C++11如何使用lambda函数模板做参数

Question

开始使用C++11特性，碰到一些问题。以下写了一个快速排序的函数模板，划分操作的比较函数使用一个std::function<int(T,T)>的参数接收，而在main中使用函数模板接收一个lambda函数[](int a, int b) -> int {return a < b;}时报错，函数不匹配。这里该如何处理？

#include <iostream>
#include <functional>

template <typename T>
void Swap(T *a, T *b) {
    T tmp = *a;
    *a = *b;
    *b = tmp;
}

template <typename T>
int partition(T array[], int low, int high, const std::function<int(T, T)> &cmp) {
    int i = low - 1;
    T x = array[high];
    for (int j = low; j < high; j++) {
        if (cmp(array[j], x)) {
            i++;
            Swap(&array[j], &array[i]);
        }
    }
    Swap(&array[i + 1], &array[high]);
    return i + 1;
}

template <typename T>
void quicksort(T array[], int low, int high, const std::function<int(T, T)> &cmp) {
    if (low < high) {
        int mid = partition(array, low, high, cmp);
        quicksort(array, low, mid - 1, cmp);
        quicksort(array, mid + 1, high, cmp);
    }
}

int main() {
    std::cout << "Hello, World!" << std::endl;
    int arr[] {5, 3, 9, 1, 6, 2};
    quicksort(arr, 0, 5, [](int a, int b) -> int {
       return a < b;
    });
    for(auto item: arr) {
        std::cout << item << "  ";
    }
    std::cout << std::endl;
    return 0;
}

报错信息：
E:\Project\DesignPattern\main.cpp: In function 'int main()':
E:\Project\DesignPattern\main.cpp:39:6: error: no matching function for call to 'quicksort(int [6], int, int, main()::<lambda(int, int)>)'
     });
      ^
E:\Project\DesignPattern\main.cpp:26:6: note: candidate: template<class T> void quicksort(T*, int, int, const std::function<int(T, T)>&)
 void quicksort(T array[], int low, int high, const std::function<int(T, T)> &cmp) {
      ^
E:\Project\DesignPattern\main.cpp:26:6: note:   template argument deduction/substitution failed:
E:\Project\DesignPattern\main.cpp:39:6: note:   'main()::<lambda(int, int)>' is not derived from 'const std::function<int(T, T)>'
     });
      ^

Answer 1

lambda转换成const std::function<int(T, T)>&前无法帮助推导出T的类型.

可以增加个辅助类先确定std::function的类型,再进行lambda的隐式转换.

tempalte<typename T>
struct type_deduce
{
  using type = T;
};

template<typename T>
void quicksort(T array[], int low, int high, const typename type_deduce<std::function<int(T, T)>>::type& cmp)

Answer 2

类型推导时不会进行用户自定义的类型转换，即不会进行从lambda表达式到std::function的转换。所以这里跪了。

你这里需要的是一个BinaryPredicate。直接写Fn &&fn即可。然后对Fn进行一下BinaryPredicate的检查。

或者把T变成一个non deduction context，比如

template <typename T>
void quicksort(T array[], int low, int high,
    const std::function<int (
        decltype(std::declval<T>()),
        decltype(std::declval<T>()))> &cmp)

或者

template <class T>
struct BinaryPredicateType {
  using type = std::function<int (T, T)>;
};

template <typename T>
void quicksort(T array[], int low, int high,
               const typename BinaryPredicateType<T>::type &cmp)