跟蹤一個專門的模板對象的枚舉成員值

Question

我有一個結構模板，遞歸定義爲一個枚舉值如下：

template <unsigned x, unsigned y> 
struct Special 
{ 
    enum { value = x * Special<x-1, y-1> :: value; } 
}; 

template <unsigned n> 
struct Special<n, 0> 
{ 
    enum { value = 1; } 
}; 

我想知道值對實例的逐步增長特別像：

Special <8, 3> spl; 

所以，我插入的通用結構定義中下面的構造：

Special() { cout << "Value, x and y are: " << value << "," << x << "," << y;} 

但是，我只得到最終值打印爲336,8,3。我如何獲得y從3逐漸減少到0的完整畫面？

Answer 1

但是，我只得到最終值打印爲336,8,3。我如何獲得y從3逐漸減少到0的完整圖像？

由於類本身不會從它們自身遞歸地派生，所以不會得到預期的輸出。當您構建Special<8,3>時，它不是從Special<7,2>派生的。因此，當您構建Special<8,3>時，Special<7,2>的構造函數不會被調用。

解決方法1：將類從下一個遞歸類

#include <iostream> 

template <unsigned x, unsigned y> struct Special; 

template <unsigned x, unsigned y> 
struct Special : Special<x-1, y-1> 
{ 
    Special() 
    { 
     std::cout << "Value, x and y are: " << value << "," << x << "," << y << std::endl; 
    } 
    enum { value = x * Special<x-1, y-1> :: value }; 
}; 

template <unsigned n> 
struct Special<n, 0> 
{ 
    Special() 
    { 
     std::cout << "Value, x and y are: " << value << "," << n << "," << 0 << std::endl; 
    } 

    enum { value = 1 }; 
}; 

int main() 
{ 
    Special<8,3> s; 
} 

解決方案2繼承：使用遞歸函數來獲取價值，而不是一個枚舉的

另一種方式獲得所需的輸出將通過遞歸函數調用而不是類中的enum，並具有用於在函數中生成輸出的代碼。

#include <iostream> 

template <unsigned x, unsigned y> 
struct Special 
{ 
    Special() { getValue(); } 

    static int getValue() 
    { 
     int value = x*Special<x-1, y-1>::getValue(); 
     std::cout << "Value, x and y are: " << value << "," << x << "," << y << std::endl; 
     return value; 
    } 
}; 

template <unsigned n> 
struct Special<n, 0> 
{ 
    static int getValue() 
    { 
     int value = 1; 
     std::cout << "Value, x and y are: " << value << "," << n << "," << 0 << std::endl; 
     return value; 
    } 
}; 

int main() 
{ 
    Special<8,3> s; 
} 

輸出在這兩種情況下：

Value, x and y are: 1,5,0 
Value, x and y are: 6,6,1 
Value, x and y are: 42,7,2 
Value, x and y are: 336,8,3