tash 128 , ne 24 , nuk shkon , kshtu qe shenojm 0
2.Tash 64 , ne 24 , nuk shkon , kshtu qe shenojm 0
3.Tash 32 , ne 24 , nuk shkon , kshtu qe shenojm 0
4.Tash 16 , ne 24 , shkon 1her , kshtu qe shenojm 1 , (24-16=8) dmth mbetja 8
5.Tash 8 , ne 8 (Mbetja nga 24-16=8) , shkon 1her , kshtu qe shenojm 1 , (8-8=0) dmth mbetja 0
6.Tash 4 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
7.Tash 2 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
8.Tash 1 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
Nga kjo perfundojm se kur okteti i trete 24 shendrrohet ne bita na jep ket :
00011000
--------------------------------
Okteti i katert 8 tash shënojmë me posht
128 , 64 , 32 , 16 , 8 , 4 , 2 , 1
OKTETI I KATERT (8)
1.Tash 128 , ne 8 , nuk shkon , kshtu qe shenojm 0
2.Tash 64 , ne 8 , nuk shkon , kshtu qe shenojm 0
3.Tash 32 , ne 8 , nuk shkon , kshtu qe shenojm 0
4.Tash 16 , ne 8 , nuk shkon , kshtu qe shenojm 0
5.Tash 8 , ne 8 , shkon 1her , kshtu qe shenojm 1 , (8-9=0) dmth mbetja 0
6.Tash 4 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
7.Tash 2 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
8.Tash 1 , ne mbetjen 0 , nuk shkon , kshtu qe shenojm 0
Nga kjo perfundojm se kur okteti i katert 8 shendrrohet ne bita na jep ket :
00001000
--------------------------------
Dmth i kemi konvertuar te 4 Oktet nje nga nje :
192:11000000
168:10101000
24:00011000
8:00001000
Tanii arrijm ne perfundimin se kur ip 192.168.24.8 e konertojm ne bita fitojm ket vler :
11000000 10101000 00011000 00001000
Pra : 192.168.24.8 = 11000000 10101000 00011000 00001000
Shpresoj qe ta keni kuptur ket tutorial , pasi qe eshte me te vertet shum i rendesishem pasi i gjith komunikimi behet ne ket menyr , dmth eshte naz e cdo komunikimi.
Dmth kur esht 1 ateher ka sinjal , kurse kur eshte 0 nuk ka sinjal