1 00:00:01,080 --> 00:00:08,820 In this video, we are going to discuss two important instructions, call and read once again to demonstrate 2 00:00:08,820 --> 00:00:11,040 the use of call and read instructions. 3 00:00:11,040 --> 00:00:16,760 I have created a simple program with the file name called a Red Dot S.. 4 00:00:17,220 --> 00:00:20,550 So let's navigate to our directly called call Dash Red. 5 00:00:20,940 --> 00:00:25,470 Inside this, I have a file called Call Dash Red Dot. 6 00:00:25,470 --> 00:00:34,380 See if you notice the program is basically making a function call to a function called func and it has 7 00:00:34,380 --> 00:00:35,700 a seven arguments. 8 00:00:36,180 --> 00:00:40,590 Once this function call is made, it basically executes the function definition. 9 00:00:41,280 --> 00:00:47,970 After the execution is completed, it returns back to the next statement in the main function, which 10 00:00:47,970 --> 00:00:48,900 is printf. 11 00:00:48,900 --> 00:00:53,600 I am done and it executes it now like we did in the previous video. 12 00:00:53,820 --> 00:01:00,450 Let's compile this and open it up in Jollibee and understand what kind of instructions are being executed 13 00:01:00,690 --> 00:01:01,890 while this is being done. 14 00:01:02,520 --> 00:01:13,430 So let's first combine this using GC Car Dash, Red Nash, National Call Dash right. 15 00:01:14,970 --> 00:01:15,630 There it is. 16 00:01:16,500 --> 00:01:20,250 Now we should have the binary car dash right now. 17 00:01:20,250 --> 00:01:21,870 Let's open it up using GTP. 18 00:01:26,490 --> 00:01:31,120 And I'm going to set up a breakpoint at the entry point of this program, which is main. 19 00:01:33,270 --> 00:01:38,070 Now let's run the program and let's go through the disassembly I'm typing run. 20 00:01:39,450 --> 00:01:41,670 We are currently inside the main function. 21 00:01:41,700 --> 00:01:47,520 We are going to execute a bunch of instructions before making a call to the function named func. 22 00:01:48,390 --> 00:01:52,400 Now, let's step aside until we reach this push instruction. 23 00:01:52,830 --> 00:01:59,960 So I am typing asi asi hitting enter once again and there it is. 24 00:01:59,970 --> 00:02:06,570 Now we are about to execute this push zero X to misconstruction which is at main plus 12. 25 00:02:07,260 --> 00:02:10,550 Let's observe the next few instructions to be executed. 26 00:02:11,070 --> 00:02:16,920 There is one push instruction, as you can see here, and there are a bunch of move instructions. 27 00:02:17,280 --> 00:02:24,510 We can probably do our business businessmen to better view this d'arrigo that is one push instruction 28 00:02:25,050 --> 00:02:29,250 and there are six more instructions which are moving some values. 29 00:02:30,030 --> 00:02:32,430 I'm not counting this more instruction for now. 30 00:02:32,880 --> 00:02:37,730 So we essentially have six more instructions and one push instruction. 31 00:02:38,160 --> 00:02:45,510 Now let's examine what these values are zero six two zero two five eight and all this. 32 00:02:46,050 --> 00:02:48,000 So I'm just opening up a calculator. 33 00:02:50,790 --> 00:02:57,930 And I'm currently in hex mode, so I'm just typing to ABC and let's check what it is in decimal. 34 00:02:58,710 --> 00:03:00,530 If you notice this, this is 700. 35 00:03:01,140 --> 00:03:07,080 So let me type cat called Red Dot. 36 00:03:07,080 --> 00:03:12,630 See if you notice 700 is the last argument. 37 00:03:13,260 --> 00:03:15,040 How many arguments do we have here? 38 00:03:15,180 --> 00:03:18,570 One, two, three, four, five, six, seven. 39 00:03:19,170 --> 00:03:22,890 So the seventh argument is being pushed onto the stack. 40 00:03:23,550 --> 00:03:26,370 Now let's take what we have after the push instruction. 41 00:03:29,620 --> 00:03:40,000 Here it is, which is zero two five eight two five eight in Hex, should be 600 in decimal. 42 00:03:40,450 --> 00:03:42,460 So let's once again check the program. 43 00:03:44,110 --> 00:03:46,690 Six hundred is the sixth argument. 44 00:03:47,050 --> 00:03:52,450 And if you notice, this sixth argument is being moved into the register online. 45 00:03:53,350 --> 00:04:01,210 This is a sub register of online, since the value being moved is not large enough that online the register 46 00:04:01,220 --> 00:04:03,590 is used instead of online. 47 00:04:03,910 --> 00:04:06,260 So this is the sixth argument to the function. 48 00:04:06,850 --> 00:04:12,170 Next, the value of zero x one of four is being moved into the register. 49 00:04:12,220 --> 00:04:20,950 Added to this is a Register of the Register R8 once again, since the value being moved is not large 50 00:04:20,950 --> 00:04:21,310 enough. 51 00:04:21,550 --> 00:04:25,940 And does that register already is used instead of Orit? 52 00:04:26,620 --> 00:04:30,280 This is the fifth argument to the function next. 53 00:04:30,310 --> 00:04:35,920 The value zero x one nine zero is being moved into the register X.. 54 00:04:36,220 --> 00:04:43,120 This is the decimal 400 and once again X is the register of R6. 55 00:04:43,690 --> 00:04:50,040 Since the value being moved is not large enough, X registry is used instead of R6. 56 00:04:50,410 --> 00:04:52,590 This is the fourth argument to the function. 57 00:04:53,140 --> 00:04:59,830 Next, the value zero one to see, which is decimal 300 is being moved into the register. 58 00:05:00,970 --> 00:05:07,490 This is the register of RDX once again, since the value being moved is not large enough, that index 59 00:05:07,520 --> 00:05:10,320 register is used instead of RDX. 60 00:05:10,630 --> 00:05:18,130 This is the third argument to the function and similarly we have zero x C8 which is being moved into 61 00:05:18,130 --> 00:05:18,750 airside. 62 00:05:19,330 --> 00:05:24,400 This is a sample register of autosite and once again the value is not large enough. 63 00:05:24,670 --> 00:05:28,330 So that IACI register is used instead of RSA. 64 00:05:28,630 --> 00:05:35,050 And this is the second argument to the function, not the last one, the value zero x five. 65 00:05:35,050 --> 00:05:45,160 They want to be a 6F zero zero five, which is a large decimal number eight being moved into the register 66 00:05:45,160 --> 00:05:53,170 ideate since the value being moved is large enough this time the register idea is directly used instead 67 00:05:53,170 --> 00:05:54,450 of a sub register. 68 00:05:54,850 --> 00:05:57,340 So this is the first argument to the function. 69 00:05:57,670 --> 00:06:01,480 If you notice the C program, here it is. 70 00:06:03,310 --> 00:06:10,270 Now, these few instructions that we have just examined should give us an idea about how arguments are 71 00:06:10,270 --> 00:06:19,510 handled when a function call is made in 1964, it processes the first six arguments of a function are 72 00:06:19,540 --> 00:06:23,200 stored in the registers of the outside. 73 00:06:23,910 --> 00:06:28,420 The are six, are eight and nine, respectively. 74 00:06:29,110 --> 00:06:34,010 From the seventh argument, they will be pushed onto the stack from right to left. 75 00:06:34,600 --> 00:06:41,080 Now, after all the arguments are handled, there is another important thing that we should notice when 76 00:06:41,080 --> 00:06:48,250 we execute the instruction admen plus fifty nine, which is here, it will go ahead and execute the 77 00:06:48,250 --> 00:06:49,510 definition of the function. 78 00:06:50,230 --> 00:06:56,020 Once the execution completes, it has to return back and execute the print statement in the main function. 79 00:06:56,660 --> 00:06:59,230 Then how does it know how to come back to the main function? 80 00:06:59,940 --> 00:07:05,100 Let's knock down what we have on the top of the stack before executing call instruction. 81 00:07:05,440 --> 00:07:09,970 Let's also note down the address of the next instruction after the call instruction. 82 00:07:10,480 --> 00:07:11,740 So let's assign. 83 00:07:14,560 --> 00:07:17,980 Let's step aside until we reach the call instruction, 84 00:07:23,320 --> 00:07:29,200 if you notice, we are about to execute this instruction after executing the next move instruction, 85 00:07:29,810 --> 00:07:36,580 but before executing this, let's take note of the address of the next instruction after the call instruction. 86 00:07:37,240 --> 00:07:38,320 I'm is copying it here. 87 00:07:42,660 --> 00:07:50,040 And I'm pasting it here and let's check what we have on the top of the stack. 88 00:07:51,200 --> 00:07:56,990 If you notice, we currently have the argument that we pushed earlier, which is zero X to be C. 89 00:07:59,790 --> 00:08:04,530 And I'm just pasting it here now, let's type aside. 90 00:08:06,140 --> 00:08:08,930 Still, the top of the stack is the same. 91 00:08:09,320 --> 00:08:14,570 Now let's execute this call instruction and let's see what happens to the top of the stack. 92 00:08:16,810 --> 00:08:23,700 Have executed the instruction, and if you now observe the top of the stack, we have a different value 93 00:08:24,180 --> 00:08:30,900 and just copying this and pasting it here once again in the notepad right after the address that we 94 00:08:30,900 --> 00:08:38,100 have copied earlier, if you notice, the address that is pushed onto the stack is the same address 95 00:08:38,100 --> 00:08:42,020 that we have gotten, which is right after the call instruction. 96 00:08:42,480 --> 00:08:50,760 So what it means is when the instruction is executed, it has pushed the address of its next instruction 97 00:08:50,760 --> 00:08:51,580 onto the stack. 98 00:08:52,290 --> 00:08:57,990 So once the definition of the function is executed, it will return to this address by pumping this 99 00:08:57,990 --> 00:09:00,820 value into the RPE register from the stack. 100 00:09:01,260 --> 00:09:05,260 So let's test our theory by completely executing this function definition. 101 00:09:05,550 --> 00:09:07,200 So let's do a single step. 102 00:09:09,570 --> 00:09:13,560 I'm typing a sign multiple times. 103 00:09:29,560 --> 00:09:34,960 All right, we are reaching the right instruction, which is the last instruction in the function funk, 104 00:09:35,320 --> 00:09:40,070 so I'm just stepping aside once again as I once again and there you go. 105 00:09:40,450 --> 00:09:40,900 We are. 106 00:09:40,900 --> 00:09:42,160 And direct instruction. 107 00:09:42,160 --> 00:09:48,940 And if you look at the next instruction after this read, it is going to be the instruction or the address 108 00:09:48,940 --> 00:09:51,840 that we have just seen on the top of the stack earlier. 109 00:09:52,630 --> 00:09:58,720 But how does the processor know that after executing this written instruction, it has to go to this 110 00:09:58,720 --> 00:09:59,260 address? 111 00:09:59,920 --> 00:10:06,580 When the direct instruction is executed, it is going to pop the value which is on the top of the stack 112 00:10:06,880 --> 00:10:10,930 into the ORAC register, and then it will execute that. 113 00:10:11,560 --> 00:10:17,920 If you notice the top of the stack at this point of time, it has the same address, which is the address 114 00:10:17,920 --> 00:10:20,340 of the next instruction to the instruction. 115 00:10:20,620 --> 00:10:27,030 And when we execute this read instruction, this address is going to be poured into the AP register. 116 00:10:27,490 --> 00:10:31,240 So let's type aside and look at that. 117 00:10:31,630 --> 00:10:38,640 AP register now contains the address of the instruction, which is right after our call instruction. 118 00:10:39,190 --> 00:10:40,900 So that's how it functions work. 119 00:10:41,380 --> 00:10:48,250 To give a recap in this lecture, we have seen how functions work and how registers and stack are used 120 00:10:48,250 --> 00:10:50,020 to manage the arguments of the function. 121 00:10:50,320 --> 00:10:54,820 And then we have also seen how the return address is placed on the stack. 122 00:10:55,360 --> 00:11:02,200 So primarily we learned a few things in this lecture, how arguments are managed using registers and 123 00:11:02,200 --> 00:11:07,690 stack when a function is called, how return value is placed on the stack when a function is called. 124 00:11:07,930 --> 00:11:15,010 And we have also learned how the return address is poured into the AP register when a function is returning 125 00:11:15,010 --> 00:11:16,140 from its execution. 126 00:11:16,870 --> 00:11:23,770 We have also understood that read instruction always pops the value from the top of the stack into RHP 127 00:11:23,770 --> 00:11:27,630 register to be able to return to the next instruction of the calling function. 128 00:11:28,300 --> 00:11:29,480 That's all for this video. 129 00:11:29,650 --> 00:11:30,760 See you in the next one.