将函数作为形参传入c++时出错

Question

我有一个测试搜索方法函数时间的工作实现，它采用:

checkSearchTime(T(*funcPointer) (T myArray[], int size, T wanted)

作为参数，我尝试对排序方法做同样的事情:

checkSortTime(T(*funcPointer) (T myArray[],int size)

得到一个错误。这是错误:

Error   C2664   'void checkSortTime<int>(T (__cdecl *)(T [],int),T [],int)': cannot convert argument 1 from 'void (__cdecl *)(T [],int)' to 'T (__cdecl *)(T [],int)'   

以下是两个文件的代码:

int main() {
memLeaks();
constexpr auto SIZE = 5;
int myArray[SIZE];
populateArrayRandom(myArray, SIZE);
PrintArray(myArray, SIZE);
//insertionSort(myArray, SIZE);
PrintArray(myArray, SIZE);

checkSearchTime(binary_search, myArray, SIZE, 12);
checkSortTime(selectionSort, myArray, SIZE); // THIS IS WHERE THE ERROR IS
//checkAllSorts(myArray, SIZE); // WOULD LIKE TO CALL THIS AFTER
}

template<typename T>   
void checkSearchTime(T(*funcPointer) (T myArray[], int size, T wanted),
T arrayArgument[], int sizeArgument, T wantedArgument) {

// Use auto keyword to avoid typing long
auto start = std::chrono::high_resolution_clock::now();
funcPointer(arrayArgument, sizeArgument, wantedArgument);
auto stop = std::chrono::high_resolution_clock::now();
auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);
//std::cout << "Microseconds: " << duration.count() << std::endl;
std::cout << "[" << duration.count() << "] Nanoseconds" << std::endl;
}
template<typename T>                                                             
void checkAllSearches(T myArray[], int size, T value) {
std::cout << "Search Implementations and Timingn";
std::cout << "Binary=";
checkSearchTime(binary_search, myArray, size, value);
std::cout << "Linear=";
checkSearchTime(linear_search, myArray, size, value);
std::cout << "JumpSearch=";
checkSearchTime(jump_search, myArray, size, value);
std::cout << "Exponential=";
checkSearchTime(exponential_search, myArray, size, value);
std::cout << "FibMonaccian=";
checkSearchTime(fibMonaccian_search, myArray, size, value);
}
template<typename T>                                                                                
void checkSortTime(T(*funcPointer) (T myArray[],int size), //tried const here, doesn't work either
T arrayArgument[], int sizeArgument) {
// Use auto keyword to avoid typing long
auto start = std::chrono::high_resolution_clock::now();
funcPointer(arrayArgument, sizeArgument);
auto stop = std::chrono::high_resolution_clock::now();
auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);
//std::cout << "Microseconds: " << duration.count() << std::endl;
std::cout << "[" << duration.count() << "] Nanoseconds" << std::endl;
}

Answer 1

正如您已经计算出的那样，排序函数(selectionSort)的返回类型必须是void。

所以，我只是把这个答案贴出来给那些可能面临类似问题的人。

改变这种

template<typename T>                                                                                
void checkSortTime(T(*funcPointer)

到下面，将解决问题:

void checkSortTime(void(*funcPointer)

Answer 2

这是因为排序方法不返回任何东西。

Answer 3

不是真正的答案，只是想让你知道:通过const T(&name)[const size]传递数组"保持数组的大小。并且保证不改变它的内容(const)。这种语法也可以在这样的模板中使用(在这个例子中，我只是返回值，没有更新为返回void):

#include <iostream>
template<typename T, std::size_t N>
T checkSearchTime(T(*funcPointer)(const T(&myArray)[N], const T& wanted), const T (&arrayArgument)[N], const T& wantedArgument)
{
return funcPointer(arrayArgument, wantedArgument);
}
template<std::size_t N>
int lookup(const int(&arr)[N], const int& value)
{
for (int i = 0; i < N; ++i) std::cout << i << " ";
std::cout << "n";
return value;
}
int main()
{
int arr[]{ 1,2,3,4,5 };
int wanted = 3;
int value = checkSearchTime(lookup, arr, wanted);
std::cout << "value = " << value;
}