Quicky Question

When you create an array, such as int m[10], m is actually a pointer, yes yes?

And because pointer arithemetic allows you to add pointers of the same data type, you can add to *m with *(m+1) to skip over one "slot", which is equal to the sizeof(int), right?

The whole m[1] thing is just a cleaner and more understandable way of doing it, yes?

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Originally Posted by Krak

When you create an array, such as int m[10], m is actually a pointer, yes yes?

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m is an array of 11 ints. However, the name m can be used as a pointer to the first element of the array.

Quote:

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And because pointer arithemetic allows you to add pointers of the same data type, you can add to *m with *(m+1) to skip over one "slot", which is equal to the sizeof(int), right?

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You aren't allowed to add pointers. I assume you meant adding values to pointers. Yes, *(m+1) accesses location [m + sizeof(int)].

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The whole m[1] thing is just a cleaner and more understandable way of doing it, yes?

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Depends what you're doing and your personal preference. Many people find a pointers/iterators approach easy to use - look at the STL for example.

>When you create an array, such as int m[10], m is actually a pointer, yes yes?<

Well, no actually, however m acts as a const reference to the first value in the array.

>And because pointer arithemetic allows you to add pointers of the same data type, you can add to *m with *(m+1) to skip over one "slot", which is equal to the sizeof(int), right?<

yes, you can use pointer arithmatic to walk through an array. However you cannot change the value of m.

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You aren't allowed to add pointers

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umm. Well you can add pointers, you cannot change the value of m because it acts as a const pointer. Adding pointers however is very possible:

Code:

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#include <iostream>

using std::cout;
using std::endl;
using std::cin;

int main() {

        int myArray[4] = {3, 5, 7, 9};
        int* myPointer;
       
        myPointer = myArray;

        for(int i = 0; i < 4; i++, myPointer++) {
                cout << *myPointer << endl;
        }

        cin.get();
        return 0;
}

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-Andy

That's not adding pointers. That's adding integral values to pointers - a completely different thing.

My bad - misunderstood what you meant. :)

> m is an array of 11 ints.

No, it's 10 ints.

> however m acts as a const reference to the first value in the array
Actually, it acts as an immutable pointer to the (mutable, unless said to be const) first element of the array.

Oops - been using VB arrays :o

http://www.eskimo.com/~scs/C-faq/s6.html