WEBVTT 00:01.210 --> 00:05.320 Hello again! In this video, we are going to look at adding elements to strings. 00:07.910 --> 00:13.130 The string class has a member function called append. If we call this on a string. 00:13.440 --> 00:16.520 it arguments is going to be added to that string, at the back. 00:16.790 --> 00:22.910 So if we have a string whose data is "Hello" and we append another string whose data is " world", then the 00:22.910 --> 00:26.210 first string ends up containing the data "Hello world". 00:28.410 --> 00:31.530 And we could also do this using the plus equals operator. 00:33.480 --> 00:39.420 So these come from the two different interfaces to the string class. The plus equals is from the original 00:39.420 --> 00:44.370 implementation, which made it look like a built-in type, and the append member function is to make 00:44.370 --> 00:46.200 it more like a standard container. 00:47.850 --> 00:50.310 The main difference is that you can do a lot more with append. 00:50.370 --> 00:54.480 It takes a lot more argument types, although it does not take a single character. 00:54.810 --> 00:57.390 So if you want that, you need the plus equals operator. 00:58.830 --> 01:04.740 As an example, you can use a substring directly without having to create a substring object. 01:05.760 --> 01:09.480 So we have a string "wow!!!!" with four exclamation marks. 01:10.320 --> 01:17.460 We have an offset three, which is the element which is the first exclamation mark and then the count 01:17.460 --> 01:17.860 is two. 01:17.880 --> 01:25.320 So we're going to append these two exclamation marks, and then the string will be "Hello world double exclamation 01:25.320 --> 01:25.620 mark". 01:28.330 --> 01:33.730 So here is that code. We are taking "Hello", and appending "world" to it. And then we take another 01:33.730 --> 01:37.330 string "Hello" and append the exclamation marks to it. 01:40.060 --> 01:43.780 And we get "Hello world" and "Hello", exclamation exclamation. 01:46.340 --> 01:50.060 Something that you'll probably find more useful is the insert member function. 01:50.540 --> 01:52.640 So again, this is part of the container interface. 01:52.700 --> 01:55.610 So all the containers have a similar function. 01:58.290 --> 02:04.590 In this case, it takes an offset, which is the index where the string is going to be inserted 02:04.590 --> 02:07.350 and the string object that it is going to insert. 02:09.330 --> 02:17.280 Two is the index of the letter 'r' and then this is going to insert "lte" before the letter 'r', so the 02:17.280 --> 02:22.950 result of that is going to be that the string will contain the data "folder". 02:24.420 --> 02:30.630 And as another example, if we have a string "care" and "omp" and then at index one. 02:31.080 --> 02:34.260 So that is before the letter a. It is going to insert "omp". 02:35.460 --> 02:37.050 So we end up with "compare". 02:39.200 --> 02:40.910 And there is that code. 02:43.490 --> 02:46.010 So we get "folder" and "compare". 02:48.450 --> 02:50.040 Some more things you can do. 02:51.030 --> 02:52.500 You can use a substring. 02:52.740 --> 03:00.960 So if we give a string object, offset, count, then it will insert that part of the string, into the string 03:00.960 --> 03:01.650 that it is called on. 03:02.070 --> 03:07.920 So if we have a string with "trombone", that is going to be the letters 'b' and 'o', and then it is going 03:07.920 --> 03:10.860 to insert that before the second letter 'x'. 03:11.490 --> 03:13.170 So we end up with "xbox". 03:13.590 --> 03:18.170 And if any of Microsoft's lawyers are watching, this is a trademark, registered copyright, 03:18.180 --> 03:19.050 all rights reserved. 03:19.380 --> 03:20.190 Please don't sue me! 03:22.120 --> 03:24.970 We can also insert a single char, multiple times. 03:28.030 --> 03:34.630 So offset one, so that is before the 'a' in "cash". We are going to insert the letter 'o' three times, so we 03:34.630 --> 03:35.940 should get "crrrash". 03:39.750 --> 03:42.630 And so here is the code for those. So we have "trombone" 03:43.750 --> 03:50.410 into "xx", we have 'r' into "cash" and then one where we use the return value from a find operation. 03:50.710 --> 03:54.430 So you remember that find will return an index. 03:54.760 --> 03:56.950 So that is the index of the first matching character. 03:56.980 --> 04:03.430 So if we could find with argument 'o', it is going to be the index of that element. And then we can pass that 04:03.430 --> 04:07.570 index as the argument to a call to insert. 04:08.140 --> 04:14.890 So "opos" will be the position of 'o'. And we are being good careful programmers, so we make sure that 04:14.890 --> 04:21.490 we have actually got a valid index before we use it, because something might change and then we would be 04:21.490 --> 04:23.140 accessing an invalid element. 04:25.340 --> 04:32.210 And then we insert the letter 'o' twice, so that will go before the first occurrence of 'o' in this string. 04:37.190 --> 04:40.900 So there we are, we get "xbox", "crrrash" and "Hellooo". 04:48.130 --> 04:51.070 We can also use iterators with insert. 04:52.890 --> 04:56.610 You remember that the end iterator it will return the last plus one character. 04:56.970 --> 05:02.280 And if we subtract one from that, we get an iterator to the last character in the string. And then 05:02.280 --> 05:05.700 we can use that iterator as the first argument to insert. 05:06.300 --> 05:11.970 And this will insert the character 'l' before the last character of the string, which is the letter 'd'. 05:12.690 --> 05:14.610 So the string will now be "world". 05:16.240 --> 05:18.820 And we can also use the end iterator directly. 05:19.180 --> 05:24.700 That is the last character plus one, so if we insert before that, it go before the last character plus 05:24.700 --> 05:27.100 one, but after the last character. 05:28.600 --> 05:36.010 So if we get this end iterator and insert 'l' twice after it, this will insert 'l' twice after 05:36.010 --> 05:38.380 the last character and we get "skill". 05:43.470 --> 05:45.120 So let's try that. 05:47.540 --> 05:51.620 And there it is. So we get "world" and "skill". 05:57.110 --> 06:03.200 Something that you need to be careful about when you're using iterators with insert statements. The insert 06:03.200 --> 06:08.840 operation may cause any iterators that you had before the operation to become invalid. 06:09.410 --> 06:13.670 For example, the string has an internal memory buffer where it stores all the data. 06:14.060 --> 06:20.150 If that memory buffer is just big enough to store all the data it has, and you add some more data, then 06:20.150 --> 06:22.130 the buffer will no longer be big enough. 06:22.610 --> 06:28.430 In that case, the string is going to allocate a new buffer and copy all the data into the new buffer 06:28.430 --> 06:29.570 and then delete the old one. 06:30.380 --> 06:36.440 So if you had an iterator to the element when it was in the old buffer, then that iterator will no 06:36.440 --> 06:39.290 longer be valid, because that buffer is no longer part of the string. 06:39.770 --> 06:42.230 So that element is not part of the string's data. 06:46.140 --> 06:51.390 So as an example, if you get an iterator to the first element of a string and then you add some more 06:51.390 --> 06:52.470 data to the string. 06:52.860 --> 06:59.490 So in this case, we are adding the letter 'd' 16 times at the back of the string, and that is enough to 06:59.490 --> 07:05.340 cause the string to be reallocate its memory buffer. Or at least, it has been on every compiler that I have 07:05.340 --> 07:05.880 tried it with. 07:07.400 --> 07:10.340 And then all the data now goes into this new buffer. 07:11.240 --> 07:16.580 You would think that just inserting some elements would not change the position of the first element. 07:17.090 --> 07:20.270 But in fact, that is now pointing to an element that is not in the buffer. 07:20.780 --> 07:23.030 And you will get a memory error when the program runs. 07:26.250 --> 07:27.990 So here is the code. We have our string. 07:28.560 --> 07:35.580 We get an iterator to the first element. And then we insert the letter 'd' 16 times at the back of the string. 07:36.720 --> 07:40.770 and then we try to insert a character at the front of the string. 07:42.090 --> 07:43.170 So let's try that. 07:46.100 --> 07:46.460 Right! 07:47.450 --> 07:49.040 Okay, so the program has crashed. 07:55.930 --> 08:02.010 So the thing to do here is to, if you like, "refresh" the iterator. So we need to get a new iterator 08:02.040 --> 08:03.730 to the first element in the string. 08:07.220 --> 08:08.660 (So I can paste that back in) 08:09.530 --> 08:14.900 So we reassign the first voter, so it is now at the new first element. 08:15.200 --> 08:16.370 And then it should all work. 08:18.900 --> 08:19.680 Yep, there we are. 08:22.470 --> 08:25.860 OK, so that is one thing you need to watch out for with iterators and insert. 08:26.430 --> 08:28.020 Okay, so that's it for this video. 08:28.410 --> 08:29.280 I'll see you next time. 08:29.290 --> 08:31.110 But meanwhile, keep coding!