How is *it++ valid for output iterators?

Question

In example code, I often see code such as *it++ for output iterators. The expression *it++ makes a copy of it, increments it, and then returns the copy which is finally dereferenced. As I understand it, making a copy of an output iterator invalidates the source. But then the increment of it that is performed after creating the copy would be illegal, right? Is my understanding of output iterators flawed?

Answer 1

The standard requires that *r++ = t work for output iterators (24.1.2). If it doesn't work, it's not an output iterator by the standard's definition.

It is up to the iterator implementation to make sure such statements work correctly under the hood.

The reason that you shouldn't keep multiple copies of an output iterator is that it has single pass semantics. The iterator can only be dereferenced once at each value (i.e. it has to be incremented between each dereference operation). Once an iterator is dereferenced, a copy of it cannot be.

This is why *r++ = t works. A copy is made of the original iterator, the original iterator is dereferenced and the copy is incremented. The original iterator will never be used again, and the copy no longer references the same value.

Answer 2

The expression *it++ does not (have to) make a copy of it, does not increment it, etc. This expression is valid only for convenience, as it follows the usual semantics. Only operator= does the actual job. For example, in g++ implementation of ostream_iterator, operator*, operator++ and operator++(int) do only one thing: return *this (in other words, nothing!). We could write for example:

it = 1;
it = 2;
*it = 3;
++it = 4;

Instead of: *it++ = 1; *it++ = 2; *it++ = 3; *it++ = 4;