The name of an array is actually a pointer to the first element in the array. Writing
myArray[3] tells the compiler to return the element that is 3 away from the starting element of myArray.
This explains why arrays are always passed by reference: passing an array is really passing
a pointer.
This also explains why array indices start at 0: the first element of an array is the element that is 0 away from the start of the array
pointer arithmetic is a way of using subtraction and addition of pointers to move around between locations in memory, typically between array elements. Adding an integer n to a pointer produces a new pointer pointing to n positions further down in memory.
Take the following code snippet:
long arr [] = {6 ,0,9,6};
long * ptr = arr ;
ptr ++;
long *ptr2 = arr + 3;
When we add 1 to ptr in line 3, we don’t just want to move to the next byte in memory, since each array element takes up multiple bytes; we want to move to the next element in the array. The C++ compiler automatically takes care of this, using the appropriate step size for adding to and subtracting from pointers. Thus, line 3 moves ptr to point to the second element of the array.
Similarly, we can add/subtracttwo pointers: ptr2 - ptr givesthe number of array elements between ptr2 and ptr (2). All addition and subtraction operations on pointers use the appropriate step size.
Because of the interchangeability of pointers and array names, array-subscript notation (the
form myArray[3]) can be used with pointers as well as arrays. When used with pointers, it
is referred to as pointer-subscript notation.
An alternative is pointer-offset notation, in which you explicitly add your offsetto the pointer
and dereference the resulting address. For instance, an alternate and functionally identical
way to express myArray[3] is *(myArray + 3).
You should now be able to see why the type of a string value is char : a string is actually an array of characters. When you set a char to a string, you are really setting a pointer to point to the first character in the array that holds the string.
You cannot modify string literals; to do so is either a syntax error or a runtime error, depending on how you try to do it. (String literals are loaded into read-only program memory at program startup.) You can, however, modify the contents of an array of characters. Consider the following example:
char courseName1[] = {’6 ’, ’. ’, ’0 ’, ’9 ’, ’6 ’, ’\0 ’ };
char *courseName2 = "6.096 ";
Attempting to modify one of the elements courseName1 is permitted, but attempting to modify one of the characters in courseName2 will generate a runtime error, causing the program to crash.