(C++) Answer of exercise \#8: Library trouble \#0

February 24, 2017 · View on GitHub

(C++) Answer of exercise #8: Library trouble #0

This is the answer of Exercise #8: library trouble.

The compiler takes you to the following code:

// search_n. Search for count consecutive copies of val. template <class _ForwardIter, class _Integer, class _Tp> _ForwardIter search_n(_ForwardIter __first, _ForwardIter __last, _Integer __count, const _Tp& __val) { _STLP_DEBUG_CHECK(__check_range(first, last)) if (count <= 0) return first; else { first = find(first, last, val); while (first != last) { _Integer n = count - 1; _ForwardIter i = first; ++i; while (i != last && n != 0 && *i == val) { ++i; --n; // <---- THIS LINE } if (n == 0) return first; else first = find(i, last, val); } return __last; } }

This must means that __n is also a const data type. This is true, as shown below:

// search_n. Search for count consecutive copies of val. template <class _ForwardIter, class /non-const/ _Integer, class _Tp> _ForwardIter search_n(_ForwardIter __first, _ForwardIter __last, /non-const/ _Integer __count, const _Tp& __val) { _STLP_DEBUG_CHECK(__check_range(first, last)) if (count <= 0) return first; else { first = find(first, last, val); while (first != last) { /non-const/ _Integer n = count - 1; _ForwardIter i = first; ++i; while (i != last && n != 0 && *i == val) { ++i; --n; // <---- THIS LINE } if (n == 0) return first; else first = find(i, last, val); } return __last; } }

So, due to the template use, __n will be of type const int, when we pass a __count of type const int! This conflict with the use of __n as a changing index. The type of __n should be the non-const type of __count.