（摘）C++ Primer Plus, Fourth Edition---Using a Dynamic Array

After you create a dynamic array, how do you use it? First, think about the problem conceptually. The statement


int * psome = new int [10]; // get a block of 10 ints



creates a pointer psome that points to the first element of a block of ten int values. Think of it as a finger pointing to that element. Suppose an int occupies four bytes. Then, by moving your finger four bytes in the correct direction, you can point to the second element. Altogether, there are ten elements, which is the range over which you can move your finger. Thus, the new statement supplies you with all the information you need to identify every element in the block.

Now think about the problem practically. How do you access one of these elements? The first element is no problem. Because psome points to the first element of the array, *psome is the value of the first element. That leaves nine more elements to access. The simplest way may surprise you if you haven't worked with C: Just use the pointer as if it were an array name. That is, you can use psome[0] instead of *psome for the first element, psome[1] for the second element, and so on. It turns out to be very simple to use a pointer to access a dynamic array, even if it may not immediately be obvious why the method works. The reason you can do this is that C and C++ handle arrays internally by using pointers anyway. This near equivalence of arrays and pointers is one of the beauties of C and C++. We'll elaborate on this equivalence in a moment. First, Listing 4.13 shows how you can use new to create a dynamic array and then use array notation to access the elements. It also points out a fundamental difference between a pointer and a true array name.
Listing 4.13 arraynew.cpp

// arraynew.cpp _ using the new operator for arrays


#include <iostream>


using namespace std;


int main()




...{


double * p3 = new double [3]; // space for 3 doubles


p3[0] = 0.2;                  // treat p3 like an array name


p3[1] = 0.5;


p3[2] = 0.8;


cout << "p3[1] is " << p3[1] << ". ";


p3 = p3 + 1;                  // increment the pointer


cout << "Now p3[0] is " << p3[0] << " and ";


cout << "p3[1] is " << p3[1] << ". ";


p3 = p3 - 1;                  // point back to beginning


delete [] p3;                 // free the memory


return 0;


}

Here is the output:

p3[1] is 0.5.

Now p3[0] is 0.5 and p3[1] is 0.8.

As you can see, arraynew.cpp uses the pointer p3 as if it were the name of an array, with p3[0] as the first element, and so on. The fundamental difference between an array name and a pointer shows in the following line:

p3 = p3 + 1; // okay for pointers, wrong for array names

You can't change the value of an array name. But a pointer is a variable, hence you can change its value. Note the effect of adding 1 to p3. The expression p3[0] now refers to the former second element of the array. Thus adding 1 to p3 causes it to point to the second element instead of the first. Subtracting one takes the pointer back to its original value so that the program can provide delete [] with the correct address.
The actual addresses of consecutive ints typically differ by two or four bytes, so the fact that adding 1 to p3 gives the address of the next element suggests that there is something special about pointer arithmetic. There is.