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?
m is an array of 11 ints. However, the name m can be used as a pointer to the first element of the array.Originally Posted by Krak
You aren't allowed to add pointers. I assume you meant adding values to pointers. Yes, *(m+1) accesses location [m + sizeof(int)].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?
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.The whole m[1] thing is just a cleaner and more understandable way of doing it, yes?
>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.
i don't think most standard compilers support programmers with more than 4 red boxes - Misplaced
It is my sacred duity to stand in the path of the flood of ignorance and blatant stupidity... - quzah
Such pointless tricks ceased to be interesting or useful when we came down from the trees and started using higher level languages. - Salem
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:You aren't allowed to add pointers
-AndyCode:#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; }
i don't think most standard compilers support programmers with more than 4 red boxes - Misplaced
It is my sacred duity to stand in the path of the flood of ignorance and blatant stupidity... - quzah
Such pointless tricks ceased to be interesting or useful when we came down from the trees and started using higher level languages. - Salem
That's not adding pointers. That's adding integral values to pointers - a completely different thing.
My bad - misunderstood what you meant.
i don't think most standard compilers support programmers with more than 4 red boxes - Misplaced
It is my sacred duity to stand in the path of the flood of ignorance and blatant stupidity... - quzah
Such pointless tricks ceased to be interesting or useful when we came down from the trees and started using higher level languages. - Salem
> 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.
All the buzzt!
CornedBee
"There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
- Flon's Law
Oops - been using VB arrays
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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