c++ - 将基于堆的 N 元数组引用传递给函数

Question

我正在尝试将 N 元数组传递给 C++ 中的函数。

为简单起见，对于一维数组，这里是:

#include <iostream>

template <size_t N> int sign(int(&arr)[N], int i)
{
    return (arr[i] >= 0) ? +1 : -1 ;
}

int main(int argn, char** argv)
{
    const int SIZE = 2;

    int (*arr1d) = new int[SIZE];
    arr1d[0] = 12;
    arr1d[1] = -1;

    std::cout << "sign 0 : " << sign(arr1d, 0) << std::endl ;     
    std::cout << "sign 1 : " << sign(arr1d, 1) << std::endl ;
}

但这不会编译:

Sign.cpp:17:50: error: no matching function for call to ‘sign(int*&, int)’
     std::cout << "sign 1 : " << sign(arr1d, 1) << std::endl ;
                                              ^
Sign.cpp:3:29: note: candidate: template<long unsigned int N> int sign(int (&)[N], int)
 template <size_t N> int sign(int(&arr)[N], int i)
                         ^~~~
Sign.cpp:3:29: note:   template argument deduction/substitution failed:
Sign.cpp:17:50: note:   mismatched types ‘int [N]’ and ‘int*’
     std::cout << "sign 1 : " << sign(arr1d, 1) << std::endl ;

我不介意不使用模板(这似乎是这里的问题)。

有什么建议吗？

---

二维情况下的相同错误:

#include <iostream>

template <size_t M, size_t N> int sign(int(&arr)[M][N], int i, int j)
{
    return (arr[i][j] >= 0) ? +1 : -1 ;
}

int main(int argn, char** argv)
{
    const int SIZE = 2;

    int (*arr2d)[SIZE] = new int[SIZE][SIZE];                           
    arr2d[0][0] = 12;
    arr2d[0][1] = -1;
    arr2d[1][0] = 32;
    arr2d[1][1] = -4;

    std::cout << "sign 0 1: " << sign(arr2d, 0, 1) << std::endl ;
}

---

Answer 1

尝试将指向数组的指针传递给您的符号函数。

template <size_t M, size_t N> 
int sign(int(*arr)[N], int i, int j)
{
    return (arr[i][j] >= 0) ? +1 : -1 ;
}

然后您可以按如下方式调用您的符号函数:

std::cout << "sign 0 1: " << sign<SIZE, SIZE>(arr2d, 0, 1) << std::endl ;