WEBVTT 00:01.140 --> 00:05.050 Let's discuss, compare and jump instructions in this video. 00:05.550 --> 00:12.380 So far, we have written our programs in assembly language to demonstrate compassion and jump instructions. 00:12.390 --> 00:17.220 I have written a very simple SI program named Jump Si. 00:18.390 --> 00:20.120 So let's take a look at the program. 00:20.520 --> 00:25.410 I have saved this file inside the direct card siempre Dasht Jambi. 00:26.040 --> 00:29.130 Inside this I have a file called Jump Dart. 00:29.130 --> 00:31.740 See, it's a very simple SI program. 00:32.460 --> 00:37.230 If you observe the program, it's a simple program with if ands condition. 00:37.830 --> 00:39.090 Let's go from the beginning. 00:39.510 --> 00:42.420 First we have initialized to variables. 00:42.720 --> 00:50.580 Variable error is initialized with value tool and the variable B is initialized with value. 00:50.580 --> 01:00.120 Three if is equal to be, the statement will be executed and the string is equal to be will be printed. 01:00.900 --> 01:07.890 If these two are not equal, the jump will be taken to this block and the string is not equal to be 01:08.130 --> 01:09.270 will be executed. 01:09.810 --> 01:12.630 This is all fine and easy to understand in C language. 01:12.990 --> 01:18.000 Let's see what is happening behind the scenes by going through the assembly version of this program 01:18.210 --> 01:19.140 using Jilib. 01:19.680 --> 01:30.690 Let's first compile this program and using GTC jump dot c bargepole jump. 01:33.520 --> 01:39.830 We should now have the binary jump, we can check that using less there you go. 01:40.240 --> 01:46.720 Now let's open it up using GDB and I'm going to set up a breakpoint at the entry point of this program. 01:47.170 --> 01:51.310 And the entry point of this program is going to be the main function. 01:51.790 --> 01:55.390 So type Jalabi dot slash jump. 01:56.690 --> 01:59.530 And let's set up a breakpoint at Main. 02:00.680 --> 02:01.250 There you go. 02:01.760 --> 02:05.370 Now let us run the program and let us go through the disassembly. 02:06.020 --> 02:09.620 I'm typing right here it is. 02:09.630 --> 02:12.270 This is the disassembly that we are going to execute now. 02:12.290 --> 02:14.540 Currently, we are inside the main function. 02:15.060 --> 02:20.870 Now, let's quickly do our business men to see the full view of the disassembly. 02:21.890 --> 02:22.470 Here it is. 02:22.490 --> 02:24.440 This is the complete main function here. 02:25.380 --> 02:29.160 Let's execute this program until we reach this instruction. 02:29.720 --> 02:34.850 We can either set up a breakpoint here or we can type site multiple times. 02:35.390 --> 02:36.590 Let me die type ASI. 02:42.450 --> 02:44.980 All right, so we have reached this instruction. 02:45.240 --> 02:52.200 We are currently at Maine plus 12, if you closely observe the next instruction to be executed, it 02:52.200 --> 02:54.900 is placing the value to onto the stack. 02:55.410 --> 02:59.870 We can spot that by looking at the register are B in the instruction. 03:00.480 --> 03:03.630 So let's step aside and observe the stack. 03:04.380 --> 03:10.430 If you examine the stack before executing this instruction, the value to doesn't exist on the stack. 03:10.770 --> 03:15.120 So let's step aside and look at that. 03:15.270 --> 03:18.420 The value of two is currently placed on the stack. 03:19.050 --> 03:25.790 Now once again, the next instruction to be executed is placing the value tree on the stack. 03:26.460 --> 03:33.300 So I'm typing aside once again and let's observe what happens on the stack after executing this instruction. 03:36.510 --> 03:40.280 Look at that, the value tree is actually placed on the stack as well. 03:41.630 --> 03:45.710 Look at that, the value tree is now placed on the stack as well. 03:46.370 --> 03:48.080 OK, next instruction. 03:49.300 --> 03:56.680 If you observe the next instruction to be executed, once again, it is referencing a value that's available 03:56.680 --> 04:03.220 on the stack, if you notice the value that it is looking for is are B, B minus zero X eight, which 04:03.220 --> 04:06.460 is essentially this, which means the value to. 04:07.180 --> 04:11.380 So it is moving the value to into the register X.. 04:12.010 --> 04:14.860 So let's first check the value of our X. 04:17.180 --> 04:27.110 Currently, this is the value of Addicks, if you say it should contain the value to let's say let's 04:27.110 --> 04:28.750 go back and check X. 04:30.420 --> 04:38.850 Dirigo Artex register now contains the value to now let's move on to the next instruction, the next 04:38.850 --> 04:46.380 instruction is comparing the value available in extra stir with some value that's available on the stack. 04:46.880 --> 04:50.040 If you notice this R, B, B, minus zero X four. 04:51.030 --> 04:56.220 Here it is, which means 063 is what it is comparing it against. 04:56.730 --> 05:02.880 So the X is currently holding two and the value of three is currently available on the stack. 05:03.120 --> 05:09.260 And this assembly program is basically checking if these two values are equal or not. 05:09.900 --> 05:13.560 And this instruction is comparing those two values. 05:14.220 --> 05:21.720 In the C program we were doing a comparison by doing if A equals to be in assembly, the instruction 05:21.750 --> 05:24.300 CMB is used for comparisons. 05:25.050 --> 05:28.440 So essentially two is being compared with three. 05:28.650 --> 05:34.860 And based on the result of this instruction, the control will be transferred into some location using 05:34.860 --> 05:37.820 the next instruction, which is Jayanthi. 05:39.180 --> 05:41.780 We will come to this Jaynie instruction in a bit. 05:41.790 --> 05:44.530 Let's first step aside and hit enter. 05:47.490 --> 05:54.060 Now if you look at the zero flag, it is not set when sampi instruction gets executed. 05:54.300 --> 05:58.200 It essentially makes a subtraction of these two values. 05:58.500 --> 06:02.000 When you do two minus three, the result is non-zero. 06:02.430 --> 06:04.950 If there isn't, a zero zero flag will be set. 06:05.190 --> 06:09.180 If the flag is set, you should see this word zero in uppercase. 06:09.900 --> 06:15.240 Since the flag is not that you are seeing it in the lowercase, but if you look at the same flag, this 06:15.240 --> 06:19.340 is set because two minus three is going to produce a negative sign. 06:20.310 --> 06:24.210 Now, what we are interested in is the next instruction, which is Jayanthi. 06:24.780 --> 06:28.860 Since these are not equal, we are going to take a jump to some location. 06:29.370 --> 06:31.230 Obviously that is the case. 06:31.740 --> 06:35.580 We will come here and execute this conditional jump. 06:35.580 --> 06:43.020 Instructions are usually followed by a comparison instruction because this takes a jump based on a specific 06:43.020 --> 06:46.320 condition and that's usually set by Campi instruction. 06:47.250 --> 06:52.800 So simply instruction in this case did a comparison between the value to entry. 06:53.190 --> 06:57.870 And since the result was not zero, it did not set zero flag. 06:58.110 --> 07:00.590 And you can see this here as well. 07:00.960 --> 07:05.940 And the reason it shows here is this jump is taken because zero flag is not set. 07:06.750 --> 07:12.150 Now let's check with this jump is taken to let us take this jump by typing assign. 07:14.090 --> 07:21.140 And if you notice, there is a call to Prenter function soon after executing two more instructions, 07:21.300 --> 07:29.990 but before that, but before that, there is an instruction which is LTA idea and some address. 07:30.500 --> 07:33.470 LTA instruction is Lord effective address. 07:33.800 --> 07:39.890 When this instruction gets executed, the contents of the memory location are not loaded, only the 07:39.890 --> 07:43.240 effective addresses is computer and placed into the register. 07:43.730 --> 07:49.250 In this case, a clear instruction moves that effective address into audio register. 07:49.730 --> 07:52.850 So let's first check what we have at this address. 07:53.240 --> 08:00.380 So I'm just copying this slash piece and I'm just pasting it here. 08:01.730 --> 08:02.270 Look at that. 08:02.390 --> 08:06.700 We have the string is not equal to be at this address. 08:07.220 --> 08:11.130 So this address is pointing to the string that's about to be printed. 08:11.690 --> 08:14.030 So why is it being moved to RBI? 08:14.600 --> 08:20.570 That's because in 1964, with architecture, the arguments are loaded into to register. 08:20.960 --> 08:28.070 This printer function that is about to be executed contains one argument, and the first argument always 08:28.070 --> 08:30.680 gets loaded into the RBA register. 08:31.010 --> 08:37.990 And that's exactly why this is getting loaded into this idea register before we execute this printer 08:38.000 --> 08:38.570 function. 08:39.050 --> 08:48.920 So let's say and if you check the RBA register, it contains an address that's pointing to this string 08:49.520 --> 08:50.300 right now. 08:50.330 --> 08:57.420 Let's say there is a call instruction, which is basically going to execute the printf function. 08:58.250 --> 09:02.900 So if we type Essi, it is going to get into the print definition. 09:03.290 --> 09:11.000 And if you type C to continue with the execution, it is going to print is not equal to B after completing 09:11.000 --> 09:12.890 the execution of print function. 09:13.430 --> 09:19.790 Now we have briefly discussed how arguments are past when a function is executed in this video. 09:20.150 --> 09:26.390 But in the next video, we are going to discuss more details about how arguments are managed when a 09:26.390 --> 09:27.530 function call is made. 09:27.800 --> 09:33.860 The idea behind this video is to show you the CMB and conditional jump instructions. 09:34.410 --> 09:35.540 That's all for this video. 09:35.870 --> 09:37.010 See you in the next one.