c++ - 使用模板函数将转换应用于具有索引的 vector

Question

我正在尝试对 vector 应用变换，同时还在变换函数中添加索引。我正在使用以下转换函数 transformWithIndex。如果我从 result.push_back(itransformationFunction(idx, element)); 中删除 idx，它会正常工作。

I understood that I have an extra argument. But I don't understand how to modify the transformation function in order to handle the IDX.

---

// Example program That can be tested
#include <iostream>
#include <string>
#include <vector>

template <typename Range, class TransformationFunction>
inline std::vector<typename std::result_of<TransformationFunction(typename Range::value_type)>::type>
transformWithIndex(const Range &iRange, const TransformationFunction &itransformationFunction) {
  std::vector<typename std::result_of<TransformationFunction(typename Range::value_type)>::type> result;
  int                                                                                            idx = 0;
  for (auto &&element : iRange) {
    result.push_back(itransformationFunction(idx, element));
    idx++;
  }
  return result;
}

int main()
{
   std::vector<int> source = {1, 2, 3};
   std::vector<int> result = transformWithIndex(source, [](int i) { return ++i; });

   return 0;
}

---

错误如下:

>  In instantiation of 'std::vector<typename
> std::result_of<TransformationFunction(typename
> Range::value_type)>::type> transformWithIndex(const Range&, const
> TransformationFunction&) [with Range = std::vector<int>;
> TransformationFunction = main()::<lambda(int)>; typename
> std::result_of<TransformationFunction(typename
> Range::value_type)>::type = int]': 21:82:   required from here 12:58:
> error: no match for call to '(const main()::<lambda(int)>) (int&,
> const int&)' 21:58: note: candidates are: 12:58: note: int (*)(int)
> <conversion> 12:58: note:   candidate expects 2 arguments, 3 provided
> 21:65: note: main()::<lambda(int)> 21:65: note:   candidate expects 1
> argument, 2 provided

Answer 1

您调用 itransformationFunction(idx, element)(2 个参数) 而你的 lambda 曾经期望一个。

将调用更改为:

transformWithIndex(source, [](int index, int &elem) { return index + 1; })

并且您的result_of 也应该被修复以包含索引:

template <typename Range, class TransformationFunction>
std::vector<typename std::result_of<TransformationFunction(std::size_t, typename Range::value_type)>::type>
transformWithIndex(const Range &iRange, const TransformationFunction &itransformationFunction) {
  std::vector<typename std::result_of<TransformationFunction(std::size_t, typename Range::value_type)>::type> result;
  int                                                                                            idx = 0;
  for (auto &&element : iRange) {
    result.push_back(itransformationFunction(idx, element));
    idx++;
  }
  return result;
}

和 std::decay_t 应该应用于处理仿函数返回引用或 const 对象的情况。