How to use sizeof(a)/sizeof(a[n])

Question

To avoid things quietly breaking if you change the array size, I suggest std::copy(a, a + sizeof(a)/sizeof(a[0]), b);. Even better, wrap the sizeof() junk in a macro -- or even betterer, use a function template instead: template <typename T, size_t N> size_t size((T&)[N]) { return N; } – j_random_hacker Sep 8 '12 at 7:29

When i I was looking into Q&A this morning I found this comment(with 4 upvotes). I'm quite new at C++. What does a+sizeof(a[0]) means here, I thought sizeof(a[0]) will return 4 which stands for a int memory byte? Many thanks in advance!!.

Answer 1

If a is an array then sizeof(a)/sizeof(a[0]) is the amount of elements in a. It is the size of whole array per size of one element.

So if a is a an array, then a + sizeof(a)/sizeof(a[0]) is a pointer to right after the end of the array.

Answer 2

That would simple be:

sizeof(a)/sizeof(a[0])

doesn't matter which element you chose (ie n or 0)