将函数作为参数传递给 C++ 时出错

Question

我有一个测试搜索方法函数时间的工作实现，它需要：

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

作为参数，我尝试用排序方法做同样的事情：

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

我收到一个错误。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)'

这是两个文件的代码：

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 Timing\n";
    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

这是因为排序方法不return任何东西。

Answer 2

如您所知，排序函数 (selectionSort) 的 return 类型一定是 void。

所以，我只是为可能面临类似问题的其他人发布这个答案。

改变这个

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

到下面，将解决问题：

void checkSortTime(void(*funcPointer)

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。并且您保证不更改其内容 (const)。这种语法也可以用在这样的模板中（在这个例子中我只是 return 值，没有更新为 return 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;
}

将函数作为参数传递给 C++ 时出错

Error passing a function in as a parameter C++

c++

header

function

void