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我試圖創建自己的「智能迭代」,我想用SFINAE有根據迭代器的標籤有些運營商:SFINAE:類成員不能重新聲明
這裏是我的代碼:
template<class Iterator, class Predicat, class Tag>
class RangeFilterIterator {
public:
RangeFilterIterator(Iterator begin, Iterator end, Predicat predicat) :
mBegin(begin), mEnd(end), mPredicat(predicat) {}
bool operator !=(RangeFilterIterator const &r) {
return mBegin != r.mBegin;
}
typename Iterator::value_type &operator*() {return *mBegin;}
RangeFilterIterator &operator++() {
while(mBegin != mEnd && mPredicat(*mBegin++));
return *this;
}
template<class = std::enable_if_t<std::is_base_of<std::random_access_iterator_tag, Tag>::value>>
RangeFilterIterator &operator+(std::size_t n) {
while(n--)
++(*this);
return *this;
}
template<class = std::enable_if_t<!std::is_base_of<std::random_access_iterator_tag, Tag>::value>>
RangeFilterIterator &operator+(std::size_t n) = delete;
private:
Iterator mBegin, mEnd;
Predicat mPredicat;
};
template<typename Container, typename Predicate>
auto RangeFilter(Container const &c, Predicate p) {
using Iterator = RangeFilterIterator<typename Container::iterator,
Predicate,
typename Container::iterator::iterator_category>;
Iterator begin(const_cast<Container&>(c).begin(), const_cast<Container&>(c).end(), p);
Iterator end(const_cast<Container&>(c).end(), const_cast<Container&>(c).end(), p);
return Range(begin, end);
}
,並在該行RangeFilterIterator &operator+(std::size_t n) = delete
我得到了錯誤:class member cannot be redeclared
。
我不喜歡模板,但我認爲,與SFINAE只有一個將被「宣佈」。我錯過了什麼嗎?否則可以做到?
我使用class tag = Tag和使用「無用函數」得到一個修復。但是,如果我爲該功能使用「同名」,我仍然遇到同樣的錯誤... –