Marco Marco - 1 year ago 95
C++ Question

Value Of Pointers

In my book, it says Pointers are addresses and have a numerical value. You can print out the value of a pointer as cout << (unsigned long)(p)
Write code to compare p,p+1,q, and q+1. Explain the results, Im not sure what the book wants me to so here's what I have. Does anyone Know if I am doing this right

int num = 20;
double dbl = 20.0;
int *p = &num;
double *q = &dbl;
cout << (unsigned long)(q) << endl;
q = q + 1;
cout << (unsigned long)(q) << endl;

cout << (unsigned long)(p) << endl;
p = p + 1 ;
cout << (unsigned long)(p) << endl;

Answer Source

Assuming it's the pointer arithmetic you have problems with, let my try to to show how it's done in a more "graphical" way:

Lets say we have a pointer variable ptr which points to an array of integers, something like

int array[4] = { 1234, 5678, 9012, 3456 };
int* ptr = array;  // Makes `ptr` point to the first element of `array`

In memory it looks something like

| 1234 | 5678 | 9012 | 3456 |
^      ^      ^      ^
|      |      |      |
ptr    ptr+1  ptr+2  ptr+3

The first is technically ptr+0

When adding one to a pointer, you go to the next element in the "array".

Perhaps now you start to see some similarities between pointer arithmetic and array indexing. And that is because there is a common thread here: For any pointer or array p and valid index i, the expression p[i] is exactly the same as *(p + i).

Using the knowledge that p[i] is equal to *(p + i) makes it easier to understand how an array can be used as a pointer to its first element. We start with a pointer to the first element of array (as defined above): &array[0]. This is equal to the expression &*(array + 0). The address-of (&) and dereference (*) operators cancel out each, leaving us with (array + 0). Adding zero to anything can be removed as well, so now we have (array). And finally we can remove the parentheses, leaving us with array. That means that &array[0] is equal to array.

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