In my defense, we've been learning java in class where you never really deal explicitly with pointers. And java doesn't even call pointers pointers, they have to call it references. I guess it's a slightly less scary term than pointers, but it's still the same thing.
But now I get to go back to c++ since we're out for break, teaching myself various things again so I don't feel like a bum.
And I just wanted to write a simple program that would traverse the elements of an array and add up all the terms. Simple right? Except that I'm pointer illiterate and it took me a while.
Anywayz here's what I got:
| #include <iostream> using namespace std; int main() { #define SIZE 3; int x[SIZE] = {1, 2, 3); int* xptr = x; int i = 0, sum = 0; cout << "ze array is: "; while (i < SIZE) { cout << *(xptr + i) << ' '; sum += *(xptr + i); i++; } cout << '\n' << x[0] is: " << x[0]; cout << '\n' << "*x is: " << *x; cout << '\n' << *(x + 1) is: " << *(x + 1); cout << '\n' << *(x + 2) is: " << *(x + 2); return 0; } |
And this is what it outputs:
| ze array is: 1 2 3 x[0] is: 1 *x is: 1 *(x + 1) is: 2 *(x + 2) is: 3 sum is: 6 |
Something I wasn't too happy about is that, for the while loop, I wanted the condition to be while (*(xptr + i) != NULL) but the code wasn't correct for some reason.. It would print out everything I wanted but then print out a bunch of gibberish at the end.
But as always, I learned something new while doing this. Such as, when you declare an array:
int x[10]
And you write this:
*x
Then *x just points to x[0]! Which is why, in the code above, I was allowed to do shenanigans like
*(xptr + i)
Since xptr always just points to x, which is really just x[0]! And (xptr + i) is x[i], and deferencing this just gets you *(xptr + i) = the value at x[i]!
Anyways, yes, easy example, but I'm still recovering from an awful week of finals
