1. ## Array/Pointer question

Hi,

Code:
```/* The pattern matching algorithm uses 2 loops.  The outer loop
is controlled by the pointer p1 which points to elelments in array
a1 where the inner loop will begin checking for a match with array
a2.  The inner loop is controlled by the integer j which is used
to compare corresponding elelments of the 2 arrays.  If a mismatch
is found, the inner loop aborts and the outer loop continues by
incrementing p1 to look for a match starting with the next element
of a1.  If the inner loop is allowed to finish, then the condition
(j = n2) will be true and the current location pointed to by p1 is
returned.  The test driver verifies that the match has indeed been

#include <iostream>
#include <stdlib.h>

using namespace std;

short* loc(short* a1, short* a2, int n1, int n2)
{ short* endl = a1 + n1;    //  ???? if a1 = a[0],
// then wouldn't this make endl point to a[9]?
// Isn't a[0] = a[0] ... a[8]??
for (short* p1 = a1; pl < endl; pl++)
if (*pl == *a2)
{ int j;
for (j = 0; j < n2; j++)
if (p1[j] != a2[j]) break;
if (j == n2) return p1;   // ??? if j == n2 (or 5), how can
// that be since the for loop only allows j < n2?
}
return 0;
}

int main()
{ short a1[9] = {11, 11, 11, 11, 11, 22, 33, 44, 55};
short a2[5] = {11, 11, 11, 22, 33};
cout << "Array a1 begins at location:\t" << a1 << endl;
cout << "Array a2 begins at location:\t" << a2 << endl;
short* p = loc(a1, a2, 9, 5);
if (p)
{ cout << "Array a2 found at location:\t" << p << endl;
for (int i = 0; i < 5; i++)
cout << "\t" << &p[i] << ": " << p[i]
<< "\t" << &a2[i] << ": " << a2[i];
}
system("pause");
return 0;
}```
// ???? if a1 = a[0],
// then wouldn't this make endl point to a[9]?
// Isn't a[0] = a[0] ... a[8]??

// ??? if j == n2 (or 5), how can
// that be since the for loop only allows j < n2?

Dumb question or not, it just really confused me when I thought I almost completely understood the way this piece of code worked. I appreciate your help.

Swaine777

2. > ???? if a1 = a[0],
> then wouldn't this make endl point to a[9]?
> Isn't a[0] = a[0] ... a[8]??

Yes, endl would point to a1[9]. And since there is no a1[9], it points one location past the end of the array. So once p1 points one location past the end of the array, the loop will finish.
You are correct: the array goes from a1[0] to a1[8].

> ??? if j == n2 (or 5), how can
> that be since the for loop only allows j < n2?

This if statement is not within the for-loop:
for (j = 0; j < n2; j++)
So the if() statement executes once the for-loop is done. So if all five elements of a2 match, this if() statement will be true, and p1 will be returned.

>Dumb question or not, it just really confused me when I thought I almost completely understood the way this piece of code worked. I appreciate your help.

There are no dumb questions. We're all here to learn.

3. Thanks a lot swoopy. Hehe, if arrays and pointers are this confusing, wait till I get to vectors! P

Swaine777