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Arrays

You must understand the concepts discussed in the previous pointers section before proceeding.
Arrays are a collection of items (i.e. ints, floats, chars) whose memory is allocated in a contiguous block of memory.
Arrays and pointers have a special relationship. This is because arrays use pointers to reference memory locations. Therefore, most of the times, pointer and array references can be used interchangeably.

Arrays :: Declaration and Syntax

A simple array of 5 ints would look like:
```
  int ia[5];
```
This would effectively make an area in memory (if availble) for ia, which is 5 * sizeof(int). We will discuss sizeof() in detail in Dynamic Memory Allocation. Basically sizeof() returns the size of what is being passed. On a typical 32-bit machine, sizeof(int) returns 4 bytes, so we would get a total of 20 bytes of memory for our array.
How do we reference areas of memory within the array? By using the [ ] we can effectively "dereference" those areas of the array to return values.
```
  printf("%d ", ia[3]);
```
This would print the fourth element in the array to the screen. Why the fourth? This is because array elements are numbered from 0.
Note: You cannot initialize an array using a variable. ANSI C does not allow this. For example:
```
  int x = 5;
  int ia[x];
```
This above example is illegal. ANSI C restricts the array intialization size to be constant. So is this legal?
```
  int ia[];
```
No. The array size is not known at compile time.
How can we get around this? By using macros we can also make our program more readable!
```
  #define MAX_ARRAY_SIZE 5
  /* .... code .... */
  
  int ia[MAX_ARRAY_SIZE];
```
Now if we wanted to change the array size, all we'd have to do is change the define statement!
But what if we don't know the size of our array at compile time? That's why we have Dynamic Memory Allocation. More on this later...
Can we initialize the contents of the array? Yes!
```
  int ia[5] = {0, 1, 3, 4};
  int ia[ ] = {0, 2, 1};
```
Both of these work. The first one, ia is 20 bytes long with 16 bytes initialized to 0, 1, 3, 4. The second one is also valid, 12 bytes initialized to 0, 2, 1. (Examples on a typical 32-bit machine).

Arrays :: Relationship with Pointers

So what's up with all this pointers are related to arrays junk? This is because an array name is just a pointer to the beginning of the allocated memory space. This causes "problems" in C, as the Limitations sub-section will show.
Let's take this example and analyze it:
```
  int ia[6] = {0, 1, 2, 3, 4, 5};                             /* 1 */
  int *ip;                                                    /* 2 */

  ip = ia;                /* equivalent to ip = &ia[0]; */    /* 3 */
  ip[3] = 32;             /* equivalent to ia[3] = 32;  */    /* 4 */
  ip++;                   /* ip now points to ia[1]     */    /* 5 */
  printf("%d ", *ip);     /* prints 1 to the screen     */    /* 6 */
  ip[3] = 52;             /* equivalent to ia[4] = 52   */    /* 7 */
```
Ok, so what's happening here? Let's break this down one line at a time. Refer to the line numbers on the side:
1. Initialize ia
2. Create ip: a pointer to an int
3. Assign ip pointer to ia. This is effectively assigning the pointer to point to the first position of the array.
4. Assign the fourth position in the array to 32. But how? ip is just a pointer?!?! But what is ia? Just a pointer! (heh)
5. Use pointer arithmetic to move the pointer over in memory to the next block. Using pointer arithmetic automatically calls sizeof().
6. Prints ia[1] to the screen, which is 1
7. Sets ia[4] to 52. Why the fifth position? Because ip points to ia[1] from the ip++ line.
Now it should be clear. Pointers and arrays have a special relationship because arrays are actually just a pointer to a block of memory!

Arrays :: Multidimensional

Sometimes its necessary to declare multidimensional arrays. In C, multidimensional arrays are row major. In other words, the first bracket specifies number of rows. Some examples of multidimensional array declarations:
```
  int igrid[2][3] = { {0, 1, 2}, {3, 4, 5} };
  int igrid[2][3] = { 0, 1, 2, 3, 4, 5 };
  int igrid[ ][4] = { {0, 1, 2, 3}, {4, 5, 6, 7}, {8, 9} };
  int igrid[ ][2];
```
The first three examples are valid, the last one is not. As you can see from the first two examples, the braces are optional. The third example shows that the number of rows does not have to be specified in an array initialization.
This seems simple enough. But what if we stored pointers in our arrays? This would effectively create a multidimensional array! Since reinforcement of material is key to learning it, let's go back to getopt. Remember the variable argv? It can be declared in the main function as either **argv or *argv[]. What does **argv mean? It looks like we have two pointers or something. This is actually a pointer to a pointer. The *argv[] means the same thing, right? Imagine (pardon the crappy graphics skills):
```
   argv
  +---+
  | 0 | ---> "./junk"
  +---+
  | 1 | ---> "-b"
  +---+
  | 2 | ---> "gradient"
  +---+
  | 3 | ---> "yeehaw"
  +---+
```
So what would argv[0][1] be? It would be the character '/'. Why is this? It's because strings are just an array of characters. So in effect, we have a pointer to the actual argv array and a pointer at each argv location to each string. A pointer to a pointer. We will go more in depth into strings later.

Arrays :: Limitations

Because names of arrays represents just a pointer to the beginning of the array, we have some limitations or "problems."

No Array Out of Bounds Checking. For example:
```
  int ia[2] = {0, 1};

  printf("%d ", ia[2]);
```
The above code would segfault, because you are trying to look at an area of memory not inside the array memory allocation.
Array Size Must be Constant or Known at Compile-time. See Arrays :: Declaration and Syntax.
Arrays Cannot be Copied or Compared. Why? Because they are pointers. See Weiss pg. 149 for a more in-depth explanation.
Array Index Type must be Integral.

Another limitation comes with arrays being passed into functions. Take for example:
```
  void func(int ia[])
  void func(int *ia)
```
Both are the same declaration (you should know why by now). But why would this cause problems? Because only the pointer to the array is passed in, not the whole array. So what if you mistakenly did a sizeof(ia) inside func? Instead of returning the sizeof the whole array, it would return the size of a pointer which corresponds to the word size of the computer.

Arrays Review

Arrays are, in simple terms, just a pointer! Remember that!
There isn't much to review. Remember arrays have limitations because they are inherently just a pointer. Wait we mentioned that already. :)