1 00:00:01,140 --> 00:00:01,810 Hi everyone. 2 00:00:01,830 --> 00:00:06,240 In this lesson we are going to learn what is dynamic memory allocation. 3 00:00:06,360 --> 00:00:08,430 So let's start today's class. 4 00:00:08,580 --> 00:00:12,730 So let us discuss what is the biggest problem with the ADI. 5 00:00:12,900 --> 00:00:17,800 So whenever you want to create an entity you have to give the size. 6 00:00:17,820 --> 00:00:23,350 So let's say I want to create an area which will contain 15 digits at the max. 7 00:00:23,370 --> 00:00:25,560 So this 50 is actually a fixed number. 8 00:00:26,310 --> 00:00:30,690 So basically we are creating a fixed sized Eddie. 9 00:00:30,750 --> 00:00:42,360 Now we can also do like this in 10 seeing and and then into a and we can definitely do like this. 10 00:00:42,550 --> 00:00:45,640 But ideally we should never do this. 11 00:00:45,740 --> 00:00:50,010 No C++ does not get that this thing will work. 12 00:00:50,050 --> 00:00:56,140 It may work on some compiler and it may not work on other compilers so there is no guarantee that this 13 00:00:56,140 --> 00:00:57,470 thing will work. 14 00:00:57,490 --> 00:01:00,210 So that's why we should never use this. 15 00:01:00,310 --> 00:01:03,040 We should never create our Eddie. 16 00:01:03,070 --> 00:01:06,260 So this is and is actually a variable. 17 00:01:06,280 --> 00:01:13,560 So basically what we are doing we are creating a variable sized Eddie and in this approach we were creating 18 00:01:13,590 --> 00:01:14,950 a fixed sized Eddie. 19 00:01:14,970 --> 00:01:20,400 So this is hundred percent correct but there are problem with this way of creating Eddie. 20 00:01:20,430 --> 00:01:24,110 So we are creating a variable sized Eddie and the value of an. 21 00:01:24,400 --> 00:01:27,650 So the value of end will be given by the user at runtime. 22 00:01:27,660 --> 00:01:33,480 So when our program will be running then the user will give what is the value of n and then we are creating 23 00:01:33,540 --> 00:01:34,890 Eddie of that size. 24 00:01:35,550 --> 00:01:38,990 So this is actually wrong and in this time. 25 00:01:38,990 --> 00:01:41,070 So this is fixed sized Eddie. 26 00:01:41,240 --> 00:01:42,500 So this is fixed as it is. 27 00:01:42,590 --> 00:01:47,520 This is compile time we know Ed the compile demo leader. 28 00:01:47,590 --> 00:01:49,620 How much memory we required. 29 00:01:49,660 --> 00:01:51,210 But this is runtime. 30 00:01:51,310 --> 00:01:54,820 So we do not know how much memory our Eddie will take. 31 00:01:54,820 --> 00:01:58,300 So there are actually two types of memory allocation. 32 00:01:58,390 --> 00:02:03,520 So there are actually two types of memory stack memory and heap memory. 33 00:02:03,520 --> 00:02:06,100 So let's see what is a stack and heap memory. 34 00:02:08,350 --> 00:02:11,560 So the first type of memory is stack. 35 00:02:11,560 --> 00:02:20,510 And the second type of memory is heap so stack memory is small and a heap memory is very big as compared 36 00:02:20,510 --> 00:02:32,200 to the stack memory now whenever you do something like into a question or you do something like into 37 00:02:32,250 --> 00:02:34,230 a 20. 38 00:02:34,230 --> 00:02:36,840 So actually these two things. 39 00:02:36,840 --> 00:02:40,530 So the memory is going to be allocated on the stack. 40 00:02:40,530 --> 00:02:41,690 Fine. 41 00:02:41,790 --> 00:02:48,290 Now even if you do this I want to create an array of ten thousand sites. 42 00:02:48,390 --> 00:02:52,230 So in this case again the memory is going to be allocated on stack. 43 00:02:52,230 --> 00:02:54,950 So what happens is whenever I a program run. 44 00:02:55,080 --> 00:03:00,240 So when I let a program start running it will take some along the stack memory and it will run the program 45 00:03:01,430 --> 00:03:04,590 in this case when you're creating a big sized area. 46 00:03:04,610 --> 00:03:10,850 So I'm creating a big error so I know at the compile family that I need a big memory I need a large 47 00:03:10,850 --> 00:03:11,490 memory. 48 00:03:11,540 --> 00:03:12,770 So what will happen. 49 00:03:12,770 --> 00:03:16,010 Our program will take comparatively a bigger stack. 50 00:03:16,040 --> 00:03:20,530 It will come it will comparatively take a bigger stick to hold this much amount of memory. 51 00:03:20,660 --> 00:03:23,170 And then our program will run. 52 00:03:23,240 --> 00:03:24,770 So I mean I'm building myself. 53 00:03:24,770 --> 00:03:28,010 So whenever a system runs our program what will happen. 54 00:03:28,010 --> 00:03:33,690 So system will take some amount of stack memory and it will run our program. 55 00:03:33,710 --> 00:03:36,880 So even if you want to create an array of 10000 size. 56 00:03:37,190 --> 00:03:41,690 So the system knows at the compile time will lead that you're creating this big error. 57 00:03:41,700 --> 00:03:43,740 So you need a bigger stack memory. 58 00:03:43,760 --> 00:03:49,900 So the system will allocate us a bigger aspect memory and it will run our program perfectly fine. 59 00:03:49,910 --> 00:04:01,070 So the problem occurs when you will do something like this in an CNN now and in again. 60 00:04:01,100 --> 00:04:05,930 So this memory is going to be located on heap only sorry on stack. 61 00:04:05,930 --> 00:04:09,430 This memory is going to be allocated on a stack only. 62 00:04:09,680 --> 00:04:14,350 The problem will occur if the user will give the value of n to be a very large number. 63 00:04:14,360 --> 00:04:16,990 Let's return to the power six. 64 00:04:17,180 --> 00:04:17,960 So what will happen. 65 00:04:17,960 --> 00:04:24,740 System will take some amount of spec memory and it will run our program and add the runtime user can 66 00:04:24,740 --> 00:04:27,690 give the value of n let's say a very large number. 67 00:04:27,710 --> 00:04:33,170 So if user give a very large number as input then basically you expect will not have that much amount 68 00:04:33,170 --> 00:04:36,450 of memory to store this much big along very. 69 00:04:36,500 --> 00:04:42,990 So this much begetting I am repeating myself the problem will occur if you will do something like this 70 00:04:45,490 --> 00:04:51,340 so our system will run our program with some amount of stack memory fine. 71 00:04:51,400 --> 00:04:56,620 Now the value of end will be decided at that time and let's say the user give the value of n to be a 72 00:04:56,620 --> 00:05:03,370 very large number then what will happen your stack will not have enough amount of memory to hold this 73 00:05:03,370 --> 00:05:04,370 Big Eddie. 74 00:05:04,420 --> 00:05:07,710 So your program may crash. 75 00:05:07,720 --> 00:05:15,370 Now the obvious solution for this problem is C heap memory is a very big memory so you can allocate 76 00:05:15,670 --> 00:05:22,840 your Eddie in this memory if you want to do something like this then look at the memory in heap because 77 00:05:22,840 --> 00:05:30,040 heap is much much larger than Stack and one more thing the value of n is given by the user at runtime 78 00:05:30,310 --> 00:05:31,810 so the stack space. 79 00:05:31,810 --> 00:05:36,710 So this stack size could not be increased during that time okay. 80 00:05:36,730 --> 00:05:42,430 We cannot increase our stack memory during runtime so the obvious solution is add brand name allocate 81 00:05:42,430 --> 00:05:46,210 the memory inside heap because heap is much much bigger than stack. 82 00:05:46,300 --> 00:05:49,630 So this added can easily be stored inside heap. 83 00:05:51,190 --> 00:05:53,590 So this is called heap memory allocation 84 00:05:57,230 --> 00:06:03,560 heap memory allocation and the second name of heat memory allocation is actually dynamic memory allocation 85 00:06:08,560 --> 00:06:09,410 so let's see. 86 00:06:09,520 --> 00:06:14,410 Let us differentiate what is dynamic memory allocation. 87 00:06:14,410 --> 00:06:24,850 So we have stack memory and we have heat memory so in stack memory when we do something like this in 88 00:06:24,870 --> 00:06:25,200 day. 89 00:06:25,230 --> 00:06:28,980 So everything is going to be stored inside my spec memory. 90 00:06:28,980 --> 00:06:37,440 So this is called static allocation starting a location means your data is going to be stored on a stack 91 00:06:37,440 --> 00:06:38,520 memory. 92 00:06:38,580 --> 00:06:39,650 Now in heat memory. 93 00:06:39,660 --> 00:06:44,700 So if you want to store something heat memory then we call it dynamic memory allocation or you can say 94 00:06:44,700 --> 00:06:54,430 heat memory allocation or dynamic memory allocation so this memory is actually the memory is allocated 95 00:06:54,460 --> 00:06:59,200 during compile time so you can see you can write memory allocated 96 00:07:01,750 --> 00:07:02,950 during compile time 97 00:07:06,480 --> 00:07:07,500 because in date. 98 00:07:07,530 --> 00:07:13,470 So this will get compiled and memory will get allocated at compile time only now a heap. 99 00:07:13,470 --> 00:07:15,780 So you want to create a variable sized Eddie. 100 00:07:15,990 --> 00:07:17,640 You want to do something like this. 101 00:07:17,940 --> 00:07:20,670 And basically you want to create a variable sized Eddie. 102 00:07:20,730 --> 00:07:24,330 So that means so the value of end will be given by the user runtime. 103 00:07:24,360 --> 00:07:31,650 So it is called runtime memory allocation random memory allocation or you can say the memory will get 104 00:07:31,650 --> 00:07:35,360 allocated at runtime. 105 00:07:35,650 --> 00:07:39,880 So this is not actually this index for allocating memory on inside heap. 106 00:07:39,880 --> 00:07:43,010 You will see this index but this is not the right index. 107 00:07:43,210 --> 00:07:47,770 Now the stack memory is much smaller than the heap memory. 108 00:07:47,770 --> 00:07:56,640 So this is less and my heap memory is very large memory so heap is a very large memory. 109 00:07:56,720 --> 00:08:00,050 So what I want to do I want to do something like this. 110 00:08:00,290 --> 00:08:06,000 So with the current knowledge that we have we only know how to allocate memory inside stack. 111 00:08:06,080 --> 00:08:09,170 We don't know how to allocate memory inside heap. 112 00:08:09,170 --> 00:08:19,150 So now let us learn dynamic allocation how to allocate memory inside heap so let's see so if you will 113 00:08:19,150 --> 00:08:22,810 do something like this entire question. 114 00:08:22,900 --> 00:08:29,110 So what will happen this AI is going to be stored inside my stack memory. 115 00:08:29,110 --> 00:08:32,730 This idea will be stored inside these tech memory fine. 116 00:08:32,770 --> 00:08:37,220 Now what I want to do I want to store something inside my memory. 117 00:08:37,330 --> 00:08:40,120 So how can I write this index. 118 00:08:40,120 --> 00:08:41,540 So what is this index. 119 00:08:41,650 --> 00:08:47,200 So what you have to write like this new and you want to create and Deja. 120 00:08:47,230 --> 00:08:52,810 So you will write new and so with this statement so new is actually a keyword new the keyword. 121 00:08:52,810 --> 00:09:01,630 So this new and what will happen you will get four bytes of memory space inside you and heap you will 122 00:09:01,630 --> 00:09:04,380 get four bytes of memory space inside your heap. 123 00:09:04,390 --> 00:09:11,230 Now let's say you want to insert 10 you want to put pen here but this memory doesn't have a name it 124 00:09:11,230 --> 00:09:12,120 doesn't have a name. 125 00:09:12,190 --> 00:09:18,050 So here if I want to do something with a so I have a name but this doesn't have a name. 126 00:09:18,220 --> 00:09:21,790 So how can we put 10 inside these four bytes. 127 00:09:21,790 --> 00:09:25,630 So actually this new thing it returns the address of this memory. 128 00:09:25,660 --> 00:09:28,260 So if I do this let's say it addresses hundred. 129 00:09:28,270 --> 00:09:30,930 So this new thing will return me the address. 130 00:09:31,090 --> 00:09:34,070 And for storing the address we started pointers. 131 00:09:34,180 --> 00:09:39,210 So I will write it like this and start I. 132 00:09:39,430 --> 00:09:43,750 So I will write it like this in START I is new and 133 00:09:47,350 --> 00:09:51,710 so what is the effect of this statement but it might really with the effect of the slain. 134 00:09:51,850 --> 00:09:57,890 So new is going to be done the address of the memory inside heap. 135 00:09:58,100 --> 00:09:59,090 So what will happen. 136 00:09:59,090 --> 00:10:07,370 So this new and what it will do so it will go inside the heap memory and it will create four bytes it 137 00:10:07,370 --> 00:10:11,470 will give me four bytes and the let's say it does is hundred. 138 00:10:11,550 --> 00:10:17,810 So this I think so this I use a pointer so what is happening this is U.S. tech. 139 00:10:17,880 --> 00:10:23,410 This is U.S. tech memory and science tech memory you have created guy Deja 2.0. 140 00:10:23,430 --> 00:10:28,860 So this is your digital point that you disk containing and it just ended and we know point does sod 141 00:10:28,890 --> 00:10:37,640 off you two bytes so add this line total bill bytes of memory are allocated out of tool you two bytes 142 00:10:37,730 --> 00:10:49,550 are allocated on stack and four bytes are allocated on the heap now if you want to put something inside 143 00:10:49,580 --> 00:10:53,230 this memory this index is quite simple do it do it like this. 144 00:10:53,230 --> 00:10:55,960 I equals ten So ten will be inserted here. 145 00:10:56,880 --> 00:11:03,010 Okay so this I is actually pointing towards this heap memory address. 146 00:11:03,240 --> 00:11:07,530 So if you want to insert then you can do it like this spot I equals 10. 147 00:11:07,560 --> 00:11:12,850 Similarly if you want to create a double so you will write it like this. 148 00:11:13,100 --> 00:11:17,280 Double Star P is new double 149 00:11:20,370 --> 00:11:21,750 again what will happen. 150 00:11:21,810 --> 00:11:23,070 So this is good heat memory 151 00:11:26,860 --> 00:11:28,970 so double is off eight bytes. 152 00:11:29,090 --> 00:11:30,520 This is a debate. 153 00:11:30,620 --> 00:11:35,120 This is U.S. tech memory inside stack. 154 00:11:35,170 --> 00:11:37,000 You are storing double pointer. 155 00:11:37,030 --> 00:11:43,970 So this is B let's say it addresses hundred so it is storing at this hundred and it is pointing towards 156 00:11:43,970 --> 00:11:50,060 this if you want to put something if you want to put something inside this box you will do it like this 157 00:11:50,150 --> 00:11:54,080 star B equals three point fold simple. 158 00:11:54,380 --> 00:11:56,150 So let's write the code and let's see 159 00:12:03,110 --> 00:12:09,160 so into start B is new keywords so new is a keyword. 160 00:12:09,350 --> 00:12:13,390 And so at this line how much amount of memory space is allocated. 161 00:12:13,670 --> 00:12:28,050 So 8 bytes on stack because you are creating AI integer pointer and for byte 1 heap now how to access 162 00:12:28,460 --> 00:12:35,280 heap memory ERP because bees containing bees containing the address of the memory so start to be closed 163 00:12:35,280 --> 00:12:38,860 down and let's say you want to print the content. 164 00:12:38,870 --> 00:12:47,080 So if you would like to see out bee it will print the address if you want to paint the content you have 165 00:12:47,080 --> 00:12:48,880 to use deet offense operator 166 00:12:51,750 --> 00:12:53,320 so Lucy out start bee 167 00:13:03,450 --> 00:13:09,760 so this is the address of the memory which is located inside heap and this is the value 168 00:13:15,720 --> 00:13:22,950 similarly what you can do it can also do it like this so double star D is new double 169 00:13:26,090 --> 00:13:36,990 similarly you can do good start C is new get so this new thing it is are dunning me address of the memory 170 00:13:37,680 --> 00:13:44,540 so again I'm repeating myself so when you do something like this many will write the statement like 171 00:13:44,540 --> 00:13:55,090 this and start B is new int So what's going to happen memory will be located inside the heap and it 172 00:13:55,090 --> 00:14:00,360 will also be located inside the stack. 173 00:14:00,370 --> 00:14:04,420 So this is just tech stack memory is smaller than heap memory. 174 00:14:04,420 --> 00:14:11,920 So new and so you will get four bytes here this is your four bytes start B. 175 00:14:11,980 --> 00:14:19,050 So this is you're in Deja point to a debate and B is pointing towards this memory location. 176 00:14:19,060 --> 00:14:26,690 So let's say if you want to put something here let's say 20 so you have to write star B equals twenty 177 00:14:27,050 --> 00:14:35,860 so first start P means D defense better first and go to the memory location and then put 20 did fine. 178 00:14:35,880 --> 00:14:37,640 So this new is actually a keyword. 179 00:14:37,650 --> 00:14:45,810 And where does the address off this memory does of the memory which is located inside heap. 180 00:14:45,960 --> 00:14:53,850 OK so now let's see how we can allocate at a dynamically how we can put Eddie inside the heap. 181 00:14:53,880 --> 00:14:56,090 So I want to create entity inside the heap. 182 00:14:56,100 --> 00:14:58,620 So let's see what will be this index. 183 00:14:58,620 --> 00:15:08,640 So this index is quite simple you have to write it like this and start e is new in 50. 184 00:15:09,300 --> 00:15:12,180 So what will happen again. 185 00:15:12,560 --> 00:15:13,670 This is you to heap memory 186 00:15:16,930 --> 00:15:21,400 so new int and 50 so 50 means forward into 50. 187 00:15:21,400 --> 00:15:27,910 That means you will get it to wondered bytes of data so you will get an net inside the heap. 188 00:15:27,930 --> 00:15:36,380 This is your daddy index zero index 49 and this new thing what it will get done it will have done the 189 00:15:36,380 --> 00:15:42,260 address of the first element it will be done the address of the first element and I have created an 190 00:15:42,270 --> 00:15:46,010 digital pointer to store that address simple. 191 00:15:46,640 --> 00:15:51,040 So now we know how to create added dynamically. 192 00:15:51,200 --> 00:15:55,040 So after knowing this we can definitely do something like this. 193 00:15:55,310 --> 00:16:04,290 So I can write into n CNN and then I will write it like this in start. 194 00:16:04,330 --> 00:16:13,030 E is new and and so now this is Connect. 195 00:16:13,060 --> 00:16:19,840 So what will happen the value of end value given by the user at runtime and at that time what will happen 196 00:16:19,870 --> 00:16:20,490 at that time. 197 00:16:20,500 --> 00:16:22,610 We can allocate memory inside the heap. 198 00:16:23,860 --> 00:16:25,870 So the value of N will be given by the user. 199 00:16:25,870 --> 00:16:29,560 And then you will create an array of sized and inside the heap. 200 00:16:29,560 --> 00:16:33,780 So let's say the value of n is 50. 201 00:16:33,880 --> 00:16:36,090 So this will be at index 50. 202 00:16:36,340 --> 00:16:40,460 So this will be an index 49 and this is your index is you. 203 00:16:41,040 --> 00:16:41,450 Okay. 204 00:16:41,550 --> 00:16:44,540 So now you can write like this. 205 00:16:44,550 --> 00:16:47,560 So this is perfectly fine. 206 00:16:47,590 --> 00:16:48,340 So what did happen. 207 00:16:48,350 --> 00:16:52,100 Program our system will take some stack memory and it will then the program. 208 00:16:52,180 --> 00:16:56,770 And if you want to allocate the memory at that time then allocate the memory inside the heap because 209 00:16:56,800 --> 00:16:59,650 each memory is much much larger than distinct memory. 210 00:16:59,650 --> 00:17:01,990 So now our problem has been solved. 211 00:17:02,020 --> 00:17:03,820 So there were two problems with daddy. 212 00:17:04,570 --> 00:17:12,190 So there does discuss the edit problems and let us discuss how discord will help us to solve that problem 213 00:17:13,940 --> 00:17:15,630 so and has proven them. 214 00:17:15,650 --> 00:17:25,590 So the first problem is let's say I have created at a offsite thousand and let's say I am doing this 215 00:17:25,670 --> 00:17:28,110 in then in. 216 00:17:28,590 --> 00:17:31,800 And then I am doing I equals zero. 217 00:17:31,800 --> 00:17:39,800 I listed in a plus place and I am doing the scene if I now let's say the value of Aniston. 218 00:17:40,190 --> 00:17:42,750 Let's say the value often given by the user is 10. 219 00:17:42,800 --> 00:17:43,840 So what would happen. 220 00:17:45,000 --> 00:17:50,700 I have a good memory four thousand elements but I will use memory only for 10 elements. 221 00:17:50,700 --> 00:17:52,190 So basically the elements. 222 00:17:52,210 --> 00:17:55,370 So how much amount of memory space I am wasting here. 223 00:17:55,410 --> 00:17:57,900 So 9 and elements multiply each element. 224 00:17:58,020 --> 00:18:00,260 So each integer will take four bytes. 225 00:18:00,300 --> 00:18:06,140 So this much amount of memory space will get wasted if I will use the static allocation. 226 00:18:06,600 --> 00:18:14,520 If I'm using this a location so this is the first problem of the areas that we have a lot of amount 227 00:18:14,520 --> 00:18:15,800 of memory wastage. 228 00:18:15,990 --> 00:18:23,420 If we are creating a fixed sized Eddie and with the help of dining memory allocation we will not waste 229 00:18:23,480 --> 00:18:24,290 any space. 230 00:18:24,830 --> 00:18:30,150 So our first problem will get resolved with the help of dynamic memory allocation. 231 00:18:30,170 --> 00:18:32,170 Now what is the second problem with daddy. 232 00:18:32,240 --> 00:18:36,290 So the second problem is let's say you are creating an array of size 10. 233 00:18:36,480 --> 00:18:41,390 Then again you will do the same thing and then you will take how many number of elements you want to 234 00:18:41,390 --> 00:18:42,150 insert. 235 00:18:42,230 --> 00:18:47,660 And again you will be equals you know less than I plus plus and you will do something like this scene 236 00:18:47,670 --> 00:18:48,860 a if. 237 00:18:49,430 --> 00:18:55,550 Now let's say the value of randomly given by the user that in time now led to the value of and is ended. 238 00:18:55,610 --> 00:18:58,510 If you give the value of hundred then what will happen. 239 00:18:58,520 --> 00:19:05,150 Your program will crash because you can add the max a store then elements and the user has given the 240 00:19:05,150 --> 00:19:09,500 value of N hundred so your program is going to be correct. 241 00:19:09,590 --> 00:19:13,550 So again if you will use a dynamic memory allocation. 242 00:19:13,910 --> 00:19:17,830 So the second problem of daddy will get resolved. 243 00:19:18,020 --> 00:19:21,130 So it is the power of dynamic memory allocation. 244 00:19:21,260 --> 00:19:23,290 What the problems associated with Daddy. 245 00:19:23,330 --> 00:19:24,370 Will get resolved. 246 00:19:24,500 --> 00:19:26,200 We will not waste any memory. 247 00:19:26,360 --> 00:19:32,300 And also we will not face problem of this type so dynamic memory allocation is really really good. 248 00:19:33,620 --> 00:19:37,730 So after creating the eddy how can it put the elements inside. 249 00:19:37,730 --> 00:19:40,760 So just like the state to get it you can do it like this. 250 00:19:40,760 --> 00:19:44,180 If 0 is 10 so then will be here. 251 00:19:44,180 --> 00:19:45,200 You can write like this. 252 00:19:45,200 --> 00:19:49,310 So each of 20 is let's say 40. 253 00:19:49,330 --> 00:19:51,420 So you can add the next 20. 254 00:19:51,830 --> 00:19:54,880 I will insert 40 so you're going definitely do it like this. 255 00:19:54,890 --> 00:19:56,170 Just like the. 256 00:19:56,330 --> 00:19:58,820 Just like the state to get a vibe because. 257 00:19:58,820 --> 00:20:04,220 So what is it if I so I'll fight is same as the start of E plus I 258 00:20:10,150 --> 00:20:13,040 so that's why you can write this index like this. 259 00:20:13,390 --> 00:20:21,380 So this is for the inside so let's say I want to find out the largest element given and every find the 260 00:20:21,470 --> 00:20:23,270 largest element inside the 80. 261 00:20:23,330 --> 00:20:27,320 So let's solve this question okay. 262 00:20:27,340 --> 00:20:30,590 So let us write the code for finding out the largest element 263 00:20:37,500 --> 00:20:48,960 so n n seen and and start a is new int and now what will happen. 264 00:20:48,980 --> 00:20:51,680 So the value of end will be given by the user at one time. 265 00:20:51,980 --> 00:20:55,250 So this edit is going to be present inside the heap. 266 00:20:55,250 --> 00:21:01,220 We are going to create that inside heap and 8 bytes will be allocated on the stack because we are getting 267 00:21:01,220 --> 00:21:02,040 a digital pointer. 268 00:21:02,510 --> 00:21:04,640 So now let us take input from the user. 269 00:21:04,670 --> 00:21:11,000 So like was 0 I lost and in a plus plus seen a 0 5 270 00:21:15,020 --> 00:21:17,620 and now let us try to find out the maximum element. 271 00:21:17,660 --> 00:21:21,370 So initially let's say the largest element is minus 1. 272 00:21:21,440 --> 00:21:26,960 For now let's assume that the user will give all the positive elements as input so I equals zero. 273 00:21:27,090 --> 00:21:29,640 I listed in a plus plus. 274 00:21:29,690 --> 00:21:31,610 So if the current element. 275 00:21:31,640 --> 00:21:38,000 So if you if I if the current element is good then the largest element then you can update your largest 276 00:21:38,000 --> 00:21:39,820 element. 277 00:21:40,430 --> 00:21:43,040 And finally you can bring your largest element 278 00:21:46,110 --> 00:21:49,890 so let's an area called okay. 279 00:21:49,900 --> 00:21:52,710 So here I forgot to add 7 column 280 00:21:57,950 --> 00:22:03,910 so there are 5 elements and the element solid C 1 2 3 5 and 4. 281 00:22:04,340 --> 00:22:08,550 So 5 is the largest element civil code is working fine. 282 00:22:08,660 --> 00:22:13,220 Now tell me how much memory is going to be allocated if I write something like this. 283 00:22:13,580 --> 00:22:21,330 So if I write and start a is new and 50. 284 00:22:21,520 --> 00:22:24,640 So how much memory will get allocated. 285 00:22:24,640 --> 00:22:25,670 So there are two parts. 286 00:22:25,690 --> 00:22:33,800 First the ad is going to be created inside the heap of 50 sites. 287 00:22:33,930 --> 00:22:37,410 So inside the heap how much memory will be allocated. 288 00:22:37,440 --> 00:22:38,920 So 50 in 2 4. 289 00:22:39,300 --> 00:22:44,750 So two hundred bytes are going to be allocated on heap and on stack. 290 00:22:44,760 --> 00:22:47,150 How much memory will be allocated. 291 00:22:47,160 --> 00:22:49,230 So we are creating a integer pointer. 292 00:22:49,680 --> 00:22:51,480 So this is a. 293 00:22:51,570 --> 00:22:52,870 So even right. 294 00:22:53,010 --> 00:22:59,900 So in totally add this line two hundred eight bytes will be allocated. 295 00:23:00,310 --> 00:23:01,500 Fine. 296 00:23:01,630 --> 00:23:06,220 Now let us see one more difference between the static and dynamic allocation and the most important 297 00:23:06,220 --> 00:23:07,680 difference. 298 00:23:07,720 --> 00:23:17,470 So for example this is used to mean and let's say this is you if and inside. 299 00:23:17,480 --> 00:23:22,520 If you are writing int i and yet if will end here. 300 00:23:22,550 --> 00:23:25,370 So there are some more line of code. 301 00:23:25,550 --> 00:23:26,500 So what will happen. 302 00:23:26,510 --> 00:23:29,720 So as soon as this I point out it will reach this line. 303 00:23:29,750 --> 00:23:30,380 What will happen. 304 00:23:30,380 --> 00:23:33,270 So the memory will be automatically deleted. 305 00:23:33,380 --> 00:23:39,330 So as soon as the scope of the variable i finish its memory will be delegated. 306 00:23:39,350 --> 00:23:41,580 So this is static allocation. 307 00:23:41,780 --> 00:23:43,550 So this is stack. 308 00:23:43,580 --> 00:23:45,170 So this a memory. 309 00:23:45,170 --> 00:23:52,320 So this i.e. variable is going to be present inside the stack and as soon as I will reached this line. 310 00:23:52,510 --> 00:23:54,210 So the memory allocated. 311 00:23:54,310 --> 00:24:00,010 So basically the four bytes allocated to the variability will get deleted automatically. 312 00:24:00,010 --> 00:24:05,080 But this is not the case with the dynamic and location. 313 00:24:05,080 --> 00:24:06,630 So if you will write it like this. 314 00:24:06,670 --> 00:24:15,900 If then if you were creating a dynamic memory if you are using dynamic communication then what will 315 00:24:15,900 --> 00:24:16,200 happen. 316 00:24:16,200 --> 00:24:17,550 So this i.e.. 317 00:24:17,640 --> 00:24:23,920 So this variable ie it doesn't have any scope so no scope is there for a variable. 318 00:24:24,150 --> 00:24:30,000 Because this is heap memory allocation dynamic memory location so it will not get deleted. 319 00:24:30,060 --> 00:24:36,510 So the biggest difference between the static memory allocation and the dynamic memory allocation is 320 00:24:38,190 --> 00:24:42,630 so if you are allocating memory statically for example in die then what will happen. 321 00:24:42,900 --> 00:24:49,440 So they will be automatic release of memory on bases of scope 322 00:24:52,510 --> 00:24:56,560 whereas in the dining memory relegation there is no concept of scope. 323 00:24:56,660 --> 00:25:03,040 So if you are doing something like this in study is new and so what you have to do. 324 00:25:03,070 --> 00:25:05,510 So there is no scope there is no concept of scope. 325 00:25:05,540 --> 00:25:08,540 So you have to manually release the memory. 326 00:25:08,730 --> 00:25:13,210 You have to manually release the memory. 327 00:25:13,480 --> 00:25:15,760 So what will happen if I will do something like this. 328 00:25:15,760 --> 00:25:16,510 For example 329 00:25:20,330 --> 00:25:21,080 vital through 330 00:25:27,590 --> 00:25:43,980 a and similarly what will happen if I will do something like this while to in start I is new and so 331 00:25:43,990 --> 00:25:49,740 what will happen if I will run this code so you entered the VI loop. 332 00:25:49,850 --> 00:25:52,210 So first of all this loop is vital. 333 00:25:52,220 --> 00:25:54,830 So that means this is going to run in finite times. 334 00:25:54,830 --> 00:25:57,650 So this is in finite loop then what will happen. 335 00:25:57,650 --> 00:26:05,540 So as soon as you reach this line 4 bytes will be allocated inside your stack memory then you will go 336 00:26:05,540 --> 00:26:06,190 above. 337 00:26:06,200 --> 00:26:12,180 So as soon as you will go above this 4 bytes idea located and then you will again reach this line. 338 00:26:12,260 --> 00:26:14,470 So again the 4 bytes will be allocated. 339 00:26:14,480 --> 00:26:16,220 Then again you will go up. 340 00:26:16,220 --> 00:26:18,390 So this 4 bytes will be deleted. 341 00:26:18,560 --> 00:26:22,220 Then you will come here again so 4 bytes will be allocated again. 342 00:26:22,220 --> 00:26:23,690 Then you will go up again. 343 00:26:23,720 --> 00:26:25,310 So 4 bytes will be deleted. 344 00:26:25,310 --> 00:26:26,620 So this will go on. 345 00:26:26,660 --> 00:26:33,020 So basically at max so at any point of time at max you will. 346 00:26:33,020 --> 00:26:40,490 You are using food by itself memory space so your program will look perfectly fine no doubt. 347 00:26:40,490 --> 00:26:46,340 Now let us consider this case what will happen if I will run Discworld so enter this. 348 00:26:46,340 --> 00:26:49,620 So as soon as you reach this line what will happen. 349 00:26:49,670 --> 00:26:57,860 So eat bites are allocated inside you and stack memory because you are creating and Deja pointer and 350 00:26:57,920 --> 00:27:00,850 4 bytes are allocated inside your heap. 351 00:27:00,850 --> 00:27:04,360 So inside heap 4 bytes are allocated. 352 00:27:04,580 --> 00:27:06,140 Now you will reach this line. 353 00:27:06,230 --> 00:27:08,620 So this 8 bytes will be dedicated. 354 00:27:08,720 --> 00:27:10,670 Then you will again go up. 355 00:27:10,670 --> 00:27:14,830 You will reach this line so again 8 bytes will be allocated for Stack. 356 00:27:14,930 --> 00:27:17,070 But this for bytes are not deleted. 357 00:27:17,150 --> 00:27:24,870 So again you are writing new int so the 4 bytes will get allocated again then you will reach this line. 358 00:27:24,870 --> 00:27:28,520 So this 8 bytes will be allocated because this is stack. 359 00:27:28,530 --> 00:27:29,370 Memory allocation. 360 00:27:29,370 --> 00:27:30,640 Then again you will go up. 361 00:27:30,750 --> 00:27:36,470 You will reach this line so again 8 bytes will be allocated on stack memory and new and so forth. 362 00:27:36,480 --> 00:27:39,430 Bytes will be allocated on heap again. 363 00:27:39,660 --> 00:27:42,630 So again you will reach this line. 364 00:27:42,630 --> 00:27:44,940 So this 8 bytes will be d allocated. 365 00:27:45,030 --> 00:27:46,440 And again you will go up. 366 00:27:46,440 --> 00:27:52,770 You will reach this line so again 8 bytes will be allocated inside to stack memory and 4 plus 4 here. 367 00:27:52,800 --> 00:27:54,410 Because new integer. 368 00:27:54,630 --> 00:27:55,590 So this will go on. 369 00:27:55,590 --> 00:27:57,090 So ultimately what will happen. 370 00:27:57,240 --> 00:28:06,540 Your memory usage is increasing your memory usage is continuously increasing and at some point of time 371 00:28:06,550 --> 00:28:07,510 what will happen. 372 00:28:07,590 --> 00:28:10,380 Your app your program will crash. 373 00:28:10,530 --> 00:28:14,250 At some point of time our program will crash. 374 00:28:14,250 --> 00:28:16,740 So basically if you will write code like this 375 00:28:21,370 --> 00:28:22,850 so if I will do something like this 376 00:28:29,080 --> 00:28:40,950 and if I really lose something like this. 377 00:28:41,090 --> 00:28:42,970 So what will happen. 378 00:28:43,070 --> 00:28:44,140 So this is your homework. 379 00:28:44,150 --> 00:28:45,010 What you will do. 380 00:28:45,050 --> 00:28:50,440 So first you will run this piece of code and you will command out this piece of code. 381 00:28:50,540 --> 00:28:55,190 So if you will read this piece of code what you will do is you will open your activity monitor or you 382 00:28:55,190 --> 00:28:59,480 will open your task manager and you will notice what is your memory usage. 383 00:28:59,480 --> 00:29:05,650 So in this case your memory usage will be constant because at max we will be using for by of memory 384 00:29:05,650 --> 00:29:07,060 space. 385 00:29:07,160 --> 00:29:13,670 Now command out this code and run this code run this three pieces off line then this three lines what 386 00:29:13,670 --> 00:29:14,850 will happen again. 387 00:29:14,930 --> 00:29:20,000 You will open your task manager or activity monitor and then you will again check your memory usage 388 00:29:20,090 --> 00:29:22,550 so your memory usage will keep increasing. 389 00:29:22,550 --> 00:29:24,940 So your memory usage will keep increasing. 390 00:29:24,950 --> 00:29:31,400 And finally your program is going to be crashed because of lack of memory so you can do this. 391 00:29:31,400 --> 00:29:33,590 So for now I am commending our discord 392 00:29:38,990 --> 00:29:43,490 so what is happening here why it is happening because we discussed about that. 393 00:29:43,550 --> 00:29:47,570 The static memory will be deleted automatically. 394 00:29:47,570 --> 00:29:52,520 But if you want or delete the heap memory you have to manually release the memory if you want to delete 395 00:29:52,550 --> 00:29:53,540 the heap memory. 396 00:29:53,630 --> 00:29:56,650 You have to manually believed that memory. 397 00:29:56,690 --> 00:29:59,370 Now how we can manually delete the memory. 398 00:29:59,570 --> 00:30:00,980 So just like a new keyword. 399 00:30:00,980 --> 00:30:02,660 We also have a keyword delete. 400 00:30:03,530 --> 00:30:04,970 So what they are doing here. 401 00:30:06,980 --> 00:30:13,560 And start B is new and so what will happen. 402 00:30:13,770 --> 00:30:20,880 This is you your heat memory D idea for bytes 403 00:30:23,660 --> 00:30:24,920 fine. 404 00:30:25,010 --> 00:30:36,820 This is U.S. tech memory these are 8 bytes so p and eight bytes so what is B is pointing towards this 405 00:30:36,820 --> 00:30:43,180 memory location and I want to delete this memory because this memory we have to lose this memory and 406 00:30:43,180 --> 00:30:48,040 this memory will get deleted automatically as soon as it will reach out of scope. 407 00:30:48,040 --> 00:30:53,420 So for deleting the memory we have a keyword delete and we have to give the address. 408 00:30:53,530 --> 00:30:59,770 So let's say that this is hundred so various address present addresses presented so you will write to 409 00:30:59,770 --> 00:31:03,560 delete b you will write delete B. 410 00:31:03,640 --> 00:31:07,140 Now after this line get executed what will happen. 411 00:31:07,210 --> 00:31:09,930 So delete B means go to this address. 412 00:31:10,060 --> 00:31:15,310 So go to this address and delete the memory so delete the memory. 413 00:31:15,460 --> 00:31:19,420 Now the memory has been deleted so this be pointer. 414 00:31:19,420 --> 00:31:21,440 So this memory will not get deleted. 415 00:31:21,490 --> 00:31:25,920 Just remember many people have this confusion that this memory will also get deleted. 416 00:31:25,930 --> 00:31:32,140 No this memory will be deleted automatically then it will reach out of scope. 417 00:31:32,140 --> 00:31:36,600 So basically we are deleting this memory simple. 418 00:31:36,610 --> 00:31:39,330 Now one thing after deleting the memory. 419 00:31:39,460 --> 00:31:41,580 Please do not try to system and again. 420 00:31:41,590 --> 00:31:43,380 So do not try to do this. 421 00:31:43,410 --> 00:31:44,010 Why. 422 00:31:44,080 --> 00:31:45,610 Because if you will try to do this. 423 00:31:45,610 --> 00:31:46,570 What they are doing. 424 00:31:46,570 --> 00:31:48,020 So this memory has been deleted. 425 00:31:48,280 --> 00:31:54,070 So you are going to this address and basically this memory is now not belonging to us. 426 00:31:54,100 --> 00:31:59,310 This memory does not belongs to us and you are trying to put 10 inside this memory. 427 00:31:59,440 --> 00:32:01,290 So this is very dangerous. 428 00:32:01,300 --> 00:32:03,670 So do not try to do this. 429 00:32:03,730 --> 00:32:08,890 So do not try to do this because this is very dangerous because after deleting the memory the memory 430 00:32:08,920 --> 00:32:10,000 does not belongs to us. 431 00:32:10,390 --> 00:32:15,540 So do not try to do something like this now one more thing. 432 00:32:15,540 --> 00:32:17,120 So we have deleted this memory. 433 00:32:17,130 --> 00:32:21,100 We can also do like this after deleting the memory after deleting the memory. 434 00:32:21,120 --> 00:32:28,260 We will never do this and we can also allocate new memory so I can write new and also I can write new 435 00:32:28,260 --> 00:32:28,700 it again. 436 00:32:28,710 --> 00:32:29,430 So what will happen. 437 00:32:29,430 --> 00:32:33,330 I will get an idea for byte and now this people point here. 438 00:32:33,360 --> 00:32:41,080 Then again I can do delete B so this memory will get deleted and then again what can I do. 439 00:32:41,100 --> 00:32:44,050 I can write. 440 00:32:44,280 --> 00:32:47,000 So this is only b not start B. 441 00:32:47,040 --> 00:32:51,430 So again I can write new intended. 442 00:32:52,030 --> 00:32:53,060 So what will happen. 443 00:32:53,140 --> 00:32:55,280 So I will get an eddy. 444 00:32:55,540 --> 00:32:56,400 Now this time. 445 00:32:56,470 --> 00:32:57,780 Pointing towards this eddy. 446 00:32:58,240 --> 00:33:03,450 So how do we dilute entity if you want to delete one element. 447 00:33:03,470 --> 00:33:06,790 This is the syntax but if you want to delete entity you have to give. 448 00:33:06,890 --> 00:33:08,690 You have to write it like this. 449 00:33:08,690 --> 00:33:11,610 So this is the syntax for deleting any. 450 00:33:11,780 --> 00:33:14,970 So this is for deleting Eddie. 451 00:33:15,020 --> 00:33:17,960 Okay so this is for single element deletion. 452 00:33:17,960 --> 00:33:24,080 So what will happen go to delete B and the meaning of this is actually so BS an idea or destroy the 453 00:33:24,080 --> 00:33:24,830 whole Eddy. 454 00:33:24,860 --> 00:33:29,270 So go to the first memory location and destroy the complete Eddy. 455 00:33:29,360 --> 00:33:30,350 So let's see in. 456 00:33:31,730 --> 00:33:33,410 So again it is very simple. 457 00:33:33,460 --> 00:33:34,040 You can do 458 00:33:36,930 --> 00:33:45,780 and start B is new int now I want to delete the memory and so delete B now after deleting the b you 459 00:33:45,780 --> 00:33:51,630 can again allocate the memory so B and again delete the memories or delete B. 460 00:33:51,690 --> 00:33:56,090 Now you can again allocate the memory so and let's say 50. 461 00:33:56,580 --> 00:34:01,730 So to delete identity you have to write this index this one you have to give this a squared records 462 00:34:01,740 --> 00:34:03,760 and delete B. 463 00:34:03,870 --> 00:34:11,170 So this is this index for deleting 80 so delete at is index 464 00:34:18,120 --> 00:34:23,830 and this is just index for deleting one single element so delete single. 465 00:34:27,440 --> 00:34:32,680 Element so this is all for this video I will see you in the next one.