我正在编写以下代码来测试模板专业化。有没有办法让指针衰减到类型,以便我可以使用模板参数推导来获取我的 M 和 N,这样我就可以获得二维数组的行和列?我知道我可以为二维数组向量向量,但这是模板专业化的练习。以下代码按原样工作,但请查看是否取消注释以下内容并注释掉当前的 T** 构造函数它不起作用。因此,构造函数当前硬编码用于删除 data_ 成员的行数,这并不理想,理想情况下我会从模板推导中获取此参数。我想知道是否有办法在下面显示的以下代码中完成这项工作。
//doesn't like this constructor
// template <unsigned int M, unsigned int N>
// Data ( T (&d)[M][N] ): data_(d), rows_(M), cols_(N) {};
#include <iostream>
template <typename T>
class Data
{
public:
Data ( const T& d ): data_(d) {};
Data ( T&& d ): data_(std::move(d)) {};
std::string getType() { return "Plain Data Type"; }
private:
T data_;
};
template <typename T>
class Data<T**>
{
public:
//doesn't like this constructor
// template <unsigned int M, unsigned int N>
// Data ( T (&d)[M][N] ): data_(d), rows_(M), cols_(N) {};
Data ( T** d ): data_(d), rows_(25) {};
~Data() {
for ( unsigned int i = 0; i < rows_; ++i)
{
delete [] data_[i];
}
delete [] data_;
}
std::string getType() { return "Pointer to Pointer Data Type"; }
private:
T** data_;
unsigned int rows_;
unsigned int cols_;
};
template <typename T>
class Data<T*>
{
public:
Data ( T* d ): data_(d) {};
~Data() { delete data_; }
std::string getType() { return "Pointer Data Type"; }
private:
T* data_;
};
int main ( int argc, char *argv[])
{
float f(9.65);
Data<int> d1(f);
std::cout << d1.getType() << std::endl;
int *i = new int(5);
Data<int*> d2(i);
std::cout << d2.getType() << std::endl;
int **j = new int*[25];
for ( int i = 0 ; i < 25; ++i)
j[i] = new int[50];
Data<int**> d3(j);
std::cout << d3.getType() << std::endl;
}
output:
Plain Data Type
Pointer Data Type
Pointer to Pointer Data Type