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Hello everyone.



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So in this lesson we will learn dynamic allocation of 2D eddies.



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So first of all how we will create one be added dynamically.



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So I will write you end and five simple but I want to create the edit.



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So you might be thinking this index will be new and let's say 10 rows and columns.



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But this index is not right.



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This is not the way we create two data dynamically.



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So what do we want to do differently.



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So we want 10 rules should be there and 20 columns should be there.



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So this is our requirement that 10 rows and 20 columns.



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So basically we can say that our identities.



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So ultimately we can say what the studio is or it is just a combination of areas.



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So I can't say there are 10 it is and each of size 20.



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So let's call it even.



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And this is it then.



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So there are identities and each of size 20.



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So what I want to do.



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So to create just one Eddie to clear just one area.



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What I have to lose.



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I will use this index.



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So if I will write new into 20 because I want my area of size 20.



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So if I want to clear just one Eddie what this new thing will have done.



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So this will return me the address of the first element.



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So basically two older than me in Star similarly if I want to create an area of size 20 I will use this



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above index.



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So this is in star be new it and 20.



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I want 20 columns.



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Basically I want to create an array of size 20.



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So again it will have done me the element it within me that does with the first element.



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So it is going to return and start similarly.



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This is going to be done in start and similarly this is also going to have done me the first element



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that was the first element.



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So this is going to have done me in start.



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Correct.



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Now what I have to do so basically I have to store all these addresses



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if you want to create a truly array and so you have to store all that.



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This is now to store all that.



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This is where to do so I will create an 80 often size 10.



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And each element will be installed.



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So each element will be in in integer point.



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So I'm going to create an area of integer pointer.



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So this is very often disappointed.



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And the size is going to be then.



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So how do you create an area of in digits.



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So first you write to do the type.



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So in our case that it diapers in Star that it debuts in Star then we type start.



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So start then we write be basically the name then we write to do then the data type.



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So in this case I'm creating additive in Tejas.



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But here I want to create Adi often being appointed so I will write new in Star data type.



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So this is not easy to type.



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And then the size is so I want to create an area of size 10 so this way if we will use this approach



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I will be able to create this Eddie.



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Now after doing this we have to create all these eddies we have to create all these areas.



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So can I write b 0 is new into 20.



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Similarly I will write B1 is new into 20s and this will go on at the last.



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I will write B nine is new end and 20.



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So these are all in beta ready.



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So these are all in D ready so that's why I have to do something like this and it is obviously that



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I will not write like this I will obviously use a loop so I will just write data what this Eddie and



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I will use this I will write this simple now after doing this for example if you want to access to for



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example after doing all the hard work if you want to access be a four and five.



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So what is our structure.



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So I was structure is this is an area of integer pointer this is in a different digit pointer.



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This is you going to be your b's binding towards the first element and for each index you have an 80



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so.



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So each index is storing the address of the first element of its Eddy.



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So you want go to photo so I will go to photo it will go to photo you will be able to find an array



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and then go to the fifth index.



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So you will go to the fifth index.



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Simple so finally what is my situation.



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So this is you're in Deja appointed



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and each index is distorting and address affinity



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each index is storing the address of vanity



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and all this work is basically all the memory is basically the heap memory so all this is heap.



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So what is stack memory so in stack memory what do we have that double point.



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So this is B since you're disappointed it really be of a debate and it is playing towards it distorting



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the address of the first element.



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So this is the structure and what is the goal.



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So first of all we have to create this integer ready so for creating this and B the address index is



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very simple.



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This will be an area of integer point and so forth that is the type that I believe this one.



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Then you have the right start operator because you want to store the address then the name of the area



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and let's say B.



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So this is new data diapers in Deja 2.0.



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So each element is an integer point that because it is going to store the index of the first element



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and let's say I want to create any of size 10.



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Similarly what we have to do so you have to hydrate or what this salary.



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So let s I read or disagree.



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So I equals zero i.e. less than 10 a plus plus.



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What are you going to do.



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You have to create this wouldn't be added.



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So for creating this one lira you will write B of A is new.



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And 20.



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So finally if you will do this.



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What will happen.



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This is seem as if you want to lose technical locations or this is B of 10 and 20.



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Okay so this is how we allocate to the area dynamically.



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So let us write the code.



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So first of all I need to add a often digit point.



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So now I started being w pointer Then new in start and led to the size is 10.



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So I want to create an eddy of 10 rows and 20 columns to the area of 10 those and 20 columns.



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Now I just have to wait wait i equals zero.



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I list then then I bless less and now be off I is new top 20.



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So I've created an array of vendors and 20 columns to the ad often losing 20 columns.



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So for example if I want to take the rows and columns input from the user so you can definitely do like



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this in and my nine rows and columns.



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So give me the value of those.



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Give me the value of columns.



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So after doing this don't happen.



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So this will be your m. you will I trade on the middle floors and this is your columns.



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Simple.



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Now after doing this you want to take input.



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So for taking the input we remembered i equals zero 9 less than name I plus press then similarly j equals



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zero g list and energy plus plus.



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And you will simply write B of A G.



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So this is how we will take input or instead of writing this whole code what you can do so you can write



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here only.



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So let us comment out this one



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so you can write.



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You can likely take input here only I do to all those that is so lightweight I am editing over those



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add is that we have created in the above line.



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So I will take you off AIG simple.



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So what is happening here.



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Okay so what is the meaning of this line.



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So you're creating an additive in disappointed



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so the sizes M so each and B's an integer point and then what they're doing you are doing all this at



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a and b if I use new ink.



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And so this is why I'm thinking towards I am creating a new Eddy.



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So after creating the idea of size n I am I creating all this eddy.



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So I related to what this edit and I am taking input b if I j.



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Similarly you would again create entity of size and then you would like it or what do you say and you



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will dig the input then you will again create an area of size n then you will rotate over this idea



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and you will take input then you will again I refer to the.



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You will again create an area of size n and do related all the area and you will take input.



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Now one thing.



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So here what we are doing.



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So we are creating basically a matrix.



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So I will have 20 rows and 20 columns.



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So basically each row has same amount of columns so each row is having 20 columns.



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But we can do what we can do so we can create a structure like so we can do something like this.



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So what is the meaning.



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So the first rule is containing only one column.



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The second lowest containing only two columns or you can say it will contain only two element.



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The third row will contain three elements.



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Basically the site the number of columns would be three and the fourth column for total will contain



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four elements.



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And the fifth row will contain five elements so we can definitely create a structure like this.



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So basically what.



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What will happen to destructively become what do you do.



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So you have Ambrose.



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Here we are giving a fixed value and we are giving and let's say you will change the value.



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So it will change the value.



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So the first rule the size is 1.



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Now the size is 2.



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Now the size is 3 now the size is 4.



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So this is the power of dynamic allocation you can create a structure like this.



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So let's do this.



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So instead of n where you can do you can write a plus one.



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So if the value of i.e. zero the number of columns will become 1 if the value if I ever become when



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the number of columns will become 2 and we will obtain a structure like this.



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So I was to actively become like this



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but most of the time.



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What will happen.



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We will simply create this matrix only each and all will have seen on tough columns each row are going



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to have same number of columns but we can definitely do like this.



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There is nothing that can stop you from doing this.



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Okay so no one can stop you from doing this.



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Okay so what else is remaining.



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So I think we have used dynamic memory allocation and dynamic memory.



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V have to delete ourself manually deletion.



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Always remember.



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So just remember one thing whenever you are using new whenever you are using new.



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Always used delete.



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So we have to delete the memory out ourself.



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So what we're doing here we are creating IPG pointed at different digit points.



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So let's just delete.



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Let's just delete our diary so delete.



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Be if you will leave B do you think this is right.



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We did delete.



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I don't think so.



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Why.



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Because what is our structure our structure is first.



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We have an area of integer pointer.



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First we haven't had in often digital pointer and each and at each index I am storing the address of



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an 80.



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So each element of the integer point that is storing the address of an 80.



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So this is my stuff.



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Basically.



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Okay so this is in digital a pointer and these are these are a lot in Delia.



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So this whole thing is actually inside a heap this whole thing is actually inside the heap and inside



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stack we do have.



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So this is your stack memory inside stack.



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I have it by itself memory and this is B and Bs containing the address of the first element.



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So if this is the first element let's say it's addresses and so B's and storing hundred or against a



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piece pointing towards the the address of the 0 8 element fine.



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Now what I want to do so I want to delete this whole memory.



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So if you will write delete B if you will write delete B then what will happen.



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So basically this Eddie will get deleted this Eddie will get deleted and then you can not delete this



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why we can not delete these eddies because once you have deleted this memory.



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Now this memory does not belong to you and we discussed in the previous lesson and the previous we do



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that after deleting this you should not try to access that memory again because that memory now does



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not belong to you.



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So what we have to do.



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So first of all we have to delete these eddies we have to delete these areas and after deleting all



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that is we will delete this.



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Finally.



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Okay so what is our approach.



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So we know this is wrong because if you delete this big eddy then there is no way for us to delete these



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small eddies.



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So our logic is very simple our approach is very simple.



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First I tripped over this eddy and called the lead on these small small eddies after deleting all the



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errors.



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We will delete this big begetting simple so let's do this.



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So before deleting the big eddy before deleting the diesel point ready we have to hydrate and we have



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to delete small small eddies so I equals zero.



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We have to delete in B.J..



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So I call you two I left in him.



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I bless bless my dad.



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Do we have to call delete function delete be off.



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We are deleting small one that is since BFI.



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Would this be if I use a -- is an eddy.



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It does point towards entity since it does entity.



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I have to write like this.



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Now this is correct.



225

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OK you can give space or you do not give space.



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Both are correct.



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So first we have to delete all the individual areas and then we will delete our in D or appointed any.



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OK so.



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Do not forget to delete the memory which you have allocated using the dining memory location.



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So always remember this.



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This is a good practice so if you have any doubt in this lesson you can ask me.



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I will see you in the next video.



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Bye bye.