Question

我需要获得仅包含双精度的列表的总和。出于某种原因，这不起作用，但是：

#include <type_traits>
#include <utility>

template< std::size_t index, typename type >
struct ref { type & value; };

template< std::size_t index, typename type >
type && get(ref< index, type > const & r)
{
    return std::forward< type >(r.value);
}

template< typename F, typename ...types, std::size_t ...indices >
decltype(auto) apply_inverse(F & f, types &... values, std::index_sequence< indices... >)
{
    struct : ref< indices, types >... {} refs{{values}...};
    constexpr std::size_t top = sizeof...(indices) - 1;
    return std::forward< F >(f)(get< top - indices >(refs)...);
}

template< typename F, typename ...types >
decltype(auto) apply_inverse(F && f, types &&... values)
{
    return apply_inverse< F, types... >(f, values..., std::index_sequence_for< types... >{});
}

#include <iostream>

int main()
{
    auto const print = [] (auto const &... value) -> std::ostream & { return (std::cout << ... << value); };
    apply_inverse(print, 1, 2, 3) << std::endl;
}

Answer 1

你应该修改你的循环如下。

double sum(DoubleList* list) {
    double sum = 0;
    for (DoubleNode *n = list->first; n != NULL; n = n->next) {
        sum += n->value;
    }
    return sum;
}

上面的示例代码假定DoubleNode 具有一个名为next的属性，其类型为DoubleNode *，它将指针存储到下一个元素。

Answer 2

您需要确保获得循环中的下一个元素。我猜测next指针是n->next：

for (DoubleNode *n = list->first; n != NULL; n = n->next) {
    sum += n->value;
}

Answer 3

更改

for (DoubleNode *n = list->first; n != NULL; n = next)

到

for (DoubleNode *n = list->first; n != NULL; n = n->next)

C中列表的所有元素的总和

3 个答案: