WEBVTT 00:00.180 --> 00:04.020 Hello again! In this video, we are going to look at removing algorithms. 00:05.450 --> 00:10.010 You can probably guess what these do, but they may not do it in the way that you expect. 00:10.370 --> 00:15.770 So these will remove elements from an iterator range, but they will only "logically" remove them. 00:16.070 --> 00:17.780 They will not actually destroy the elements. 00:18.320 --> 00:22.310 So the removed elements are still physically present in the container. 00:24.420 --> 00:28.740 The remove algorithm call takes the iterator range and some object. 00:29.880 --> 00:35.250 Each element which is equal to that object, is moved to the back of the container, so the removed 00:35.250 --> 00:36.570 element still exists. 00:37.080 --> 00:38.850 Its value is no longer defined. 00:39.360 --> 00:43.050 So accessing it, or trying to access, it is undefined behavior. 00:44.550 --> 00:48.330 The other elements which are behind it are moved forwards, to make room for it. 00:48.690 --> 00:51.150 So the container still has the same number of elements. 00:54.240 --> 00:59.910 The return value from remove is an iterator to the first removed element. 01:01.930 --> 01:05.680 So let's imagine we have a vector with the elements 3, 1, 4, 1, 5, 9. 01:06.160 --> 01:11.620 So it looks like this. We have the control block with the element count, six. And a pointer to the 01:11.620 --> 01:15.100 data, and the data has 3, 1, 4, 1, 5, 9. 01:16.120 --> 01:21.700 If we call remove() and we give the entire vector as the iterator range and the argument 1, 01:22.210 --> 01:24.640 this will remove the elements with value 1. 01:25.360 --> 01:30.580 So these two elements are going to be moved to the back of the container and all the elements behind 01:30.580 --> 01:32.260 them are going to be moved forwards. 01:34.220 --> 01:35.990 So the vector is going to look like this. 01:36.230 --> 01:41.180 We have the elements which are not equal to 1 at the front of the vector, and the elements which are 01:41.180 --> 01:43.580 equal to 1 are at the back of the vector. 01:43.850 --> 01:47.900 So these are the removed elements. And their value is no longer defined. 01:49.280 --> 01:51.110 The element count is still 6. 01:51.110 --> 01:55.790 The vector still thinks it has 6 elements. The return value 01:55.790 --> 02:00.230 from remove() will be an iterator to this one, the first removed element. 02:01.580 --> 02:02.810 So let's try this out. 02:03.290 --> 02:08.630 We have our vector with those elements. Then we print out the data and size. 02:09.080 --> 02:10.160 Then we call remove(). 02:10.820 --> 02:16.010 We give the entire container as the estimated range, and 1 as the value to be removed. 02:16.850 --> 02:18.320 And then we print out the result. 02:19.070 --> 02:20.120 So let's see what happens. 02:22.100 --> 02:24.710 So there's the original vector, with 6 elements. 02:25.340 --> 02:29.510 After calling remove(), the vector still has 6 elements and we can see that 02:29.510 --> 02:32.660 all the elements, which are not 1, have been moved forwards. 02:33.140 --> 02:36.050 And then these two actually have the same value. 02:36.380 --> 02:41.090 So it looks as though this implementation has done a slight optimization, by not actually overwriting 02:41.090 --> 02:41.690 these elements. 02:43.550 --> 02:45.640 So these elements could be five and nine. 02:45.680 --> 02:48.620 They could be one one, or they could be complete garbage. 02:49.070 --> 02:53.660 Who knows! If we want to actually, physically, remove the elements from the container, 02:54.050 --> 02:57.140 we need to call the erase() member function for that container. 02:59.230 --> 03:06.780 erase() will take an iterator, which is the first element to be erased and a "last plus one" iterator, 03:06.900 --> 03:13.230 which is one after the last element to be erased. When we call remove(), we get an iterator 03:13.230 --> 03:16.470 returned, which is an iterator to the first removed element. 03:16.920 --> 03:23.400 So if we pass that as the first argument to erase(), it will start erasing from this element, and it will 03:23.400 --> 03:24.690 continue until that one. 03:24.900 --> 03:28.530 So this will erase all the remaining elements in the vector. 03:29.850 --> 03:33.000 And then we have a vector, which contains 3, 4, 5 and 9. 03:33.690 --> 03:36.300 So this will physically remove the elements from the container. 03:38.210 --> 03:42.350 So the vector will now look like this. We just have the elements which are not equal to 1: 03:42.830 --> 03:44.000 3, 4, 5 and 9. 03:44.780 --> 03:48.230 The elements which were equal to 1 have been destroyed. 03:48.240 --> 03:53.120 They are no longer in the container, and the vector now has 4 elements instead of 6. 03:55.030 --> 03:59.500 So here is the code we had before. We call remove() and we get the returned iterator. 04:01.180 --> 04:08.200 Then we pass that iterator as the first argument to erase(), and then we print out the vector again. 04:08.710 --> 04:10.480 So let's see what we get this time. 04:13.000 --> 04:15.250 So there is the original vector with 6 elements. 04:15.310 --> 04:19.270 There is the vector after calling remove(). And then we call erase(). 04:19.810 --> 04:23.170 So the vector now has 4 elements, which are 3, 4, 5 and 9. 04:23.590 --> 04:29.740 So the elements with value 1 have disappeared. They have actually been destroyed and are no longer part of the container. 04:30.520 --> 04:32.080 OK, so that is it for this video. 04:32.380 --> 04:35.440 I will see you next time, but until then, keep coding!