WEBVTT 00:00.150 --> 00:00.750 Hello again! 00:01.080 --> 00:03.560 In this video, we are going to look at bitsets. 00:04.920 --> 00:12.240 C has operators which manipulate integers at the bit level, so things like bit-wise AND, or the 00:12.240 --> 00:13.500 left shift operator. 00:14.400 --> 00:16.590 These were inherited by C++. 00:16.980 --> 00:24.060 And like most of the things that C++ inherited from C, these are low level, and not easy to use safely. 00:25.050 --> 00:30.360 In C++11, we have a bitset class in the standard library. In the 00:30.360 --> 00:31.080 header. 00:31.680 --> 00:37.020 This is an abstraction which represents groups of bits. So we can work with them, and manipulate them, 00:37.110 --> 00:42.030 much more easily, and safely, than we can using the old C legacy. 00:43.540 --> 00:45.430 bitset is a template class. 00:45.730 --> 00:51.210 So we need to give the number of bits as the type parameter, when we create a bitset objects. 00:52.200 --> 01:00.040 I have used 8, but you could use 7 or 9 or 13, or any number within reason, really. We can initialize 01:00.040 --> 01:07.810 a bitset from a string; from an integer, including a hex literal; and from C++14, we can also use 01:07.870 --> 01:12.670 binary literals, with an optional separator, so we can see where the bytes are. 01:15.850 --> 01:20.020 We can print out a bitset object directly, just by sending it to the stream. 01:20.530 --> 01:25.810 So that is the overloaded left shift operator for input and output, not the one for bit shifting. 01:26.410 --> 01:28.570 And that will display all the bits in the object. 01:30.190 --> 01:34.090 If we want to access the underlying data, we can call a member function. 01:34.090 --> 01:37.540 We can get this as an integer, or as a string. 01:39.770 --> 01:46.250 There is a size() member function, which gives the number of bits in the object. And we can use subscript notation 01:46.250 --> 01:47.870 for accessing individual bits. 01:48.350 --> 01:52.700 So that is just like accessing array elements by using indexes. 01:54.940 --> 01:58.300 The index zero will give the Least Significant Bit. 01:58.300 --> 02:03.430 So that is the one at the right. The one on the left is called the Most Significant 02:03.430 --> 02:04.210 Bit, by the way. 02:04.690 --> 02:09.220 So, index zero will display the value of that bit. 02:10.690 --> 02:16.120 If we want extra safety, we can also have bounds checking, by using the test() member function. 02:16.780 --> 02:20.740 This will return the bit whose index corresponds to its argument. 02:21.460 --> 02:25.090 If the argument is out of range, then we get an exception thrown. 02:30.140 --> 02:31.670 So let's try that out. 02:31.670 --> 02:34.040 We create some bitset objects. 02:34.940 --> 02:37.970 Then we display their values in various ways. 02:39.320 --> 02:44.780 We display the number of bits in this object, and then we iterate through them and print them out. 02:45.770 --> 02:47.720 And then finally we try it with bounds 02:47.720 --> 02:54.110 checking. Eight is not a valid subscript, so this should throw an exception, which we catch. 02:54.620 --> 02:58.160 So we should see a message printed out, when we run this program. 03:00.380 --> 03:01.190 So there we are. 03:01.670 --> 03:03.710 That is the value of these bitsets. 03:04.430 --> 03:11.660 We have 8 bits in the object. And they are in reverse order, because we started with index zero, and 03:11.660 --> 03:12.950 finished with index seven. 03:13.880 --> 03:15.350 Eight is not a valid index. 03:15.590 --> 03:17.000 So we get the exception thrown. 03:20.320 --> 03:23.510 We can also perform all the usual operations that we do with ints. 03:24.250 --> 03:32.920 So we can invert all the bits, we can do AND, OR and EXOR, and also left shift and right shift. 03:33.340 --> 03:36.250 And those will cause but shifting, and not input-output. 03:36.670 --> 03:39.040 We need to use brackets here, because of the operator 03:39.040 --> 03:39.940 precedence rules. 03:42.090 --> 03:42.810 So there we are. 03:42.810 --> 03:48.480 We have inverted the bits in b1, and then we have performed all these other operations. 03:51.360 --> 03:58.140 There are member functions which will modify bitsets. These can either modify all the bits, or just 03:58.140 --> 04:03.270 one bit. If we call set() with no arguments, that will set all the bits to true. 04:03.900 --> 04:07.690 If we provide a index, then it will set the bit with that 04:07.710 --> 04:08.790 index to true. 04:09.030 --> 04:13.890 And there is also a version where we can pass false as the second argument, if we want to set it to 04:13.890 --> 04:14.310 false. 04:15.480 --> 04:18.450 reset() with no arguments will make all the bits false. 04:18.990 --> 04:20.850 Or we can do that for just one bit. 04:21.660 --> 04:23.220 We can also invert all the bits. 04:23.220 --> 04:25.530 So all the bits which have value true, 04:25.530 --> 04:27.930 will now have value false, and vice versa. 04:28.470 --> 04:31.350 And we can do that for all the bits, or just one 04:31.350 --> 04:32.160 argument bits. 04:34.730 --> 04:38.090 So in this function, we are going to use this pattern of bits. 04:38.600 --> 04:42.350 We will make a copy of it each time, so we do not get confused about what we have changed. 04:43.730 --> 04:47.300 So we call the same member functions as we did on the slide. 04:47.300 --> 04:51.260 So we call set() with no arguments, with one argument, and so on. 04:52.010 --> 04:54.800 Then reset() and then flip(). 04:57.880 --> 05:00.060 So we have set all the bits to true there. 05:00.070 --> 05:03.760 We have set them to false there, and then we have inverted the bits there. 05:07.820 --> 05:11.750 There are also some member functions we can use for checking the entire bitset. 05:12.080 --> 05:15.200 If all the bits are true, then all() will return true. 05:16.010 --> 05:20.450 If at least one bit in the object is true, then any() will return true. 05:21.290 --> 05:24.680 And if all the bits are false, then none() will return true. 05:25.400 --> 05:30.350 And there is also a count() member function, which will return the number of bits which are true. 05:32.470 --> 05:34.210 So we have a similar program 05:34.210 --> 05:41.470 again. We have these bits. We have another bit set object in which all the bits are true, and one 05:41.470 --> 05:42.800 in which they are all false. 05:43.480 --> 05:47.290 And then we just call these member functions for those three objects. 05:52.430 --> 05:54.410 So, are all the bits set? 05:54.830 --> 05:59.090 So that will be false for this object, true for this one, and false for this one. 05:59.900 --> 06:00.770 Are any bits set? 06:00.830 --> 06:01.790 That is true for this one, 06:01.790 --> 06:04.310 and this one, but false for that one. And so on. 06:05.000 --> 06:08.600 And we have five bits set here, eight here, and zero here. 06:09.560 --> 06:11.000 Okay, so that is it for this video. 06:11.390 --> 06:12.200 I will see you next time. 06:12.200 --> 06:14.330 But until then, keep coding!