 Solution

For the implementations above the value isThe first half-iterator returns the odd-position characters while consuming the even-position characters as its required skips. By the time it is exhausted, it has consumed the entire shared underlying iterator. The second half-iterator consequently has no entries left when concatenation reaches it. Sharing one source does not create two independent traversals. Neither constructor consumes anything, so it also does not shift the first output before iteration begins. The source's implementation references here should be read as parts (d) and (e); the printed reference to (c) and (d) is a cross-reference slip.