IPv4 4 transcript

We are going to learn conversion from hexadimal to binary. Now a few things that we need to understand. Uh when we say binary we say base 2 and for decimal base 10 for hexadimal 16. Why is that? Because if you look at here in binary you can have only two numbers like either zero or one. So we call it two. When it is decimal you can have from 0 to 9. Numbers from 0 to 9. So you see if you count from 0 to 9 it is 10. For hexadimal you have 16 numbers. 0 to 9 10 numbers.

A to F six numbers. All together 16. When you are doing the hexadimal the decimal value for a b c d e f is 10 11 12 13 14 15 respectively. This is something that you need to memorize. It is super easy like 0 to 9 and after 9 a b c d 10 11 12 13 14 15. Okay. Now let us take a number 45 a a e. Now first thing we need to understand one hexadesimal digit contains four bits. So there are how many hexadesimal digits? 1 2 3 4.

So four hexadimal digits will contain 16 bits. Now if you look at here we need to cons we need to get the binary for four hexadimal digits. Now for four first you need to convert the hexadimal into decimal. And we know how to convert decimal to binary. So the hexad decimal of 4 is= 4. The hexa decimal of 5 is = 5. a= if you look at here 10. E is = 14. Okay. Now convert. This is this is the conversion that we learned from the um another class the converting from decimal to binary.

Converting from decimal to binary. So four 2 divide uh fouride by two caution is two remainder is zero. 2 / 2 caution is 1 remainder is zero. 1 / two caution is zero remainder is one. Now look at here very carefully. 1 0 0 1 0 0 and put zero at the beginning. Put zero at the beginning. In the same way if you do five you get 1 0 0 and you put zero at the beginning. Now if you look at here 0 1 0 0 1 0 1 0 1 0 0 1 0 1. I will show you another way to find it.

Uh this is another way to do it. So if you look at here we have four four 1 2 3 4. uh because this four is um uh in order to like in order to form a hexadimal we need four bits. So 1 2 3 4 a hexa decimal 1 2 3 4 e hexa decimal 1 2 3 4. Now when you get four bits the table is 1 2 4 8. Okay. 5 1 2 4 8 The formula is very simple like 2 to the^ 0 2 ^ 1 2 the power 2 2 the^ 3. Okay. So this is the formula with the binary table we did earlier the conversion from um binary to decimal.

Now if you look at here let's come here and see. So for four we need to find the binary. Let's say our pointer is at four and our column value is 8 4 2 1. So when we write this column we we write from right to left. But when we will do start solving when we will start to sol find the binary we'll go from left to right. Very important when you write this table 1 2 4 8 16 32 64 you write from right to left. Okay. So this is nothing but 2 the^ 0 2 the^ 1 2 the^ 2 2 the^ 3 2 the^ 4.

Okay. But when you start solving the problem, let's say 64, 32, 16, 8, 4, 2, 1, you start from right to left. Start from right to left. Okay? Do from here, here, here, here, whatever. Okay? Now, how we start solving the problem? If you look at here very carefully, the pointer value is four and our column values are 8 4 2 1 respectively. So we start from left to right. The pointer is four and the column value is 8. We start from left to right.

So the column value starts from 8. Now you need to ask is four greater than or equals to the column value. If the pointer value is greater than or equals to the column value. If the pointer value is greater than or equals to column value, we put one. So we can write here if pointer is greater than or equals to column is one or else zero. Okay. Now if you look at here 4 is greater than or equals to 8. False is zero. So zero.

If the pointer value is still four, column value will become four. Greater than or equals to four. Yes, it's equal to four. So we will write here one. Now very important once you write here one your pointer value will change. So you the comparison that you have done with you need to subtract that. What did I say? The value that you compared with the pointer value we need to subtract that. So the pointer value was four.

The column value was four. Subtract four minus 4 is zero. When the pointer value becomes zero, you already reach to the solution. Put rest of them zero. So if you look at here 0 1 0 1 0 0. Okay. Now here the pointer value becomes zero. So rest of them will be zero. So rest we have only one. We'll put here zero. Now multiply 0 with 8, 1 with four, 0 with two, 0 with 0, whatever you got four. So this is the binary value.

Let us do another one. The pointer value is five. The column value will start with eight. Is five greater than or equals to 8? If five greater than or equals to 8, no. If it is no, then zero. If five greater than or equals to four, yes. If yes, then one. Now what you need to do once we get one, we need to change the pointer value. The pointer value will become 5 - 4 is 1. The new pointer value is one. We already have done column until four.

Now column value until four. Now come the column value will be two. Is one greater than or equals to two? Zero. Now pointer value is one. The column value will become one. Is uh greater than or equals? Right? So if it is greater than or equals if it is true then the value is one and after that the pointer value will get changed. So subtract the pointer value from the compared value with uh with the subtract the compared value with the pointer value then you'll get zero.

Once it is zero you reached the solution. So if you look at here 0 1 sorry 1 1 0. So if you look at here so you can write here the first one is zero then the second one is one. So if you look at here the first one is zero the second one is one third one is zero the fourth one is one. Okay. Now the same way if you want to do a first of all you need to find the decimal value of a is 10. So you already know how to find the um one you know that how to find the binary but we are going from uh fix a decimal.

However we need to find the decimal value because we need the pointer value. The pointer value is 10. We'll start the column value with 8. compare is 10 greater than or equals 0. That's right. Then the binary is 1. Now you need to change the pointer value. 10 minus 8 you compared with 8. You get two. The new pointer value is two. The column value is four. Is 2 greater than or equals to four? No. So it will become zero.

Now pointer value is still two because last time we got zero the column value will become two. to greater than or equals to true then the pointer value is one. Now the pointer value whenever you get the binary one you need to change the pointer value pointer 2 - 2 is zero. Okay. So once you reach zero so that is where is the solution. So it will become zero. Okay. Now let's take a number 192 which is a decimal number.

So if you do the 192 you will get this one. Okay follow the previous procedure you will find the binary which is this one. Now let's do this one. So if you look at here this is an IP address 192 168. So you know that this is a this is a decimal. So for this IP address 192 you have got eight bits. This is the first octed. This is the second octed. is the first octed. This is the second octed. This is the third octed. And this is the fourth octed.

Now there are eight binaries. So you look at here eight binaries. So you need to draw the table with eight binaries. The formula is 2 to the^ 0 1 2 3 4 and 7. And if you write the value, the value is 1 to 48 16 32 64 128. Okay, you can memorize this one. Now is 128 sorry is the pointer value which is 192 pointer value 192 is greater than or equals to column value is 128. Yes. So the value is one. So you got here one.

Now if the pointer value will change 192 minus 128 is 64. Now the pointer value is 64. If the pointer value is 64 which is greater than or equals to 64 because the next column value is 64. Yes. So the binary is one. You put here one. The pointer value will become 64 minus 64 which is zero. Once the pointer value becomes zero, you already reach the solution. Here everything will be rest of the things will be zero. Now if you multiply 1 mult* 128 sorry is 128 1 * 64 is 64.

If you add them up you'll get 192 as decimal. Now here you see you did the same comparison 192 greater than 128 then the uh what what is the is then you get 64 64 - 64 okay we have already done that over there okay now in this step so if you look at here very carefully once you get uh here once you got the result. So if you write 45 AE this is the one that you need to write as binary 0 1 0 0 1 0 1 okay 1 0 1 1. So you need to write them together then which will be equivalent to 45 a e Now from hexadimal to decimal.

Last time we got hexadimal to binary, hexadimal to binary. So this is the binary one 0 1 01 for 45 a. Now from um hexa decimal to decimal we'll always start from right to left right to left E A5 4 right to left the value the decimal value is 14 this is the decimal value for each individuals for each individuals 14 10 15 uh 5 4. Now you need to start with the power of 16 16 0. So you have 16 0 16 ^ 1 16 ^ 2 16 ^ 3. So write them sequentially.

After that you multiply 14 with 16 ^ 0 10 into 16 ^ 1 5 into 16 ^ 2 4 into 16 ^ 3. Once you have done the multiplication, if you add them, you'll get 178 38. 178 38. So for this is the decimal. This is the hexadimal value and this is the decimal value. Start from right to left. Write the individual values and after that multiply each of them individuals with 16 to the power. Add them is the total decimal value. Now I will show you another way where you convert hexadimal into binary.

Okay? Where you convert hexadimal into binary first and after that from that binary you convert to the decimal and after that from that binary you convert to the decimal. Okay. Now the first thing is what I said here we directly converted hexadimal to decimal and you need to know like you need to do this conversion okay 16 to the power. Now we are going to convert hexadimal into binary. We already know the hexadimal into binary conversion how it is done.

Now if you look at here how many hexadimal digits we have here four that means we have here 16 bits. So for eight bits we had a table right 2 to the^ 0 2 the^ 1 2 3 4 5 6 7 until 7. For eight bits we had this table. For 16 you need more eight. So it will be 2 to the^ like more eight. Now for eight we had 1 2 3 4 5 6 7 8 like 1 2 3 4 5 6 7 8 until 2 ^ 7 I do here is until 2 to the^ 7. Now we have more 8. So 8 + 8 16 but they're together.

Then 1 2 3 then 9 10 11 12 13 14 15 16. If you put the binaries accordingly, you will find the binaries this way. Now you do the multiplication. 0 * 2 the^ 15 which will be so 2 ^ 15 is 32768. 0 * 2 the^ 15 is 0. Then 2 ^ 14 is 16384. 1 * 2 ^ 14 is 16384. Then 0 * 2 ^ 13 2 ^ 13 is 1 81 92 2 ^ 12 is 4096 2. So if you look at here 1 2 4 8 16 32 64 128 256 5124 2048 496 81 90 it always got um it always gets double right.

So now you put the binaries do the multiplication with that and you'll get something like this and if you add all of them all the 16 values you'll find 1783 8. So this is another way of doing it depending on depending on um depending on um what number you have okay how many hexodimal digits you have you can calculate okay whichever way you prefer I I I did show you in both the ways okay Um,