WEBVTT 00:00:00.540 --> 00:00:05.040 We now know how to push objects at the specific position in the array, but 00:00:05.050 --> 00:00:09.880 there is also a way to remove them. Erase function will delete the element 00:00:09.890 --> 00:00:15.200 at the specific index, but we also have to provide this position by using 00:00:15.200 --> 00:00:20.450 the begin iterator. I want to erase the first element, so I will just pass 00:00:20.450 --> 00:00:22.060 the iterator itself. 00:00:22.940 --> 00:00:24.160 Let's compile this. 00:00:25.440 --> 00:00:29.440 The first element had an ID of 1, and it seems that it was 00:00:29.440 --> 00:00:31.650 successfully removed from the array. 00:00:32.540 --> 00:00:36.630 Move assignment operator was invoked four times because these four 00:00:36.630 --> 00:00:40.500 subjects have to be moved to the start of the array to fill in the 00:00:40.500 --> 00:00:42.950 blank left by the erased element. 00:00:43.840 --> 00:00:46.800 But we can also remove a range of elements by providing 00:00:46.800 --> 00:00:49.260 the starting and ending position. 00:00:49.740 --> 00:00:54.980 I want to start at the index of 1 and stop at the index of 3. Erase 00:00:54.980 --> 00:00:58.920 function will remove the element at the starting index, but it will 00:00:58.920 --> 00:01:01.950 not remove the element at the ending index. 00:01:03.240 --> 00:01:04.459 Let's compile this. 00:01:05.940 --> 00:01:10.380 As you can see, only the second and third element were removed from the vector 00:01:10.390 --> 00:01:15.690 because the starting index was 1. Index 3, which defines the end of the 00:01:15.690 --> 00:01:19.960 specified range, is not included in this deletion process. 00:01:20.340 --> 00:01:24.080 You can see that the move assignment was invoked two times. 00:01:24.090 --> 00:01:27.900 This is because vector moved these last two elements and 00:01:27.900 --> 00:01:30.450 placed them on top of the deleted ones. 00:01:31.740 --> 00:01:35.970 Now we have a way of removing elements at the specific position, but what 00:01:35.970 --> 00:01:40.460 if we want to do something more challenging? For example, let's say that I 00:01:40.460 --> 00:01:45.590 have this vector of integers, and I want to get rid of all of the 6s. We 00:01:45.590 --> 00:01:49.530 could create a for loop, which will go over all of the elements, and when 00:01:49.530 --> 00:01:54.370 it finds a 6 at the specific index, it will pass that position to the 00:01:54.380 --> 00:01:55.450 erase function. 00:01:55.840 --> 00:01:58.030 But there is a better way to do this. 00:01:58.040 --> 00:02:01.660 We can use the remove function from the algorithm library. 00:02:03.340 --> 00:02:05.950 The first two arguments specify the range, 00:02:05.960 --> 00:02:10.449 a part of the vector that we want this remove function to operate on. 00:02:11.140 --> 00:02:14.430 Usually, we want to do this on the whole vector, so we can 00:02:14.430 --> 00:02:16.960 pass the begin and the end iterator. 00:02:17.640 --> 00:02:22.650 The third argument is the value that we want to remove, in this case 6. 00:02:23.540 --> 00:02:27.620 The remove function doesn't actually move anything. It just moves the wanted 00:02:27.620 --> 00:02:32.800 elements over the unwanted ones, but the leftovers are still there in the 00:02:32.800 --> 00:02:36.160 vector. The size of the vector will not change. 00:02:36.940 --> 00:02:38.350 Let me show you what I mean. 00:02:39.040 --> 00:02:43.540 So this is how the vector looks like at the beginning. Remove 00:02:43.540 --> 00:02:46.820 function will start to look through the elements and check 00:02:46.820 --> 00:02:49.050 which one of them are unwanted. 00:02:49.440 --> 00:02:54.660 The first element is 2. It is not 6, so this one is wanted, 00:02:55.740 --> 00:02:57.460 The second element is 7. 00:02:57.470 --> 00:03:01.870 This one is also not 6, so nothing has to move. We want all of 00:03:01.870 --> 00:03:04.760 the unwanted elements at the end of the array. 00:03:05.440 --> 00:03:09.810 Now we get to the third element, which is 6. And as we know, 6 is 00:03:09.820 --> 00:03:15.640 unwanted. But this is still okay because the unwanted number is at the 00:03:15.640 --> 00:03:18.050 end of this part of the array we covered. 00:03:18.440 --> 00:03:21.650 However, once we get to the number 8, which is wanted, 00:03:21.660 --> 00:03:25.160 the unwanted number 6 is no longer at the end. 00:03:25.540 --> 00:03:29.380 So the remove function will move this number 8 over to 00:03:29.380 --> 00:03:33.410 number 6. Move might not be the right word. 00:03:33.420 --> 00:03:37.760 If this was an object with move semantics, it would indeed move it. 00:03:37.770 --> 00:03:40.770 But since these are just primitive integers, 00:03:40.780 --> 00:03:44.960 it will actually copy the value 8 over this integer 6. 00:03:45.740 --> 00:03:50.220 This other element with the integer of 8 is also unwanted because 00:03:50.220 --> 00:03:53.210 this is just the leftover from the rearrangement. 00:03:53.220 --> 00:03:57.950 So the function will treat this number just as the unwanted number 6. 00:03:58.340 --> 00:04:01.060 The next element is 1, which is wanted. 00:04:01.740 --> 00:04:04.920 Since this 8 is a leftover and unwanted, 00:04:04.930 --> 00:04:08.280 we need to copy this wanted number 1 over it because 00:04:08.290 --> 00:04:11.160 unwanted numbers always have to be at the end. 00:04:12.440 --> 00:04:14.960 The next element is the unwanted 6. 00:04:14.960 --> 00:04:19.360 And since this 1 is also unwanted, we don't do anything. 00:04:19.370 --> 00:04:22.360 All of the unwanted elements are at the end. 00:04:22.740 --> 00:04:23.230 Finally, 00:04:23.230 --> 00:04:27.340 this number 5 is wanted, so we can make a copy of it and 00:04:27.340 --> 00:04:30.260 put it over the last unwanted number. 00:04:31.240 --> 00:04:34.500 And this is how the final vector looks like. If you 00:04:34.500 --> 00:04:36.080 ignore these last two elements, 00:04:36.080 --> 00:04:39.410 you can see that all of the wanted numbers from the original array are 00:04:39.410 --> 00:04:45.970 correctly placed, and all of the 6s are removed. But these leftovers are 00:04:45.970 --> 00:04:51.640 still here at the end, and we don't need them. Fortunately, the remove 00:04:51.640 --> 00:04:55.700 function will return the iterator that points to the start of the unwanted 00:04:55.700 --> 00:05:01.170 part of the array, in this case, the position of this number 6. We can use 00:05:01.170 --> 00:05:05.970 this return value to pass this function as a first argument of the erase 00:05:05.970 --> 00:05:06.550 range. 00:05:07.740 --> 00:05:10.440 The ending iterator will be the end of the array. 00:05:10.450 --> 00:05:14.720 So, in other words, these unwanted leftovers will be deleted 00:05:14.720 --> 00:05:17.460 from the vector by the erase function. 00:05:18.340 --> 00:05:22.180 This combination of remove and erase function is so common that 00:05:22.180 --> 00:05:25.060 we call this technique erase‑remove idiom. 00:05:25.740 --> 00:05:30.030 You use this idiom whenever you want to eliminate elements that fulfill a 00:05:30.030 --> 00:05:36.560 certain criteria, in this case, the number 6. Since this idiom is so common, 00:05:36.570 --> 00:05:42.930 C++ 20 provides us with another function named erase, but don't confuse this 00:05:42.930 --> 00:05:45.960 erase function with the vector's erase function. 00:05:46.340 --> 00:05:49.840 The erase function we used belongs to the vector class while 00:05:49.840 --> 00:05:53.470 this new function belongs to the global scope, and it is 00:05:53.470 --> 00:05:56.060 defined inside of the vector header. 00:05:57.140 --> 00:06:00.030 This new function is combining the vector's erase 00:06:00.030 --> 00:06:02.290 function and the remove function. 00:06:02.300 --> 00:06:06.350 It basically implements the erase‑remove idiom on its own. 00:06:06.840 --> 00:06:10.660 All we need to provide is the whole vector and then the value we 00:06:10.660 --> 00:06:14.910 want to remove from that vector. Erase function will assume that we 00:06:14.910 --> 00:06:18.360 want to work with the whole array, and it will first reorganize the 00:06:18.360 --> 00:06:22.450 vector like the remove function and then erase the leftovers like 00:06:22.450 --> 00:06:24.160 the old erase function. 00:06:24.540 --> 00:06:27.760 This is so much easier than using two functions and 00:06:27.760 --> 00:06:30.260 passing iterators all over the place. 00:06:31.240 --> 00:06:33.370 Now that we know how these functions work, let's 00:06:33.370 --> 00:06:35.260 use them on the subject's vector. 00:06:35.640 --> 00:06:38.410 I want to work on the whole vectors, so I will pass 00:06:38.410 --> 00:06:40.860 both iterators to the remove function. 00:06:41.240 --> 00:06:45.520 The third argument is a value, but this is a little bit more complicated 00:06:45.520 --> 00:06:49.360 because subjects are not a primitive data types like integers. 00:06:49.740 --> 00:06:53.400 Fortunately, there is a complimentary remove_if function, which 00:06:53.400 --> 00:06:55.750 does the same thing as the remove function. 00:06:55.760 --> 00:06:58.580 But instead of passing the value you want to remove, 00:06:58.590 --> 00:07:01.260 you pass a function that returns a boolean. 00:07:01.640 --> 00:07:05.150 This function will be a lambda because we only need it here. 00:07:06.040 --> 00:07:09.060 Every element from this vector will be passed down to this 00:07:09.060 --> 00:07:12.290 function, so I will make sure that I pass it by reference 00:07:12.290 --> 00:07:14.460 to avoid unnecessary copies. 00:07:15.340 --> 00:07:18.560 I want to remove all of the subjects with the ID of 2. 00:07:18.860 --> 00:07:23.460 So if a subject has the ID of 2, I will return true. 00:07:24.240 --> 00:07:27.480 This is the list of IDs before the remove_if function. 00:07:27.490 --> 00:07:31.550 Can you try to guess what this vector will look like after we execute the 00:07:31.550 --> 00:07:35.910 function? Following the same logic from the previous example, we know that the 00:07:35.910 --> 00:07:39.480 first element is wanted because it has an ID of 1. 00:07:39.730 --> 00:07:41.780 The next element is unwanted. 00:07:41.790 --> 00:07:47.020 But since it's at the end, we don't care. However, 3 is wanted, so we need 00:07:47.020 --> 00:07:51.590 to call the move assignment operator to move the subject 3 over to subject 00:07:51.590 --> 00:07:56.040 2. Since moving invalidates old subjects, 00:07:56.050 --> 00:08:01.750 this element is left with an empty object with the ID of 0. ID 2 00:08:01.750 --> 00:08:05.160 is unwanted, so we move on to the last element. 00:08:06.240 --> 00:08:11.030 Since unwanted elements have to be at the end, we move assign these four over 00:08:11.030 --> 00:08:16.080 the empty subject because that is the first unwanted subject in the array. 00:08:16.080 --> 00:08:19.880 This invalidates the object at the last position. 00:08:19.890 --> 00:08:25.560 So the final ID list is 1, 3, 4, 2, and 0. 00:08:26.440 --> 00:08:32.260 I can prove this by compiling this program, and it all works as we expected. 00:08:32.270 --> 00:08:35.150 It only needs to do two move assignments to 00:08:35.150 --> 00:08:37.559 reorganize elements in the right way. 00:08:38.340 --> 00:08:42.210 Now we can pass the result of this function to the vector's erase function. 00:08:42.220 --> 00:08:47.540 But instead of doing that, let's remove this and use the new erase function. 00:08:47.550 --> 00:08:52.050 I just need to pass in the vector and the condition presented with lambda. 00:08:52.940 --> 00:08:57.260 When working with lambdas, we can use the complimentary erase_if function, 00:08:57.260 --> 00:09:01.360 which combines the vector's erase and the remove_if algorithm. 00:09:02.240 --> 00:09:05.460 And if I now compile this, you can see that the elements with 00:09:05.460 --> 00:09:08.750 the ID of 2 are also deleted from the vector. 00:09:09.640 --> 00:09:12.230 Okay, that is all I have to say on vectors. 00:09:12.240 --> 00:09:15.060 If you have any questions, ask them in the discussion.