c++ - 传递不同的函数作为参数

Question

假设我有这个类:

template<class K, class Compare>
class findMax {
    K* keyArray; // Supposed to be an array of K.
    int size;

public:
      findMax (K n, Compare init); // Here I would like to initialize the array.
      ~findMax ();

      K& Find(K key, Compare cmp); // Return an elemnt in the array according to a condition.
      void printArray(Compare print); // Print the array according to a condition.

};

我想要每个 cmp当我实现构造函数时，函数会有所不同，Find和 printArray .
例如:

template<class K, class Compare>
findMax<K, Compare>::findMax(int n, Compare init) {
    keyArray = new K [n];
    init(keyArray, n);
}

哪里init是我在源文件中实现的一个函数，例如:

// Init will initialize the array to 0.
void init (int* array, int n) {
    for (int i=0; i<n; i++)
        array[i] = 0;
}

不过，我希望能够将不同的函数发送到 Find例如，在两个元素之间进行比较。我无法弄清楚如何，因为当我创建一个新的 findMax 时对象，例如 findMax<int, UNKNOWN> f ，我用什么代替 UNKNOWN ?

Answer 1

试试这个-

#include <iostream>
#include <functional> 
using namespace std;

template<class K, class Compare>
class findMax {
    K* keyArray; // Supposed to be an array of K.
    int size;

public:
      findMax (K n, Compare init){init();}; // Here I would like to initialize the array.
      ~findMax (){};
    template<typename Compare1>
      K& Find(K key, Compare1 cmp){ cmp();}; // Return an elemnt in the array according to a condition.
      template<typename Compare2>
      void printArray(Compare2 print){print();}; // Print the array according to a condition.

}; 
int main() {
    findMax<int,std::function<void()>> a{5,[](){cout<<"constructor"<<endl;}};
    a.Find(5,[](){cout<<"Find"<<endl;});
    a.printArray([](){cout<<"printArray";});
    return 0;
}