1 00:00:02,550 --> 00:00:03,300 Hi, everyone. 2 00:00:03,390 --> 00:00:04,620 Welcome back to the cause. 3 00:00:05,100 --> 00:00:08,400 So today we will learn a new concept called appointers. 4 00:00:09,810 --> 00:00:11,880 So Boynton's is a very interesting topic. 5 00:00:12,630 --> 00:00:13,680 So let's start. 6 00:00:15,810 --> 00:00:17,230 And I question. 7 00:00:17,620 --> 00:00:18,480 So what will happen? 8 00:00:18,540 --> 00:00:20,330 I will get food by itself. 9 00:00:20,330 --> 00:00:21,150 Memory Speace. 10 00:00:24,690 --> 00:00:27,480 Dan is stored here now in the next line. 11 00:00:27,540 --> 00:00:33,480 If I try and print the value of I so want to live my output, my output will be 10. 12 00:00:35,320 --> 00:00:38,870 But how a system knows Veis, actually, I stood. 13 00:00:39,900 --> 00:00:40,590 Any guesses? 14 00:00:41,550 --> 00:00:44,190 So we have something called Simbel Table. 15 00:00:47,240 --> 00:00:52,050 The symbol table is generated by a compiler at the completion time. 16 00:00:53,260 --> 00:00:55,400 So symbol table is actually a map. 17 00:00:56,630 --> 00:00:57,410 It will store. 18 00:00:57,540 --> 00:01:02,680 There is a variable fly and the variable eye has got a memory base of 700. 19 00:01:03,320 --> 00:01:06,020 So this has a memory address of seven ended. 20 00:01:06,590 --> 00:01:09,560 So when I am printing the value of I, what will happen? 21 00:01:09,860 --> 00:01:13,030 The compiler will search I in the symbol table. 22 00:01:13,460 --> 00:01:14,810 Yes, there is a variable. 23 00:01:14,850 --> 00:01:19,280 I and the variable I is present at memory address seven ended. 24 00:01:19,670 --> 00:01:25,910 So I will go to I mean sys system will go to seven hundred and it will print ten. 25 00:01:27,700 --> 00:01:27,980 OK. 26 00:01:28,520 --> 00:01:35,580 So as soon as you create a variable, for example, Entergy, the variable G has to make an entry into 27 00:01:35,640 --> 00:01:36,440 Simbel Table. 28 00:01:36,500 --> 00:01:38,880 Suppose the addresses 794. 29 00:01:40,220 --> 00:01:45,950 Similarly, if you create an either variable galaxy, so she will make an entry in decimal table, suppose 30 00:01:45,950 --> 00:01:47,600 that does is it ended? 31 00:01:48,440 --> 00:01:48,680 OK. 32 00:01:48,920 --> 00:01:54,170 So as soon as you create a variable, that variable has to make an entry in the symbol table. 33 00:01:55,470 --> 00:01:58,010 Now, symbol table stored many information. 34 00:01:58,340 --> 00:02:03,800 For example, it will store the name of the variable and the memory address apart from it. 35 00:02:03,920 --> 00:02:04,900 It will also store. 36 00:02:05,300 --> 00:02:06,170 What is the type. 37 00:02:06,950 --> 00:02:15,200 Suppose I of integer and it will also store like what is the scope of I and many other informations. 38 00:02:16,010 --> 00:02:16,280 OK. 39 00:02:17,210 --> 00:02:21,410 So now what I want is I want to print this address. 40 00:02:21,500 --> 00:02:22,310 Seven hundred. 41 00:02:23,060 --> 00:02:25,520 I want to know veis actually I stored. 42 00:02:25,940 --> 00:02:27,730 I want to print its address. 43 00:02:28,580 --> 00:02:31,550 So we have something called address of Operator. 44 00:02:32,150 --> 00:02:33,700 So address off I. 45 00:02:34,850 --> 00:02:40,550 So if we will write C out address ofay I will get this address. 46 00:02:40,670 --> 00:02:41,420 Seven entered. 47 00:02:42,320 --> 00:02:42,650 OK. 48 00:02:43,010 --> 00:02:44,240 So let's see. 49 00:02:47,840 --> 00:02:51,440 So I have a myriad of file pointers in product CPB. 50 00:02:52,920 --> 00:02:54,060 Now I have a variable. 51 00:02:54,270 --> 00:02:55,940 And day it goes down. 52 00:02:59,070 --> 00:03:01,210 Now let us spend the I. 53 00:03:01,920 --> 00:03:03,560 So she yelled at myself. 54 00:03:03,960 --> 00:03:04,170 I. 55 00:03:12,780 --> 00:03:18,820 So this is the memo that herself off I so if you remember, I have already told you that the this is 56 00:03:18,840 --> 00:03:20,580 stored in hexadecimal Lemba system. 57 00:03:20,910 --> 00:03:22,500 So this is in hexadecimal. 58 00:03:22,590 --> 00:03:23,310 No system. 59 00:03:27,290 --> 00:03:32,550 Now, what I more know is so now I want to store address off I. 60 00:03:34,100 --> 00:03:37,150 So we need a data type that was stored at this off. 61 00:03:37,490 --> 00:03:37,730 I. 62 00:03:39,110 --> 00:03:43,310 So it gad flawed. 63 00:03:46,850 --> 00:03:50,270 Any of these three gone to the address of a very. 64 00:03:51,750 --> 00:03:53,820 So we have something called pointers. 65 00:03:54,540 --> 00:04:02,460 So at this point, the point is actually a variable that will store the address of other variables. 66 00:04:02,790 --> 00:04:05,490 So a variable that will store. 67 00:04:07,430 --> 00:04:09,680 Address of the variables. 68 00:04:12,860 --> 00:04:14,920 This is the standard definition of a point. 69 00:04:15,810 --> 00:04:18,500 OK, so Blinda. 70 00:04:20,310 --> 00:04:21,590 B equals. 71 00:04:21,900 --> 00:04:23,190 And percent of I. 72 00:04:23,760 --> 00:04:28,100 So point B, B is A pointer that will store the address of I. 73 00:04:28,860 --> 00:04:29,160 OK. 74 00:04:29,610 --> 00:04:31,230 So this is actually wrong. 75 00:04:32,160 --> 00:04:33,420 We can't write like this. 76 00:04:33,450 --> 00:04:36,570 I would tell you in very short time why this is wrong. 77 00:04:37,620 --> 00:04:44,640 So suppose B's containing 700, so P is containing 700. 78 00:04:44,790 --> 00:04:46,050 So this is my variability. 79 00:04:47,680 --> 00:04:49,350 This is variable rarely addresses. 80 00:04:49,470 --> 00:04:51,320 So it ended and value is ten. 81 00:04:52,020 --> 00:04:53,600 So be is storing 700. 82 00:04:53,700 --> 00:04:56,460 That means I can access 10 from here. 83 00:04:58,200 --> 00:05:05,470 OK, so what I want know is go to memory or does it say 100 and tell me what is it there? 84 00:05:07,020 --> 00:05:07,300 OK. 85 00:05:07,770 --> 00:05:11,250 So what I want now is go to memory or does seven. 86 00:05:11,820 --> 00:05:14,040 And tell me what is written there. 87 00:05:15,090 --> 00:05:19,470 So can you tell me the answer is you can't my. 88 00:05:21,310 --> 00:05:24,400 Because you don't know how many bites to read. 89 00:05:26,940 --> 00:05:35,280 This is so unhindered, so a when so under two and seven under three. 90 00:05:35,700 --> 00:05:38,280 So one bite, one bite, one bite and one bite. 91 00:05:39,330 --> 00:05:45,040 And suppose the number is so this is my number and memory. 92 00:05:45,280 --> 00:05:46,010 So it ended. 93 00:05:46,590 --> 00:05:50,450 So I want to go to memory of this 700. 94 00:05:50,580 --> 00:05:52,290 And tell me what is written there. 95 00:05:52,920 --> 00:05:55,890 And I am telling you that you can't answer this question. 96 00:05:56,200 --> 00:05:56,400 Why? 97 00:05:56,880 --> 00:05:59,310 Because you don't know how many bites to read. 98 00:06:01,130 --> 00:06:05,870 So suppose I told you somehow you got to know that we have to read for Bite's. 99 00:06:06,990 --> 00:06:07,220 OK. 100 00:06:07,570 --> 00:06:14,710 So as you somehow we got to know that we have to do it for right now, can you tell me how to intubate 101 00:06:14,710 --> 00:06:15,610 those four bites? 102 00:06:17,060 --> 00:06:20,480 What I want to say is, suppose I give you four bites. 103 00:06:21,520 --> 00:06:23,470 Now, tell me the meaning of this photo by its. 104 00:06:24,950 --> 00:06:26,760 You can tell why. 105 00:06:27,500 --> 00:06:33,050 Because these four bites might be representing four different characters. 106 00:06:33,860 --> 00:06:36,460 These four bites might be one fluid. 107 00:06:37,760 --> 00:06:44,240 These four bites might be representing half size of a double half size of a double. 108 00:06:44,280 --> 00:06:49,400 Basically, I give you first four bites, first four bites of eight by. 109 00:06:51,350 --> 00:06:53,300 So these four bites can be anything. 110 00:06:57,050 --> 00:07:03,380 So we are facing two difficulties here with the help of this representation, with the help of point 111 00:07:03,380 --> 00:07:04,100 thirty to type. 112 00:07:06,020 --> 00:07:10,610 And there are two difficulties are first, I don't know how many bites to read. 113 00:07:12,320 --> 00:07:18,850 Second, if I know that I have to root for bites, then I don't know how to interpret those four bites. 114 00:07:19,730 --> 00:07:23,540 Those four bites might be four characters when fluid half size over the bowl. 115 00:07:23,600 --> 00:07:24,530 Basically anything. 116 00:07:25,850 --> 00:07:28,970 So there are two difficulties in this representation. 117 00:07:29,300 --> 00:07:31,610 Hence, this representation is wrong. 118 00:07:33,160 --> 00:07:33,480 OK. 119 00:07:35,530 --> 00:07:40,660 So when we write for into B equals M percent Towfigh. 120 00:07:41,290 --> 00:07:49,660 So once I have the address of I in point B in the next lane, there is no way for me to see what is 121 00:07:49,720 --> 00:07:52,920 exactly stored at memory at this seven hundred. 122 00:07:53,440 --> 00:07:56,140 That information is just not available to me. 123 00:07:57,360 --> 00:08:00,920 OK, so how do we create pointers? 124 00:08:02,090 --> 00:08:03,860 So we create pointers like this. 125 00:08:08,220 --> 00:08:09,540 Now, how do read this line? 126 00:08:11,120 --> 00:08:15,140 B is a pointer pointing to an integer. 127 00:08:16,160 --> 00:08:19,580 So we will read from right to left. 128 00:08:21,250 --> 00:08:21,970 One more time. 129 00:08:22,900 --> 00:08:26,880 B is a pointer pointing to an integer. 130 00:08:27,610 --> 00:08:29,470 Now this all votes the problem. 131 00:08:30,370 --> 00:08:33,790 So the first problem was, I don't know how many bytes to read. 132 00:08:34,540 --> 00:08:36,250 So how many bytes? 133 00:08:38,260 --> 00:08:38,970 To read. 134 00:08:40,310 --> 00:08:47,830 Now I know how many bikes read because I have it here, so I know that I will go to memory at this seven 135 00:08:47,840 --> 00:08:50,330 ended and I will read for Bite's. 136 00:08:51,810 --> 00:08:56,910 Seven zero three and seven zero one and seven zero two. 137 00:08:57,720 --> 00:09:02,870 So the first problem is solved because of this in Beja. 138 00:09:03,090 --> 00:09:08,490 So I wouldn't go to memory this I attended and I will read forward by it because in BGA is up for debate. 139 00:09:09,570 --> 00:09:16,590 Now, the second problem was how to interpret those bytes, how to interpret. 140 00:09:18,840 --> 00:09:23,780 And this problem is also soiled because now I know that it is ending, T.J.. 141 00:09:25,230 --> 00:09:29,290 So I will interpret this for bites as an integer. 142 00:09:30,960 --> 00:09:34,650 Now, both the problems are solved with this representation. 143 00:09:35,220 --> 00:09:37,390 So this representation is much, much better. 144 00:09:37,770 --> 00:09:42,300 And we will represent our pointers and we will make pointers using this approach. 145 00:09:42,870 --> 00:09:46,560 So the above approach, this approach is never used. 146 00:09:48,450 --> 00:09:50,980 OK, so let's see. 147 00:09:56,480 --> 00:10:02,990 So I have a point B, which is pointing to anybody here at least, or at this off I. 148 00:10:04,260 --> 00:10:06,890 Now let us try to print the value of P. 149 00:10:10,310 --> 00:10:15,130 So virtually my output, basically, Addazio, Fi and B bothof seem. 150 00:10:20,040 --> 00:10:21,030 So you can see here. 151 00:10:22,090 --> 00:10:27,720 Output of board and personal air doesn't fly and be both the same. 152 00:10:28,450 --> 00:10:30,220 OK, so memory addresses seem. 153 00:10:32,800 --> 00:10:34,960 Now, let us learn something more interesting. 154 00:10:41,430 --> 00:10:48,600 So let us talk about size of appointer, basically the size of appointer is fixed. 155 00:10:49,890 --> 00:10:55,470 So if you have a pointer which is containing that dustoff, an integer. 156 00:10:55,650 --> 00:11:02,490 So and I and you also have a pointer which will store that does have a character. 157 00:11:02,970 --> 00:11:07,080 So Canstar, let's say P2 is storing that dustoff. 158 00:11:07,200 --> 00:11:22,770 See, now, if I will bend the size of B and if I will try and print the size of B2, both output will 159 00:11:22,770 --> 00:11:24,180 be same size of both. 160 00:11:24,180 --> 00:11:31,010 The point does not seem if you will print the size of I. 161 00:11:31,620 --> 00:11:34,170 I will get forward by it because integer is of four. 162 00:11:34,170 --> 00:11:34,530 Right. 163 00:11:35,100 --> 00:11:39,420 And if you will print the size of C, you will get one byte. 164 00:11:40,110 --> 00:11:43,820 Martin, case of point does size will be seen. 165 00:11:44,460 --> 00:11:44,970 Let's see. 166 00:11:55,690 --> 00:12:00,720 Now, let us create a character pointer. 167 00:12:00,850 --> 00:12:02,310 So Gastar, let's see. 168 00:12:02,680 --> 00:12:09,460 BE2 is distorting the address of variable C. 169 00:12:21,120 --> 00:12:25,590 So the output of this line and the output of this line will be same. 170 00:12:26,370 --> 00:12:26,880 Let's see. 171 00:12:31,610 --> 00:12:33,800 So basically, four bites and four bites. 172 00:12:35,800 --> 00:12:37,600 So let me give and line here. 173 00:12:50,120 --> 00:12:52,770 So the size of I integer pointer is forward. 174 00:12:53,070 --> 00:12:55,800 And size of a connector pointer is also forward. 175 00:12:57,600 --> 00:13:03,300 Even though the size of India was four bites and the size of a character is one bite, but the size 176 00:13:03,300 --> 00:13:04,710 of the point that is seem. 177 00:13:06,590 --> 00:13:08,630 Now, let us try to understand why this happened. 178 00:13:11,310 --> 00:13:15,450 Suppose days, a 200 square yard house. 179 00:13:18,080 --> 00:13:22,640 So this is a small house, let's call it get. 180 00:13:24,170 --> 00:13:27,920 And we have 800 screamy art house. 181 00:13:30,790 --> 00:13:32,070 Let Scollard and. 182 00:13:35,830 --> 00:13:37,810 Now, let us talk about their addresses. 183 00:13:39,940 --> 00:13:47,950 So if this house is big and this houses small, that doesn't mean that this house address will be very 184 00:13:47,950 --> 00:13:51,940 long and this house address will be very small. 185 00:13:53,390 --> 00:13:54,960 The address is of both. 186 00:13:55,470 --> 00:13:57,280 The houses will be very similar. 187 00:13:58,200 --> 00:13:58,460 OK. 188 00:13:59,680 --> 00:14:10,390 So basically when we arrived and Star B equals and by Santa Fe, I just have to store the starting address. 189 00:14:11,130 --> 00:14:12,150 So be restored. 190 00:14:12,370 --> 00:14:12,940 700. 191 00:14:14,950 --> 00:14:21,790 Similarly, if you have a pointer, Gastar C equals M percent of C, so C will store just this starting 192 00:14:21,790 --> 00:14:22,840 at this later there. 193 00:14:22,900 --> 00:14:23,780 This is seven nine. 194 00:14:23,850 --> 00:14:27,640 Before we just have to store the starting address. 195 00:14:29,130 --> 00:14:35,940 That's why the size of the point that is fixed and can and this simple analogy will help you to understand 196 00:14:35,940 --> 00:14:36,450 it better. 197 00:14:37,860 --> 00:14:42,960 If a house is very big, that doesn't mean it's a dress will also be very big there. 198 00:14:43,280 --> 00:14:44,490 This will be compatible. 199 00:14:45,000 --> 00:14:48,390 That's why the size of the pointer is fixed in my system. 200 00:14:48,420 --> 00:14:49,320 It is four bytes. 201 00:14:49,980 --> 00:14:53,730 So this vary from compiler to compiler and from system to system. 202 00:14:54,880 --> 00:14:56,380 So this is it for this video. 203 00:14:56,440 --> 00:14:58,300 If you have any doubt, feel free to ask. 204 00:14:58,510 --> 00:14:59,770 I will see you the next one. 205 00:14:59,980 --> 00:15:00,510 Thank you.