WEBVTT 00:00.560 --> 00:08.300 In the realm of computer systems, memory serves as a crucial component that acts as a storage repository 00:08.300 --> 00:11.060 for both data and instructions. 00:11.480 --> 00:18.860 As we have previously previously explored, there exists specialized storage elements called registers 00:18.860 --> 00:23.420 that enable high speed access to critical information. 00:23.420 --> 00:30.470 So while registers provide rapid data retrieval, it's important to acknowledge that memory access operates 00:30.470 --> 00:33.290 at a comparably slower pace. 00:33.290 --> 00:41.540 So this discrepancy in access speed necessitates a careful balance between the usage of registers and 00:41.540 --> 00:45.590 memory to optimize overall system performance. 00:45.590 --> 00:53.300 So registers represent the pinnacle of the memory hierarchy due to the their immediate accessibility 00:53.300 --> 00:54.560 by the processor. 00:54.560 --> 01:02.460 So they facilitate rapid execution of instructions by providing quick access to frequently used data. 01:02.490 --> 01:10.480 However, the abundance of registers is inherently limited due to the physical and design considerations. 01:10.500 --> 01:20.010 This limitation prompts the integration of various memory levels, each with distinct access speeds, 01:20.160 --> 01:21.720 capacities and costs. 01:21.720 --> 01:32.970 So diving deeper, we we encounter the theoretical limit of memory size, which boasts around 264 addresses. 01:33.450 --> 01:35.820 Let's actually open the grommet. 01:37.280 --> 01:40.100 And as always, throw things on the screen. 01:40.520 --> 01:41.540 So. 01:43.670 --> 01:45.710 Uh, 264 addresses. 01:45.710 --> 01:54.440 So to put this into perspective, this limit translates at staggering, like 16 exabytes. 01:54.950 --> 01:56.660 Exabytes. 01:57.820 --> 01:59.030 Which is a lot. 01:59.050 --> 02:01.280 It's 6000 exabytes of memory. 02:01.300 --> 02:09.520 While this figure appears remarkably large, it is vital to note that practical implementation constraints 02:09.520 --> 02:14.980 prevent the utilization of such a colossal memory capacity. 02:14.980 --> 02:22.060 And while the theoretical memory capacity seems virtually boundless or the real world, design considerations 02:22.060 --> 02:27.340 inevitably impose limitations on its effective utilization. 02:27.340 --> 02:33.230 So various factors contribute to these constraints. 02:33.250 --> 02:39.280 Physical space requirements, manufacturing capabilities, power consumption and cost considerations 02:39.280 --> 02:46.300 collectively influence the decision making process when determining memory sizes for a given system. 02:46.330 --> 02:53.200 Thus, the pursuit of an ideal balance between memory capacity and practicality becomes an intricate 02:53.200 --> 02:54.110 endeavor. 02:54.130 --> 02:58.810 So now what we're going to do is we will open the our. 02:59.850 --> 03:02.160 Chrome and see. 03:13.260 --> 03:17.640 What a 16 exabytes equals here. 03:23.290 --> 03:24.190 16. 03:26.570 --> 03:28.130 Bites here. 03:53.230 --> 04:00.270 Let's actually turn the exabytes to gigabytes and 116 exabytes to gigabytes. 04:00.280 --> 04:04.600 And as you can see here, this is something like this gigabytes here. 04:05.840 --> 04:13.490 And realizing the memory subsystem that harnesses the full potential of the theoretical memory limit 04:13.490 --> 04:15.080 is a formidable challenge. 04:15.080 --> 04:23.270 So as the memory capacities increase, new architectural innovations must emerge to address the associated 04:23.270 --> 04:24.260 complexities. 04:24.260 --> 04:31.760 So these challenges encompass efficient memory, addressing data integrity, error correction mechanisms 04:31.760 --> 04:36.620 and ensuring uniform access speeds across the memory hierarchy. 04:36.650 --> 04:42.500 Overcoming these obstacles is vital to harnessing memories to potential. 04:42.530 --> 04:50.750 So in conclusion, memory emerged as a cornerstone of modern computing systems functioning as a repository 04:50.750 --> 04:56.040 for data and instructions vital to processor operations. 04:56.060 --> 05:04.070 And while registers offer expedited access, the memory hierarchy accommodates the need for scalable 05:04.070 --> 05:05.610 storage solutions. 05:05.610 --> 05:13.440 Memory access governed by the addressing mechanisms navigates through different memory types, each 05:13.440 --> 05:16.830 tailored to specific access requirements. 05:16.830 --> 05:30.060 So the theoretical memory limit presents an A1 sparring figure of 604 sorry for this 264 addresses showcasing 05:30.060 --> 05:33.290 the man's potential for memory capacity. 05:33.300 --> 05:38.400 However, practical design factors such as physical limitations, manufacturing capabilities and economic 05:38.400 --> 05:44.180 considerations steer memory capacities towards more manageable levels. 05:44.190 --> 05:52.200 As we push the boundaries of computing, the pursuit of optimal memory architecture becomes a dynamic 05:52.200 --> 06:00.000 process requiring innovative solutions to address challenges associated with expanding memory capacities. 06:00.030 --> 06:06.830 By understanding the intricate interplay of memory, technical intricacies and practical constraints, 06:06.830 --> 06:13.580 we pave the way for enhanced computing experiences that leverage the true power of memory. 06:13.700 --> 06:23.300 Now we will develop an application again here 64 bit and an Nasm application here. 06:23.300 --> 06:24.530 Let's create a new project. 06:24.530 --> 06:27.980 Delete this this We will name this project as. 06:29.620 --> 06:37.240 Uh, memory or actually, let's make me name it our the 16 exabyte. 06:37.270 --> 06:39.430 Let's actually name it x a byte here. 06:39.460 --> 06:45.990 X a byte so we can remember it takes a byte more how much it, how big is it? 06:46.000 --> 06:51.610 So we will create a section data section data here. 06:51.610 --> 06:53.500 So we will define numbers. 06:53.500 --> 06:57.130 We will define byte size variable, be num with the value. 06:57.160 --> 06:58.390 One, two, three. 06:58.420 --> 07:05.730 We num db one, two, three and we will define a word size variable w num with value. 07:05.740 --> 07:07.960 One, two, three, four five. 07:07.990 --> 07:14.200 We num w one, two, three, two, one, two, three, four, five. 07:14.350 --> 07:23.620 And we will define an array of five words and initialized to zero w array. 07:24.600 --> 07:28.830 Times and 5W0. 07:29.900 --> 07:31.370 And dinam. 07:32.900 --> 07:36.050 ID one, two, three, four, five. 07:36.230 --> 07:38.300 Here we define the. 07:39.880 --> 07:42.910 Let's actually make it an uppercase. 07:42.910 --> 07:48.670 So the D this is a double word size variable name with value. 07:48.700 --> 07:51.130 One, two, three, four, five. 07:51.160 --> 07:56.110 We will define the num one to num one. 07:56.380 --> 07:58.960 The Q here as well. 07:58.960 --> 08:00.730 One, two, three, four, five. 08:00.760 --> 08:01.510 We are here. 08:01.510 --> 08:06.460 We are defining the quadword size variable q num one with value. 08:06.460 --> 08:12.310 One, two, three, four, five and text one db. 08:13.680 --> 08:14.110 A, B. 08:14.130 --> 08:15.540 C here. 08:15.570 --> 08:16.260 Zero. 08:16.470 --> 08:22.110 Here we are defining a null terminated string variable text one with value. 08:22.140 --> 08:22.950 A, b, c. 08:26.370 --> 08:29.340 And the num to. 08:30.110 --> 08:31.370 The queue here. 08:31.400 --> 08:37.190 Here we will define a quadword size variable name with the value of P. 08:37.640 --> 08:44.180 It's actually 3.14 and 1592. 08:44.900 --> 08:45.470 60 for. 08:46.540 --> 08:46.960 Uh, no. 08:46.960 --> 08:47.290 Yeah. 08:47.980 --> 08:52.210 Uh, 92, 65 four. 08:52.360 --> 08:57.880 And after that, we will define a null terminated string variable text to with the value. 08:57.910 --> 09:01.450 See the E here. 09:03.870 --> 09:05.280 At the null termination as well. 09:05.280 --> 09:06.240 And that's it. 09:06.540 --> 09:08.290 We will go to section B here. 09:08.310 --> 09:09.450 Section. 09:10.620 --> 09:11.400 Action. 09:12.480 --> 09:16.140 And now what we're going to do is we will define four variables. 09:16.680 --> 09:18.600 We will reserve four variables here. 09:19.110 --> 09:20.130 We defined already. 09:20.160 --> 09:22.200 We are defining variables in data section. 09:22.470 --> 09:27.210 So now we will reserve one byte of space for variable. 09:27.240 --> 09:33.030 Be var in this section, be var rest be. 09:36.580 --> 09:37.240 Recipe. 09:38.620 --> 09:39.550 And one. 09:40.330 --> 09:43.450 And we will also define not defined. 09:43.480 --> 09:47.110 We will reserve the four bytes of space. 09:51.770 --> 10:00.710 We will reserve four bytes of space variable for base for variable D, var in the base section. 10:02.620 --> 10:03.820 And W. 10:05.870 --> 10:08.900 We will reserve space for an array of. 10:10.070 --> 10:11.540 Uh, ten words. 10:12.700 --> 10:17.380 Named W4 in the S section. 10:18.370 --> 10:18.700 As well. 10:18.700 --> 10:22.020 We will define a reserve keyword. 10:22.450 --> 10:23.440 A rest. 10:24.200 --> 10:25.790 Q three. 10:25.940 --> 10:33.800 Here we are reserving space for an array of three quad words named Q R in the spaces section as well. 10:33.800 --> 10:38.400 And this is the last reserve that we defined in this section. 10:38.420 --> 10:42.440 Now we will continue to the section text. 10:44.090 --> 10:47.240 And here we will define the. 10:48.310 --> 10:48.820 Global. 10:48.850 --> 10:49.450 Maine. 10:49.480 --> 10:50.530 Global. 10:50.920 --> 10:51.700 Maine. 10:54.600 --> 10:55.140 Main. 10:55.780 --> 10:57.580 And in this main here. 10:59.450 --> 11:00.350 RPM. 11:00.740 --> 11:04.940 We are pushing the value of P onto the stack. 11:04.970 --> 11:14.830 We will move the value of RSP into P, which is also setting up the stack frame. 11:15.200 --> 11:15.440 P. 11:17.500 --> 11:27.850 RWP rsp and we will load the memory of address this binom into rax. 11:28.900 --> 11:29.440 Leah. 11:31.040 --> 11:31.720 Racks. 11:34.890 --> 11:36.770 To be numb. 11:38.690 --> 11:42.080 And we will move the racks. 11:43.190 --> 11:46.970 So we will load the value stored at Benham. 11:47.900 --> 11:48.740 Interacts. 11:48.740 --> 11:58.880 But here we load the memory address of p'num interacts and now we are loading the value stored at p'num 11:58.880 --> 12:01.220 into Rax here. 12:05.700 --> 12:06.180 Venom. 12:08.510 --> 12:15.080 And here we will again load the value stored at memory, address stored and Bynum interacts. 12:15.080 --> 12:16.460 I will explain this further. 12:17.420 --> 12:19.730 This is new to us. 12:25.250 --> 12:26.630 So move. 12:27.860 --> 12:28.710 Bois. 12:30.600 --> 12:31.380 PVR. 12:32.820 --> 12:33.660 Racks. 12:36.550 --> 12:37.870 No, actually, yeah, we. 12:38.840 --> 12:40.280 We need to move the. 12:41.670 --> 12:43.860 Move the racks to. 12:47.390 --> 12:48.530 Leah move. 12:48.680 --> 12:49.190 And now. 12:49.340 --> 12:50.810 Yes, this is true. 12:52.220 --> 12:53.580 So what we did here, we. 12:53.640 --> 13:01.400 We loaded the value stored at memory address, which is stored in Benham into racks here again. 13:01.430 --> 13:06.080 Now, what we're going to do is we will move the PVR. 13:09.620 --> 13:09.970 We were. 13:09.980 --> 13:14.990 As you can see, we are using this section which we reserved in previous year in this section. 13:15.470 --> 13:16.060 Section. 13:16.070 --> 13:23.740 Now we will store the value in rax at the memory address of Bois. 13:25.070 --> 13:27.380 Rags and. 13:28.720 --> 13:30.550 We completed three move instruction here. 13:30.550 --> 13:35.670 Now we will use the Lia here to rax. 13:36.190 --> 13:41.950 We are loading the memory address of bvar into rax here. 13:43.980 --> 13:44.730 Racks. 13:45.660 --> 13:46.530 PVR. 13:46.560 --> 13:52.310 But remember, here we store the value in racks at the memory address before. 13:52.350 --> 13:58.740 Actually, if we write this command, if I write the comments here to this code, you will understand 13:58.740 --> 13:59.700 it better. 14:00.810 --> 14:03.960 Posh, the less actually it will be quicker. 14:03.990 --> 14:10.380 Don't worry posh the value of RVP onto the stack and. 14:13.250 --> 14:14.900 And here we are. 14:15.340 --> 14:16.430 Uh, let's see. 14:17.000 --> 14:19.160 We are moving the move. 14:19.630 --> 14:23.180 Move the value of RSP. 14:23.990 --> 14:26.280 Let's make it uppercase, RVP. 14:26.600 --> 14:32.000 So move the value of RSP into RVP, which is. 14:32.000 --> 14:34.790 This is the setting up the stack frame. 14:35.270 --> 14:37.160 And here. 14:38.300 --> 14:40.640 We are not. 14:41.810 --> 14:47.990 The memory address of numb into. 14:48.870 --> 14:49.710 Racks. 14:52.260 --> 15:00.890 And here we are telling the assembly to load the load, the value stored. 15:02.280 --> 15:06.180 Add venom into racks. 15:10.820 --> 15:14.930 Here we are telling the assembler to load. 15:15.820 --> 15:18.310 The value stored. 15:19.780 --> 15:20.260 At. 15:22.240 --> 15:24.070 Saw that on here and. 15:24.070 --> 15:24.460 Yeah. 15:24.490 --> 15:29.120 Load the value stored at the memory address. 15:29.140 --> 15:37.060 Memory address stored in penam into rax. 15:39.880 --> 15:48.010 And after that we we are telling the assembly to store the value in racks. 15:50.350 --> 15:55.780 Racks at the memory Address of Beaver. 15:58.400 --> 16:01.310 And after that we are telling. 16:02.020 --> 16:06.850 The assembly to load the load the memory address. 16:06.880 --> 16:15.160 Load the memory address of PVR, PVR into Rax. 16:15.340 --> 16:19.210 And now we will continue our code writing here. 16:19.210 --> 16:21.670 So we will use another instruction. 16:22.870 --> 16:23.770 Now we are. 16:25.670 --> 16:30.110 Julia here the racks to w numb. 16:32.460 --> 16:41.820 This is for load the memory load the memory address of w num into rax. 16:42.930 --> 16:44.820 Again, we will know not. 16:44.880 --> 16:48.900 Yes, we will load the value stored at w num into rax. 16:52.860 --> 16:53.730 Racks. 16:55.000 --> 16:56.380 W nam. 17:00.830 --> 17:01.910 The load. 17:01.920 --> 17:03.920 The value. 17:05.570 --> 17:10.460 Or load the so here rax w num we wrote here. 17:10.460 --> 17:15.050 So we loaded the memory address of the num into rax. 17:15.050 --> 17:19.640 But now what we're going to do is we are going to load the value. 17:21.250 --> 17:23.740 Go to the value part. 17:25.120 --> 17:33.700 In w num into rax and here we will use the Lia again because we need to tell the Assembly to load the 17:33.700 --> 17:37.000 memory address of text one. 17:38.100 --> 17:39.390 In two racks. 17:39.390 --> 17:39.980 Right. 17:39.990 --> 17:42.180 So racks. 17:44.160 --> 17:45.200 Text one. 17:45.840 --> 17:57.190 I will also add the comments load the memory memory address of text one into Rax and now we will tell 17:57.190 --> 18:01.440 the assembly to load the memory address of text one into. 18:02.480 --> 18:03.410 Racks. 18:16.880 --> 18:18.110 Thanks one. 18:20.110 --> 18:23.440 Load the memory address. 18:30.020 --> 18:31.910 Of text one. 18:34.560 --> 18:36.060 Interacts. 18:39.470 --> 18:43.760 And here you might be asking why the comments here is same. 18:44.300 --> 18:49.070 So you will learn that in when we get the output from this program. 18:50.140 --> 18:51.820 So don't worry about it. 18:54.430 --> 19:03.090 And after that we will need to load the value as the second character of Text one into Rax. 19:03.100 --> 19:06.610 In this case, the second character is here. 19:07.570 --> 19:07.930 So. 19:08.910 --> 19:09.640 Move there. 19:09.710 --> 19:10.260 Move. 19:14.020 --> 19:15.640 Now we will use the Mu again. 19:16.560 --> 19:17.550 Two racks. 19:21.340 --> 19:23.410 Text one plus one. 19:23.410 --> 19:25.720 Here we are telling the. 19:27.180 --> 19:30.870 The memory address of text. 19:31.500 --> 19:38.250 Now load the value at the second character of text. 19:38.280 --> 19:42.060 One one interacts. 19:44.710 --> 19:46.570 And we will use again. 19:47.650 --> 19:50.470 La Rocks here. 19:51.130 --> 19:56.110 And now, as we always do now, we will need to load the memory address of the second character of Text 19:56.110 --> 19:57.340 one into Rax. 19:57.430 --> 19:59.560 In order to do that, we will use this braces here. 19:59.610 --> 20:02.290 So text one plus one. 20:03.100 --> 20:04.030 What's the price? 20:04.150 --> 20:06.690 Now we are telling the Lord the. 20:08.020 --> 20:09.170 Load the memory. 20:09.180 --> 20:11.730 Memory address. 20:12.820 --> 20:14.560 Of text one. 20:17.180 --> 20:24.090 Memory address of text one plus one into Rax. 20:24.110 --> 20:25.970 Now we need to just write. 20:28.110 --> 20:29.520 Actually not here. 20:29.520 --> 20:30.960 So he wrote. 20:30.990 --> 20:32.160 MOV rax text one. 20:32.160 --> 20:36.240 So we've done this right and we use the Leah here. 20:38.320 --> 20:45.790 Although the memory address of the second character memory address of now loads the memory address of. 20:50.800 --> 20:58.570 The of the second character of text one interacts. 21:01.390 --> 21:05.320 And now we just need to write five more instructions. 21:05.320 --> 21:11.230 And we are done with this code here and in, in next lecture, we will analyze it, debug it and so 21:11.230 --> 21:19.660 on because the you will learn a lot from this code here about, uh, manipulating and using memories 21:19.660 --> 21:21.730 in assembly language. 21:21.730 --> 21:27.670 So after here we will need to use, uh, move here. 21:27.670 --> 21:29.650 So move rax. 21:32.240 --> 21:33.230 Next one. 21:33.260 --> 21:36.500 Here we are telling the assembler to load. 21:37.740 --> 21:40.410 The value stored. 21:41.790 --> 21:47.040 At the memory address of text one. 21:48.190 --> 21:51.100 Text one interacts. 21:52.120 --> 21:57.820 And after that we will use the move again to racks to load the value stored at the memory address of 21:57.820 --> 22:00.190 text one plus one into rax. 22:00.430 --> 22:01.300 So. 22:02.380 --> 22:03.580 Uh, text. 22:05.040 --> 22:07.350 X one plus one. 22:09.040 --> 22:12.100 You're telling the Lord the value. 22:13.040 --> 22:13.730 Thought. 22:14.970 --> 22:18.270 At the memory Memory address. 22:21.610 --> 22:24.520 Of text one plus. 22:27.480 --> 22:30.870 Text one plus one interacts. 22:33.890 --> 22:38.270 And now we just we will need to restore create the exit codes here. 22:38.270 --> 22:42.200 So RSP rbp. 22:42.860 --> 22:45.050 This is for restore. 22:46.590 --> 22:47.790 Restore. 22:48.750 --> 22:57.180 The value of RSP from which we need to do this for cleaning clean. 22:58.660 --> 23:01.330 Clean up the stack frame. 23:02.110 --> 23:04.720 And after that we will. 23:05.820 --> 23:06.390 Pop. 23:08.410 --> 23:10.390 Value of from the stack. 23:11.890 --> 23:14.860 Up the value of rb P from the stack. 23:16.370 --> 23:17.720 Write it up. 23:17.740 --> 23:22.240 The value of P from the stack. 23:23.420 --> 23:26.180 And now we will return from the function. 23:28.260 --> 23:31.020 Return from the function. 23:32.100 --> 23:33.210 So that's it. 23:34.440 --> 23:40.080 You need to develop this following program which exhibits a lack of discriminable output here. 23:40.080 --> 23:45.180 But we will employ a debugger to meticulously traverse each program instruction. 23:45.210 --> 23:51.630 This is a favorable option in the context of this employment of CSM. 23:51.810 --> 23:56.490 So I will explain this code in next lecture and meeting you in the next lecture.