1 00:00:00,001 --> 00:00:06,121 [MUSIC] Okay, let's continue our discussion of how manipulate data and 2 00:00:06,121 --> 00:00:12,826 memory in the C language. our next topic is going to be how we talk 3 00:00:12,826 --> 00:00:17,386 about arrays. basically arrays are adjacent locations 4 00:00:17,386 --> 00:00:21,280 in memory that store the same type of data object. 5 00:00:21,280 --> 00:00:26,248 So here I'm in this example declaring an array of integers. 6 00:00:26,248 --> 00:00:33,082 it's name is big_array and there is 128 integers in that array. 7 00:00:33,082 --> 00:00:35,475 that number between those square brackets. 8 00:00:35,475 --> 00:00:40,815 this is going to allocate 512 bytes of adjacent memory starting at some 9 00:00:40,815 --> 00:00:43,639 location. What location? 10 00:00:43,639 --> 00:00:46,518 Doesn't really matter. The one I'm showing here is pretty 11 00:00:46,518 --> 00:00:51,280 arbitrary but it will be easy for us to use in our examples. 12 00:00:51,280 --> 00:00:54,990 but basically the compiler will take care of that and will decide where in memory 13 00:00:54,990 --> 00:01:00,050 that data will go. now our pointer arithmetic can be used 14 00:01:00,050 --> 00:01:04,060 for indexing in this array. So let's take a look at some examples 15 00:01:04,060 --> 00:01:06,837 here. First of all, let me start with just 16 00:01:06,837 --> 00:01:12,688 declaring a pointer to an integer type. and we're going to call that pointer 17 00:01:12,688 --> 00:01:15,220 array_ptr. Okay? 18 00:01:15,220 --> 00:01:20,252 Now, the very next statement says, let's take big array, and assign it to array 19 00:01:20,252 --> 00:01:24,055 ptr. Well, because that's an array, and I've 20 00:01:24,055 --> 00:01:28,172 declared it as, as such. C knows that I don't mean the value big 21 00:01:28,172 --> 00:01:31,480 array, because there's 128 different values. 22 00:01:31,480 --> 00:01:36,155 What I really mean is. The address of the first location of that 23 00:01:36,155 --> 00:01:39,412 array. So what's going to be assigned to that 24 00:01:39,412 --> 00:01:43,060 pointer is the address of that first location. 25 00:01:43,060 --> 00:01:46,880 That is exactly what the second statement is going to do. 26 00:01:46,880 --> 00:01:51,290 You'll notice in this case what I've said is go to the zeroth element of the array, 27 00:01:51,290 --> 00:01:55,882 the first one. And get me its address by using the 28 00:01:55,882 --> 00:01:59,272 ampersand. So, of course, that, that will be the 29 00:01:59,272 --> 00:02:02,816 same address. In the next case, I have now used an 30 00:02:02,816 --> 00:02:06,774 index of 3. That really corresponds to the 4th 31 00:02:06,774 --> 00:02:12,020 integer in the array. There is one at index 0, one at index 1. 32 00:02:12,020 --> 00:02:16,010 One at index 2. And the fourth one at index 3. 33 00:02:16,010 --> 00:02:19,200 So how far down the, the array is that one? 34 00:02:19,200 --> 00:02:22,940 Well, this is where that pointer arithmetic becomes important. 35 00:02:22,940 --> 00:02:27,924 Since integers are of size 4 bytes. What I really mean is at 12 bytes passed 36 00:02:27,924 --> 00:02:33,480 the start of the address is where that fourth element will begin. 37 00:02:33,480 --> 00:02:38,150 So the address that I just generate here is ff000c. 38 00:02:38,150 --> 00:02:44,355 That c corresponds to 12 in decimal. 3 times 4 bytes. 39 00:02:44,355 --> 00:02:48,970 Okay? Now, that's exactly the same thing as the 40 00:02:48,970 --> 00:02:52,526 next statement. And why is that? 41 00:02:52,526 --> 00:02:58,034 Because, in this case, what I've done is first taken a, a, a pointer to the first 42 00:02:58,034 --> 00:03:01,380 element. And then said, add 3. 43 00:03:01,380 --> 00:03:04,979 But remember, c knows that if we're adding to a pointer, we don't mean just 44 00:03:04,979 --> 00:03:09,837 add the value 3. But rather, the 3 times the size of the 45 00:03:09,837 --> 00:03:15,078 data element. Or in this case, an int 3 times 4. 46 00:03:15,078 --> 00:03:20,410 Okay, so now you can kind of see why C does this in this pointer arithmetic. 47 00:03:20,410 --> 00:03:25,160 What about the next statement? Here, I'm just saying big array plus 3. 48 00:03:25,160 --> 00:03:29,093 Well again, that's the s-, same situation as the very first line, where we've taken 49 00:03:29,093 --> 00:03:33,370 big array to indicate just the first address of the array. 50 00:03:33,370 --> 00:03:38,280 And the plus 3, again, is to an address. We're adding 3 to an address. 51 00:03:38,280 --> 00:03:40,872 So we're going to multiply that by the size of the data item. 52 00:03:40,872 --> 00:03:44,160 Alright, in this next example is a bit more complex. 53 00:03:44,160 --> 00:03:47,698 You'll notice that it's using 2D reference stars, one on the left hand 54 00:03:47,698 --> 00:03:53,040 side and one on the right hand side. The one on the right hand side is saying 55 00:03:53,040 --> 00:03:57,840 get the value at array pointer, and use that as an address, and go get the value 56 00:03:57,840 --> 00:04:02,900 there. Well, what is the array pointer been set 57 00:04:02,900 --> 00:04:06,700 to? Well, we had just set it to this address 58 00:04:06,700 --> 00:04:11,551 that ended in 0c. So we're going to go into that location, 59 00:04:11,551 --> 00:04:16,383 get the value stored there, add 1 to it. Okay, and then where are we going to go 60 00:04:16,383 --> 00:04:19,330 put it. Where are we going to put the result? 61 00:04:19,330 --> 00:04:23,173 Well, the result, it says, to go put in this, at the same place where we got the 62 00:04:23,173 --> 00:04:28,292 value. So it says, go to the address pointed to 63 00:04:28,292 --> 00:04:36,126 by address by array_ptr. That star in front, again, is saying the 64 00:04:36,126 --> 00:04:43,460 reference, the value in array ptr and go to that location. 65 00:04:43,460 --> 00:04:50,790 Well again the value in array _ptr is 0c, so we going to go to that location again. 66 00:04:50,790 --> 00:04:54,158 So what does this accomplish? Well, what this has accomplished is to 67 00:04:54,158 --> 00:04:59,390 increment the value stored in that fourth element of the array. 68 00:04:59,390 --> 00:05:04,370 At index 3. but it does not change the value of the 69 00:05:04,370 --> 00:05:08,720 array_ptr variable. That stayed the same. 70 00:05:08,720 --> 00:05:13,188 And it's still pointing to that address 0c. 71 00:05:13,188 --> 00:05:17,590 Okay, what about our next statement. You'll notice here we're trying to 72 00:05:17,590 --> 00:05:23,560 reference the 130th element of the array. well, our array was only 128. 73 00:05:23,560 --> 00:05:26,990 What happens here? Well turns out C doesn't care. 74 00:05:26,990 --> 00:05:31,080 It just applies the same point of arithmetic rules it did before. 75 00:05:31,080 --> 00:05:35,606 In this case multiplying 130 by 4 and adding that to the starting address of 76 00:05:35,606 --> 00:05:42,660 the array, and will just give us an address that points to that location. 77 00:05:42,660 --> 00:05:44,990 That turns out to be out of the bounds of our array. 78 00:05:44,990 --> 00:05:48,720 We've gone too past the end of the array. But C doesn't check for that. 79 00:05:48,720 --> 00:05:52,900 And will see later on, other languages do, but C does not. 80 00:05:52,900 --> 00:05:57,609 And that's one of the dangers in using C. Now, the general rule here is that if I 81 00:05:57,609 --> 00:06:04,190 have an expression that looks like this. that is the same as saying bigger A plus 82 00:06:04,190 --> 00:06:09,095 I because just bigger A by itself is the starting address. 83 00:06:09,095 --> 00:06:14,120 And plus i gets multiplied by the size of the element of the array before its added 84 00:06:14,120 --> 00:06:19,650 to that address. So that it implicitly computes this 85 00:06:19,650 --> 00:06:24,359 expression. Which you'll notice is the same thing by 86 00:06:24,359 --> 00:06:29,104 saying the starting address of the array plus i times the size of the element of 87 00:06:29,104 --> 00:06:34,168 the array. Okay, and in C we can use the size of 88 00:06:34,168 --> 00:06:41,132 function to tell us what size the element of the array is. 89 00:06:41,132 --> 00:06:47,630 Okay, let's take a look now at strings rather than intergers. 90 00:06:47,630 --> 00:06:52,364 A C-style string, a character string, is represented by an array of light. 91 00:06:52,364 --> 00:06:57,815 Each byte has a code in it two digit hex code and one of the more popular ones is 92 00:06:57,815 --> 00:07:03,760 the ASCII code that uses one byte for each character. 93 00:07:03,760 --> 00:07:08,830 And here you see a map of most of the characters in the ASCII character set. 94 00:07:08,830 --> 00:07:13,978 And the way C's, indicates the end of a string, is with a byte that only had the 95 00:07:13,978 --> 00:07:18,956 value 0 in it. So a string like Harry Potter for example 96 00:07:18,956 --> 00:07:23,430 would be represented by the following sequence of bytes. 97 00:07:23,430 --> 00:07:27,912 You'll notice that there is a byte for the space, between the two names. 98 00:07:27,912 --> 00:07:33,530 the, the value, the code for a blank is 32x. 99 00:07:33,530 --> 00:07:37,138 And, a 0 at the end, signifying the end of the string. 100 00:07:37,138 --> 00:07:40,743 So, the string, Harry Potter, is a 13 byte array. 101 00:07:40,743 --> 00:07:45,607 There are 12 characters including the blank, plus, a 0 at the end, to indicate 102 00:07:45,607 --> 00:07:49,750 the end. Now, the reason we put a 0 at the end, 103 00:07:49,750 --> 00:07:57,260 often referred to as a null 0, is, so that we can tell the end of the string. 104 00:07:57,260 --> 00:08:00,844 Alright, how would be compute the string length if we were given the starting 105 00:08:00,844 --> 00:08:04,910 point of the string? We would move 1 byte at a time down the 106 00:08:04,910 --> 00:08:09,280 string and checking for 0. When we find the 0, we would know we were 107 00:08:09,280 --> 00:08:11,697 at the end. And that's how we would compute the 108 00:08:11,697 --> 00:08:15,400 string length. Okay. 109 00:08:15,400 --> 00:08:18,780 Because we're dealing with 1 byte at a time, we don't have to worry about issues 110 00:08:18,780 --> 00:08:23,598 of little Indian, or big Indian, when we think about character strings. 111 00:08:23,598 --> 00:08:27,360 So, if we had the character string 1, 2, 3, 4, 5, it would actually be represented 112 00:08:27,360 --> 00:08:33,140 in precisely the same in little Indian machines, or big Indian machines. 113 00:08:33,140 --> 00:08:37,556 Now, later on we'll learn about unicode characters that have more bytes per 114 00:08:37,556 --> 00:08:41,224 character. unicode is a newer standard than ASCII 115 00:08:41,224 --> 00:08:45,633 that allows us to represent all the character sets around the world. 116 00:08:45,633 --> 00:08:49,345 And we need a lot more codes to do that for all the different alphabets that 117 00:08:49,345 --> 00:08:53,272 exist. So they can reach up to 4 bytes per 118 00:08:53,272 --> 00:08:59,605 character and Java has, Java as well as C have libraries for unicode. 119 00:08:59,605 --> 00:09:04,160 But Java commonly uses 2 btyes per character that's sort of it's default. 120 00:09:04,160 --> 00:09:08,300 So we'll see later the complexity of this but for now strings are 1 bit per 121 00:09:08,300 --> 00:09:12,115 character. Okay let's take a look at the C function 122 00:09:12,115 --> 00:09:16,140 that we can use to print out data that we store in memory. 123 00:09:16,140 --> 00:09:23,180 By printing it out one character at a time or 1 byte at a time as 2x digits. 124 00:09:23,180 --> 00:09:28,940 Alright, so here's a simple function written in C and it's called show bytes. 125 00:09:28,940 --> 00:09:35,160 It doesn't return a value so we've put a void in front of it to indicate that. 126 00:09:35,160 --> 00:09:38,480 And then it takes two arguments. The first one is the address of the 127 00:09:38,480 --> 00:09:42,732 starting location and memory. So, you'll notice it has a star there 128 00:09:42,732 --> 00:09:48,860 because it says this is a pointer to to a location a memory is not a value. 129 00:09:48,860 --> 00:09:51,604 And to what it kind of thing is it pointing, well it's pointing to 130 00:09:51,604 --> 00:09:57,310 character. And then see a char is a 1 byte quantity. 131 00:09:57,310 --> 00:10:01,770 So in this case, we're just saying we're point to a byte in memory, right. 132 00:10:01,770 --> 00:10:05,670 And then the second argument is an integer that is the length of, of bytes 133 00:10:05,670 --> 00:10:10,030 that we want to print, the number of bytes we want. 134 00:10:10,030 --> 00:10:14,440 Okay, the body of the function is just the simple four loop that goes from 1 to 135 00:10:14,440 --> 00:10:18,290 length or 0 to length minus 1 and increments the index by 1 each time 136 00:10:18,290 --> 00:10:23,116 around. That is what that i plus plus indicates 137 00:10:23,116 --> 00:10:27,008 increment by 1. The body of the loop is a, is a simple 138 00:10:27,008 --> 00:10:32,338 print statement that, we'll see the format of in a second, but it prints two 139 00:10:32,338 --> 00:10:37,810 values. The first is an address, the second a 140 00:10:37,810 --> 00:10:40,890 data value. You'll notice that we're using start, 141 00:10:40,890 --> 00:10:43,540 that pointer. So this is going to be an address. 142 00:10:43,540 --> 00:10:48,757 Then we're adding i, the index to it. Since we're doing pointer arithmetic, 143 00:10:48,757 --> 00:10:52,453 what do we multiple i by? Well, since it's only pointing to a 144 00:10:52,453 --> 00:10:56,864 character. We multiply by 1 because characters are 1 145 00:10:56,864 --> 00:11:00,045 byte. And then we print the data stored at that 146 00:11:00,045 --> 00:11:05,770 address by de-referencing that address. That's why we have the star on this side. 147 00:11:05,770 --> 00:11:09,102 So here we have the address. And here we have the data at that 148 00:11:09,102 --> 00:11:13,560 address. Our print format screen says that we 149 00:11:13,560 --> 00:11:20,973 start off by printing a pointer. Okay, that would be that first element 150 00:11:20,973 --> 00:11:24,780 here, that address. So we're going to print a pointer. 151 00:11:24,780 --> 00:11:31,092 Then we're going to insert a tab. Then we're going to print the, the number 152 00:11:31,092 --> 00:11:35,995 0 and the letter x. And we're going to do that to conform to 153 00:11:35,995 --> 00:11:42,659 C notation for hexadecimal values. And then the percent 0.2x is a special 154 00:11:42,659 --> 00:11:49,199 notation for saying, print the hexadecimal value of 2 digits, followed 155 00:11:49,199 --> 00:11:55,173 by a new line. That hexadecimal value of 2 digits is 156 00:11:55,173 --> 00:12:00,870 just the value stored at that address. Okay, the second element here in our 157 00:12:00,870 --> 00:12:04,180 print state. Okay, so that's a pretty nice little 158 00:12:04,180 --> 00:12:07,696 function. And we can call it from with anything 159 00:12:07,696 --> 00:12:12,824 actually, even integers not just not just characters. 160 00:12:12,824 --> 00:12:16,541 Because, when we call it here showing an example of another function that's 161 00:12:16,541 --> 00:12:21,390 going to show integers. We can call show bytes by casting the 162 00:12:21,390 --> 00:12:27,118 address of the integer as a character, address. 163 00:12:27,118 --> 00:12:30,050 Okay? This is C casting. 164 00:12:30,050 --> 00:12:33,344 What we've done is said, we have an address here, but it's the address of an 165 00:12:33,344 --> 00:12:36,591 integer. Can you please treat that as the address 166 00:12:36,591 --> 00:12:39,828 of a character and then pass it to the function? 167 00:12:39,828 --> 00:12:42,931 Okay? And then the length here is just the size 168 00:12:42,931 --> 00:12:46,852 of the integer. So here we would expect to see 4 bytes 169 00:12:46,852 --> 00:12:51,270 that are going to be printed out by the function. 170 00:12:51,270 --> 00:12:54,530 Let's take a look at some sample executions of this function. 171 00:12:54,530 --> 00:13:01,534 Here we see the function in integer 1,2,3,4,5 being placed in a, and when we 172 00:13:01,534 --> 00:13:08,538 call will we'll first print out that statement and then call the function, 173 00:13:08,538 --> 00:13:15,422 show int. So here's that first print statement, 174 00:13:15,422 --> 00:13:20,370 directly from above that prints out the value. 175 00:13:20,370 --> 00:13:25,060 And then you'll see that inside of show int, we've of course called show bytes. 176 00:13:25,060 --> 00:13:31,822 And it went for 4 bytes printing out each one in turn, successful bytes at cc, cd, 177 00:13:31,822 --> 00:13:36,178 ce and cf. And the rest of that address, you know, 178 00:13:36,178 --> 00:13:40,710 is pretty arbritrary. Again, it's just where that that integer 179 00:13:40,710 --> 00:13:42,974 a happened to be. And the, the integers. 180 00:13:42,974 --> 00:13:47,460 the, the values that got printed are 39, 30, 0, and 0x. 181 00:13:47,460 --> 00:13:54,390 Which we know, from previous examples, are the rep-, the hex representation of 182 00:13:54,390 --> 00:13:57,780 that value. Okay? 183 00:13:57,780 --> 00:14:00,570 So this is indicating that our machine is little Indian. 184 00:14:00,570 --> 00:14:03,460 Because we saw the least significant byte first. 185 00:14:03,460 --> 00:14:07,660 And that's how we can write a simple program to check the[UNKNOWN] of our 186 00:14:07,660 --> 00:14:09,213 machine.