I'm having a hard time grasping this code
if i DoCode:#define N 10 int a[N], i, sum = 0, *p = a; for (i = 0; i < N; i++) sum += p[i];
wouldn't I reference each element as *p[i] or to get address as &p[i]? Why is the dereference skipped? I would understand that p[i] in the example would be the address and not the actual value. Help me understand this.Code:a[10] = {1,2,3,4,5,6,7,8,9,10} int *p = a;
You set a pointer to array a.I can not understand your question
No, a[i] would give you an int. You're getting the ith element of an array of ints, which is an int.
a is an array of 10 integers. This type is automatically convertible to a pointer to int.
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
The best way to understand it, is to add a print statement inside the loop.
Include EVERYTHING you're not sure of, in the print statement - even if it's wrong and won't even compile. Find out WHY any bad part of it, won't compile, and find out EXACTLY what every little expression of the array, will give you.Code:for(i=0;i<N;i++) { printf("a[%d] is: p+i is: %p *(p+i) is: %d\n",a[i],p+i, *(p+i)); //maybe print up some sums? }
Learning by doing (exploring yourself), is a great way to learn something. Some educators even use the maxim "Learning by doing is not A way to learn, it is the ONLY way to effectively learn."
That's one of the great things about programming - you can explore on your own, so much, very easily.
P.S. Be certain to nail down the zero based indexing that C arrays use, also!
