std::istream_iterator<> with copy_n() and friends

Question

The snippet below reads three integers from std::cin; it writes two into numbers and discards the third:

std::vector num         


        

Answer 1

Unfortunately the implementer of copy_n has failed to account for the read ahead in the copy loop. The Visual C++ implementation works as you expect on both stringstream and std::cin. I also checked the case from the original example where the istream_iterator is constructed in line.

Here is the key piece of code from the STL implementation.

template inline
    _OutIt _Copy_n(_InIt _First, _Diff _Count,
        _OutIt _Dest, input_iterator_tag)
    {   // copy [_First, _First + _Count) to [_Dest, ...), arbitrary input
    *_Dest = *_First;   // 0 < _Count has been guaranteed
    while (0 < --_Count)
        *++_Dest = *++_First;
    return (++_Dest);
    }

Here is the test code

#include 
#include 
#include 
#include 
#include 

int _tmain(int argc, _TCHAR* argv[])
{
    std::stringstream ss;
    ss << 1 << ' ' << 2 << ' ' << 3 << ' ' << 4 << std::endl;
    ss.seekg(0);
    std::vector numbers(2);
    std::istream_iterator ii(ss);
    std::cout << *ii << std::endl;  // shows that read ahead happened.
    std::copy_n(ii, 2, numbers.begin());
    int i = 0;
    ss >> i;
    std::cout << numbers[0] << ' ' << numbers[1] << ' ' << i << std::endl;

    std::istream_iterator ii2(std::cin);
    std::cout << *ii2 << std::endl;  // shows that read ahead happened.
    std::copy_n(ii2, 2, numbers.begin());
    std::cin >> i;
    std::cout << numbers[0] << ' ' << numbers[1] << ' ' << i << std::endl;

    return 0;
}


/* Output
1
1 2 3
4 5 6
4
4 5 6
*/