code>,有两个可行的重载:When you call l.assign(10, 5);, there are two viable overloads:void assign(size_type n, const int& val)template <>void assign(int first, int last)当我们说非模板函数时优先于模板函数,这只有在两者的转换顺序无法区分时才成立。但是在这种情况下,函数模板将完全匹配(您的两个参数均为 int ,无需转换),而非模板则必须进行提升(必须将 10 从 int 升级到 size_t )。因此,这就是首选函数模板重载的原因。When we say that non-template functions are preferred to template functions, that is only true if the two have indistinguishable conversion sequences. But in this case, the function template will match exactly (both of your arguments are int, no conversion necessary), while the non-template will have to undergo promotation (have to promote 10 from int to size_t). So that's why the function template overload is preferred.关于如何修复它,您只需要使模板 not 成为可行的重载即可。 。这涉及为输入迭代器编写type_trait,使用 void_t 并不困难:As to how to fix it, you just need to make the template not a viable overload. That involves writing a type_trait for input iterator, which using void_t is not hard:template <typename... >using void_t = void;template <typename T, typename = void>struct is_input_iterator : std::false_type { };template <typename T>struct is_input_iterator<T, void_t< decltype(std::declval<T>() == std::declval<T>()), decltype(std::declval<T>() != std::declval<T>()), decltype(*std::declval<T>()), decltype(++std::declval<T>()), decltype(std::declval<T>()++) >> : std::true_type { };然后要求 is_input_iterator :template <typename _InputIterator, typename = std::enable_if_t<is_input_iterator<_InputIterator>::value>>void assign(_InputIterator first, _InputIterator last);还有很多其他方式可以做这种事情,我碰巧喜欢 void_t 。无论采用哪种方式,都必须确保该模板根本不可行。There are lots of other ways to do this sort of thing, I just happen to like void_t. Regardless of which way you do it, you have to ensure that the template simply isn't viable. 这篇关于forward_list:分配(_InputIterator __first,_InputIterator __last)/分配(size_type __n,const _Tp&amp; __val)的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持! 上岸,阿里云!
08-18 14:50
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