WEBVTT 00:01.260 --> 00:05.100 Hello again! In this video, we're going to have an introduction to iterators. 00:07.010 --> 00:11.510 If you've done a traditional C++ course, you've probably seen code like this. So you have an array 00:11.510 --> 00:15.080 of characters, or ints, or objects of some kind. 00:16.710 --> 00:20.850 If you use the name of the array on its own, that's a pointer to the first element. 00:21.330 --> 00:23.280 So we have p pointing to the first element. 00:24.210 --> 00:30.420 And then if you add the number of elements, you get a pointer which points just after the last element. 00:31.890 --> 00:33.930 So that does point to memory, but not to memory 00:33.930 --> 00:35.040 that is part of the array. 00:35.970 --> 00:40.320 Then you have this pointer, you can [dereference] it to get the first element. 00:40.830 --> 00:45.060 Then you increment it to move on to the second element. And you keep on doing that. 00:45.540 --> 00:51.900 And then when this pointer is equal to the last plus one element, you terminate the loop, so you 00:51.900 --> 00:53.760 stop dereferencing and incrementing it. 01:00.440 --> 01:05.100 So the final value of the pointer is not the address of any element in the actual array. 01:05.660 --> 01:10.370 If you use that pointer, then you get undefined behaviour, and I'm sure you've all seen that! 01:11.720 --> 01:15.290 So this pointer represents an imaginary element after the last element. 01:20.020 --> 01:24.890 When we're using standard string, we could use the same approach, because that's implemented as an 01:24.890 --> 01:29.830 array. We can get the address of the first elements and use a pointer to that and increment it. 01:30.370 --> 01:32.050 But that's not really the right way to do things. 01:33.010 --> 01:37.720 It'll work for string and vector, which are implemented as arrays, but there are other containers in 01:37.720 --> 01:40.510 the standard library which are not, and that approach won't work. 01:41.110 --> 01:42.610 So let's learn the proper way to do it. 01:42.940 --> 01:43.960 For library containers. 01:44.560 --> 01:47.260 We use something called an iterator, and it works the same way. 01:47.680 --> 01:49.900 So we have some standard string. 01:50.770 --> 01:54.910 We get an iterator instead of a pointer, which represents the first element. 01:55.630 --> 01:59.350 This is a member of the string object. 01:59.380 --> 02:00.520 It's a class itself. 02:01.240 --> 02:05.230 It may be implemented around a pointer, but it doesn't really matter. We don't need to know how it 02:05.230 --> 02:05.680 works. 02:07.730 --> 02:13.160 And then we compare the current position of the iterator to the last plus one element. 02:14.030 --> 02:18.710 And then if it's all okay, we can dereference the iterator and that'll give us the data in the 02:18.710 --> 02:22.250 current element. And then we increment it to move on to the next element. 02:22.820 --> 02:26.710 So it has exactly the same form as the loop that we had before. 02:26.720 --> 02:28.850 It's just we're using iterators instead of pointers. 02:32.710 --> 02:37.960 So the question is, what do we do to get these iterators? And the string provides two member functions 02:37.960 --> 02:43.930 which will do that. The begin member function will return an iterator to the first element and the end 02:43.930 --> 02:48.340 member function will return an iterator to the last plus one element. 02:49.480 --> 02:51.430 So that's the element off the end of the string. 02:53.030 --> 02:58.040 And, as with pointers, that's not a valid iterator. If you dereference it, you get undefined 02:58.040 --> 02:58.550 behaviour. 02:59.360 --> 03:03.680 And also these iterators are specific to the object that you call the member function on. 03:04.220 --> 03:06.860 If you try to mix them with iterators to other objects, 03:07.250 --> 03:08.690 again, you get undefined behaviour. 03:09.380 --> 03:14.210 So if you have begin() from one string object and and end() from a different string object, your program 03:14.540 --> 03:15.020 is probably going to crash. 03:20.470 --> 03:26.110 So this is what it might look like in memory. So we have the string object header with the element 03:26.110 --> 03:32.530 count and the pointer to the data. Then we have this memory block, which contains the data. When we 03:32.530 --> 03:33.140 call begin, 03:33.160 --> 03:35.590 we get an iterator to this first element. 03:39.040 --> 03:42.790 Then we dereference the iterator, to get the data in this element. 03:43.570 --> 03:47.260 And when we increment the iterator, it moves on to the next element. 03:49.520 --> 03:54.590 And then we have this non-existent element, the last plus one element. 03:55.370 --> 03:59.450 And this corresponds to the return value from the end member function. 04:04.520 --> 04:05.840 So now we can finish off the loop. 04:06.170 --> 04:12.670 We say that the iterator equals the result of calling begin() on this string, and then we want to compare 04:12.680 --> 04:14.540 it to the result of calling end(). 04:17.200 --> 04:22.570 So begin will be the start of the string and the iterator will be incremented as we go through it. 04:23.170 --> 04:28.210 If the iterator is equal to the return value from end(), we know that we've seen all the data and we need 04:28.210 --> 04:29.980 to stop and terminate the loop. 04:32.060 --> 04:37.730 And one useful feature of this is that, if there's no data in the container, if it's empty, then begin() 04:37.730 --> 04:39.650 and end() will return the same value. 04:40.070 --> 04:44.480 So this will fail and the loop body does not get executed at all. 04:47.350 --> 04:49.180 So here is our code for this. 04:50.290 --> 04:51.760 So we have our string. 04:52.510 --> 04:55.360 We call begin to get the iterator to the first element. 04:55.720 --> 04:58.330 And then we compare it to the last plus one element. 05:02.610 --> 05:05.540 So we get the elements in the string, H e l l o. 05:10.150 --> 05:12.760 And just to show that nothing happens if it's empty. 05:16.770 --> 05:17.400 So there we are. 05:18.420 --> 05:19.530 We don't get any output. 05:19.800 --> 05:21.450 The loop doesn't get executed. 05:26.640 --> 05:33.930 And in fact, we can simplify this by writing it as a for loop so we can do.. for iterator equals 05:33.930 --> 05:34.290 begin() 05:38.500 --> 05:40.780 While the iterator is not equal to the end of the string. 05:47.810 --> 05:49.520 Then we increment it. 05:55.550 --> 05:58.430 Nope, made a mistake somewhere. Ah, extra semi-colon. 06:03.070 --> 06:04.470 (The danger of copying and pasting!) 06:04.800 --> 06:06.060 Okay, so you can see that works. 06:06.360 --> 06:08.310 That's exactly the same, equivalent loop. 06:11.300 --> 06:13.550 And we can also do this with a vector as well. 06:13.700 --> 06:18.380 In fact, we can do it with any container, but we didn't know any others, apart from string and vector at 06:18.380 --> 06:18.890 the moment. 06:20.060 --> 06:22.460 So we define our vector. 06:24.560 --> 06:26.660 Then we get an iterator to the first element. 06:27.830 --> 06:32.000 So the int parameter is actually part of the type for the object, so we need to include that. 06:33.410 --> 06:37.840 Then we can get our iterator to the first element. Then we compare it against the last 06:37.860 --> 06:38.300 plus one 06:38.300 --> 06:40.220 iterator and... 06:41.910 --> 06:43.620 There we go, three one four one five nine. 06:44.730 --> 06:46.650 Okay, so that's it for this video. 06:47.010 --> 06:47.820 I'll see you next time. 06:47.820 --> 06:49.860 But meanwhile, keep coding!