[Happy Coding] 函数返回数组指针和数组引用

在boost::array中看到这样的代码：

    template <typename T, std::size_t N>
    T(&get_c_array(boost::array<T,N>& arg))[N]
    {
        return arg.elems;
    }
    
    // Const version.
    template <typename T, std::size_t N>
    const T(&get_c_array(const boost::array<T,N>& arg))[N]
    {
        return arg.elems;
    }

get_c_array是boost里面自由函数，意在返回boost::array中存储的数组的 引用。

函数返回数组引用的签名如下（注意特别的&符号）

T (&func(...))[N]

这个签名来源于数组引用的签名： T (&arrayR)[N]。

我们很熟悉数组指针的声明方式：T (*arrayP)[N]。

测试程序：

template <int N>
int (*return_arrayP())[N]
{
    static int array[] = {2, 3, 4};
    return &array;
}

template <int N>
int (&return_arrayR())[N]
{
    static int array[] = {2, 3, 4};
    return array;
}

int main()
{
    static const int N = 3;
    // test return pointer to array
    std::cout << "test return pointer to array\n";
    int (*arrayP)[N] = return_arrayP<N>();
    for (int i = 0; i < N; i++) {
        std::cout << (*arrayP)[i] << ", ";
    }
    std::cout << "\n";

    // test return reference to array
    std::cout << "test return reference to array\n";
    int (&arrayR)[N] = return_arrayR<N>();
    for (int i = 0; i < N; i++) {
        std::cout << arrayR[i] << ", ";
    }
    std::cout << "\n";

    // boost::array return reference to array
    // to downcast to T*
    std::cout << "use boost::array to return reference to array, and then downcast to T*\n";
    int* arrayb = NULL;
    boost::array<int, N> oneArray = {1, 2, 3};
    arrayb = boost::get_c_array(oneArray);
    for (int i = 0; i < N; i++) {
        std::cout << arrayb[i] << ", ";
    }
    std::cout << "\n";