WEBVTT 00:00.320 --> 00:06.320 Hello, my name is Typhoon and this is the last lecture of our section, and many arithmetic instructions 00:06.320 --> 00:10.040 are available, but we are going to show a selection of them. 00:10.040 --> 00:13.700 And the others are similar to what you learned here. 00:13.700 --> 00:21.770 So before we investigate arithmetic instructions and keep in mind that we use the printf with more than 00:21.770 --> 00:24.680 two arguments, so we need an additional register. 00:26.410 --> 00:27.310 The first one. 00:27.310 --> 00:31.240 The first argument goes into RTI. 00:31.660 --> 00:38.720 The second one into RSI and the third one into RDX. 00:38.770 --> 00:46.720 So this is how Printf expects us to provide the arguments in Linux and you will learn more about that 00:46.720 --> 00:53.770 later when we talk about the calling conventions, which you will learn that in next sections. 00:54.010 --> 00:57.160 And here there is a instructions here. 00:57.160 --> 01:03.640 The first instruction is Add, which can be used to add signed or unsigned integers. 01:03.730 --> 01:09.250 The second operand source is added to the first operand destination and the result is placed in the 01:09.250 --> 01:11.050 first operand destination here. 01:11.050 --> 01:14.800 And the destination operand can be a register or a memory location. 01:14.800 --> 01:19.870 So the source can be an immediate value, a register or a memory location. 01:19.870 --> 01:25.840 The source and destination cannot be a memory location in the same instruction, and when the resulting 01:25.840 --> 01:33.260 sum is too large to fit in the destination, the F flag is set for signed integers. 01:33.260 --> 01:37.280 So for unsigned integer, the o f flag is then set. 01:37.280 --> 01:38.000 So. 01:38.630 --> 01:48.530 When the result is zero, the z f flag is set to one and when the result is negative, the F flag is 01:48.530 --> 01:49.190 set. 01:52.210 --> 01:57.340 And the subtraction with sub is similar to the Add instruction. 01:58.380 --> 02:03.480 Uh, we can see it right here, adding and subtracting. 02:03.750 --> 02:05.670 So here the. 02:06.860 --> 02:14.120 To increment a register or a value in a memory location with one use the I and C. 02:14.150 --> 02:14.900 Instructions. 02:14.900 --> 02:22.700 Similarly, the a C can be used to decrement a register or value in a memory. 02:22.760 --> 02:24.110 Location with one. 02:24.110 --> 02:27.860 So the arithmetic shift instructions are a special breed. 02:27.860 --> 02:34.400 So here you will see that at the end of the code here somewhere here. 02:34.400 --> 02:38.990 So here if you shift left here. 02:38.990 --> 02:43.160 So the shift left is Sal is in fact multiplying. 02:43.190 --> 02:47.270 So if you shift left one position, you are multiplying by two. 02:47.300 --> 02:49.310 Every bit is shifted. 02:49.310 --> 02:53.690 One place to the left and zero is added to the right. 02:53.690 --> 03:02.720 So take the binary number one shift left one place and you obtain binary 10 or 2 in decimal representation. 03:02.720 --> 03:11.670 So you are getting from 0 to 2 and shift one shift left, one place again and you have a binary one 03:11.670 --> 03:15.180 zero 0 or 4 in decimal representation. 03:15.180 --> 03:19.140 So if you shift left two position, you multiply by four. 03:19.170 --> 03:22.560 But what if you want to multiply by six? 03:22.560 --> 03:31.230 So you shift left two times and then add two times the original source in that order. 03:31.350 --> 03:34.290 So shift right as r. 03:34.290 --> 03:36.300 S a r here. 03:36.950 --> 03:41.180 Uh, is similar to a shift left, but it means dividing by two. 03:41.330 --> 03:47.920 So every bit is shifted one place to the right and an additional bit is added to the left. 03:47.930 --> 03:48.740 So. 03:49.790 --> 03:50.180 Here. 03:50.180 --> 03:51.620 There is a complication. 03:51.620 --> 03:58.910 However, if the original value was negative, the leftmost bit would be one, and if the shift instruction 03:58.910 --> 04:06.950 added two added a zero, the bit at the left, the value would become positive and the result would 04:06.950 --> 04:07.760 be wrong. 04:07.760 --> 04:16.460 So in the case of negative value as a R will add a one bit to the left and in the case of positive value, 04:16.490 --> 04:21.140 zero bits will be added to the left and this is called sign extension. 04:21.140 --> 04:28.460 So by the way, a quick way to see if a hexadecimal number is negative is to look at byte seven. 04:28.460 --> 04:33.620 This is the leftmost byte continuing from byte zero, which is the rightmost byte. 04:33.620 --> 04:34.970 So the number is negative. 04:34.970 --> 04:45.440 If byte zero starts with an eight, nine, A, B, C, D, E, or F, but you need to take into account 04:45.470 --> 04:48.380 all eight bytes. 04:48.380 --> 04:51.720 For example, the byte zero x. 04:52.290 --> 04:53.340 Uh, the device. 04:53.340 --> 05:00.930 The 12 is still a positive number because the leftmost byte, which is not shown here, is zero. 05:00.930 --> 05:03.450 And there's also no arithmetic shift instructions. 05:03.450 --> 05:06.120 They will be discussed in next lecture. 05:06.330 --> 05:14.990 And next here we multiply integers for multiplying the unsigned integers you can use mul here. 05:15.210 --> 05:21.390 Um, now we will hear, uh, mul for unsigned multiplication. 05:21.390 --> 05:24.420 So I think, uh, so the in the product shift. 05:24.420 --> 05:24.980 Right. 05:24.990 --> 05:26.730 Increment number. 05:27.780 --> 05:30.300 Total integer quotient and product. 05:30.300 --> 05:30.870 Yes. 05:35.860 --> 05:41.810 And here but we use the I'm u l. 05:41.860 --> 05:49.570 This is assigned multiple action which offers more flexibility so I can take one, 2 or 3 operands. 05:49.570 --> 05:54.640 And in our example we use one operand, as you can see here. 05:55.800 --> 05:58.500 And here. 06:00.510 --> 06:08.250 And the as I said in our example, we use this one operand and the operand following the I mul instruction. 06:08.300 --> 06:14.490 I mul instruction is multiplied with the value in rax right. 06:14.490 --> 06:20.400 So you may expect that the resulting product is stored in Rax, but that is not entirely correct. 06:20.400 --> 06:22.650 So let's illustrate an example here. 06:22.650 --> 06:28.110 So you can verify that when you multiply, for example, a two digit number with a three digit number, 06:28.140 --> 06:31.380 the product has 4 or 5 digits. 06:31.380 --> 06:40.560 So when you multiply 4 to 8 bit digit with 30 bit digit, you will obtain 77, 77 bit digit or 78 bit 06:40.560 --> 06:40.860 digit. 06:40.860 --> 06:44.730 And that value does not fit in 64 bit register. 06:44.730 --> 06:45.030 Right. 06:45.030 --> 06:52.950 So to cope with this, the instruction I moved here Am will store the lower 64 bits of the resulting 06:52.950 --> 06:58.530 product in Rax and upper 64 bits in RDX. 06:58.530 --> 07:00.520 And this can be very deceptive. 07:00.520 --> 07:03.070 Okay, so now let's experiment a little bit here. 07:03.070 --> 07:10.810 So go back to the source code here, modify the number one so that it contains here. 07:12.110 --> 07:13.070 Number one. 07:13.070 --> 07:14.300 One, two. 07:14.980 --> 07:16.210 Three, four. 07:16.210 --> 07:20.260 One, 234567890. 07:20.260 --> 07:20.890 And again. 07:20.890 --> 07:23.230 One, two, three, four, five, six, seven. 07:24.010 --> 07:34.870 That's it yeah seven and now and modify the number 2 to 100 and the product will just fit in racks so 07:34.870 --> 07:38.470 you can check the here you would. 07:39.870 --> 07:45.270 As you can see, we got an error, but missing node Unix like implies executable stack. 07:47.510 --> 07:50.780 Yeah, this is just a warning and not an error here. 07:50.780 --> 07:55.730 And as you can see here, we got very different result here. 07:57.710 --> 07:59.690 And if it started the budget. 08:03.620 --> 08:13.130 And in order to debug this, we need to we will put a break before the I mole instruction. 08:14.410 --> 08:16.000 The started here. 08:16.000 --> 08:25.330 And we will, uh, here we will put a break before the I instruction here, which is for the multiplying. 08:25.360 --> 08:26.110 That's it. 08:26.830 --> 08:27.730 Now. 08:29.060 --> 08:29.510 Let's go. 08:29.510 --> 08:32.810 And here we stopped it now. 08:33.600 --> 08:33.990 Ugh. 08:34.020 --> 08:34.710 Now. 08:36.030 --> 08:40.200 The result of the multiplication here will be. 08:43.720 --> 08:48.910 That's actually and that's the result of the multiple action here will be this number. 08:48.910 --> 08:54.520 And as you can see in racks after the instruction is executed. 08:54.610 --> 08:55.240 Right. 08:56.610 --> 08:58.440 Uh, the new modified. 08:58.640 --> 09:06.000 Now, uh, actually, let's actually modify number two into 10,000. 09:07.300 --> 09:07.640 Obsession. 09:07.700 --> 09:09.610 Let's stop the debugging. 09:11.210 --> 09:13.070 It still have that breakpoint here? 09:13.940 --> 09:14.990 One, two, three. 09:15.020 --> 09:17.420 Now it's in 10,000. 09:17.840 --> 09:18.920 The bugging is started. 09:18.920 --> 09:20.720 Let's go to email here. 09:20.840 --> 09:28.670 And with this, um, after, uh, of course, after setting the restart, after setting the 10,000 in 09:28.670 --> 09:31.970 number to stop the debugging and then start it again. 09:32.820 --> 09:40.320 And after look after the racks after executing moved here like this. 09:41.150 --> 09:43.040 And here. 09:43.100 --> 09:43.600 Um. 09:45.740 --> 09:52.240 So you can see that the product is a large negative value here, right? 09:52.250 --> 09:57.290 So this is because the most significant bit in Rax is a one. 09:57.290 --> 10:02.220 And CSM concludes that this must be a negative number. 10:02.240 --> 10:09.500 Also, Printf thinks that Rax contains a negative number because Rax contains a one bit in the leftmost 10:09.500 --> 10:12.430 position, so it's assumed to be negative. 10:12.440 --> 10:15.140 So be careful with printf.