WEBVTT 00:00.400 --> 00:01.160 Hello, everyone. 00:01.600 --> 00:01.840 Here. 00:01.840 --> 00:04.320 Welcome to our lecture on binary subtraction operations. 00:04.800 --> 00:10.640 In this lecture we will explore subtraction in computers, specifically focusing on binary arithmetic. 00:10.680 --> 00:15.680 The concept of borrowing the role of flags in indicating certain conditions to run computations. 00:16.120 --> 00:18.080 And yeah let's dive right in. 00:18.520 --> 00:22.560 We'll begin with a simple yet essential example to grasp binary subtraction. 00:22.600 --> 00:29.120 Now let's subtract 14 from four using four bit binary numbers here. 00:29.800 --> 00:35.000 So decimal four in binary is 100. 00:35.400 --> 00:40.800 So 0100 and decimal 14. 00:40.840 --> 00:41.880 In binary is 00:42.040 --> 00:47.960 11101110. 00:49.400 --> 00:56.040 Now let's perform the subtraction and you will see why we can't represent negative numbers in unsigned 00:56.080 --> 00:56.760 integers. 00:57.520 --> 01:00.600 Zero zero is 001. 01:00.600 --> 01:02.200 We will borrow from here. 01:02.880 --> 01:05.480 So also we will write the borrowing here. 01:05.480 --> 01:06.480 So zero. 01:07.320 --> 01:10.130 Here we will have one, one, one. 01:10.970 --> 01:14.370 We have zero, one and zero. 01:15.130 --> 01:16.890 The borrowing is zero. 01:17.930 --> 01:23.730 So this is weird because it is what? 01:23.770 --> 01:26.730 01100110. 01:27.410 --> 01:30.970 It says that it equals to six, but it's not. 01:31.650 --> 01:35.810 Initially, the result appears as 0110. 01:35.850 --> 01:38.290 In binary, which equals six in decimal. 01:38.330 --> 01:42.410 However, we know subtracting 14 from four yields what? 01:42.890 --> 01:44.570 Minus ten. 01:45.370 --> 01:47.890 But we do not have minus ten here. 01:48.610 --> 01:50.130 So why does this happen? 01:50.210 --> 01:53.010 Because we are limited to four bits. 01:53.210 --> 01:55.770 We can accurately represent the negative result. 01:55.770 --> 02:04.450 The computer uses a special indicator called the carry flag to recognize and and handle such situations. 02:04.930 --> 02:09.530 And since we are using unsigned, we cannot use negative numbers with unsigned here. 02:10.250 --> 02:11.250 And you will see why. 02:11.930 --> 02:18.580 In computer processors, especially indicator known as the carry flag See is abbreviated as. 02:18.620 --> 02:21.580 Carry flag in computer processor. 02:22.340 --> 02:28.420 So this is kind of signal whether a subtraction operation has needed borrowing from beyond the available 02:28.460 --> 02:28.940 bits. 02:29.340 --> 02:39.540 And in this case carry flag was one since which indicates the borrowing was necessary, suggesting the 02:39.540 --> 02:49.100 calculated result isn't fully accurate due to the bit limitations here and carry flag zero. 02:49.140 --> 02:52.580 Of course, indicates no borrowing was required here. 02:52.900 --> 02:59.420 In our example, the carry flag would set to one like this, clearly indicating that our computed result 02:59.420 --> 03:03.860 exceeded the bit limitations and borrowing was involved. 03:03.900 --> 03:09.460 Now let's briefly see how this concept translates into assembly language instructions which processor 03:09.860 --> 03:14.900 use directly and assembly language typically includes specific instructions for addition and subtraction 03:14.940 --> 03:17.780 operations, often updating flags. 03:18.260 --> 03:25.860 In assembly we have the flag adds, so sets the carry flag to zero if there's a no final carry and sets 03:25.900 --> 03:30.620 carry like to one if the overflow occurs, right? 03:31.580 --> 03:39.860 In this case, we have nothing to do with ads here since as you know, ads is what ads the number. 03:40.540 --> 03:43.060 In this case we are using subtraction. 03:43.060 --> 03:45.980 So let's explain the subs here. 03:46.420 --> 03:50.020 We explained ads in previous lecture subs. 03:50.740 --> 03:54.900 So subs here sets the carry flag zero. 03:54.940 --> 04:02.380 If nobody borrowing is necessary and it sets the carry flag to one if borrowing occurs here. 04:03.060 --> 04:16.260 So example for the generic subtraction instructions assembly we have sub the destination sub destination 04:17.100 --> 04:19.380 source is source one and source two. 04:20.180 --> 04:30.780 So which basically means subtract source two from source one and save the result in destination and 04:30.940 --> 04:33.180 flags will update accordingly.