Hope To Get More Information in this web.

Thank you very much

Savenaca yauyau

Thanks for tutorials, BTW.

for (char *pnPtr = szName; pnPtr <= szName + nArraySize; pnPtr++)

Could you explain it, please?

My advise to you would be to run your program using a debugger, and step through the code. Put a watch on nVowels and then watch it count up — when it becomes 4, you’ll at least know what’s causing it.

Your site is wonderful, thanks for providing it. I was editing some of the examples you provide to see what happens with the output and I noticed something I cannot explain. When I substitute the following into your last example (Mollie/Vowels) I get the output 3, which makes sense.

const int nArraySize = 14;
char szName[nArraySize] = “Lazy BrownFox”;

But When I change the code to the below, the result becomes 4, which I do not understand. It is as if the program is counting the last whitespace character as a vowel. Can you explain this to me?

const int nArraySize = 15;
char szName[nArraySize] = “Lazy Brown Fox”;

BTW, I am using Cygwin with Dev-C++ and the GNU compilers.

Thanks and best regards,
Pete

Arrays and pointers are not strictly identical. Although arrays are implemented using pointers, arrays also contain slightly more information. When you create an array, the compiler has to keep track of how many elements the array has. Otherwise, the compiler would not know how many elements to delete when it came time to delete[] the array.

With a non-dynamically allocated array, it’s possible to see the actual size of the array:

int anArray[5]; cout < < sizeof(anArray);

This program prints out 20 (assuming 4-byte integers). This is the size of the array.

On the other hand:

int *ptr = anArray; cout < < sizeof(ptr);

prints 4. This is the size of the pointer.

When you access an array through a pointer, you’re accessing it in a slightly limited manner, because you lose access to the size and type information of the original array. This can be a pain (because it means you have to store the size of the array in a separate variable) but that’s about the extent of the practical consequences.

Note that for dynamically allocated arrays:

int *pnArray = new int[5]

the new operator returns a pointer to the array, and that’s all you ever have to work with. You never get access to the actual array object.

Functionally, in terms of how you access an array, there is no real difference between accessing the elements of an actual array and accessing the elements of an array through a pointer.

So technically, it is correct that arrays and pointers aren’t identical. Pointers give you access to a limited subset of the array information. Practically speaking, it’s generally not even worth worrying about.

I’ve been learning C from your tutorial - and I think it’s really great, so thanks.
However I just wanted to point (speaking of pointers ;-)) out, that the guy behind these two articles
http://www.cplusplus.com/articles/siavoshkc1.html
http://www.cplusplus.com/articles/Arrptr.html
would have to disagree with you on whether arrays and pointers are the same. Personally, I don’t care if he is right, since the code presented here still works intentionally.

Best regards
Jacob