C++遞歸模板類型推理

Question

我很想學習一點關於模板元編程的知識。在下面的代碼中，我試圖找到一個無符號整數類型，它足夠大，以便在編譯時指定N位，並使用一些模板遞歸。

template <typename T> 
struct NextIntegralType 
{ 
}; 

template <> 
struct NextIntegralType<void> 
{ 
    typedef unsigned char type; 
}; 

template <> 
struct NextIntegralType<unsigned char> 
{ 
    typedef unsigned short type; 
}; 

...More type 'iteration' here... 

template<size_t BITS, typename T> 
struct FindIntegralType2 
{ 
    typedef std::conditional<BITS <= sizeof(typename T::type)*8, T, FindIntegralType2<BITS, NextIntegralType<typename T::type>>> _type; 
    typedef typename _type::type type; 
}; 

template<size_t BITS> 
struct FindIntegralType 
{ 
    typedef typename FindIntegralType2<BITS, NextIntegralType<void>>::type type; 
}; 

當我聲明一個變量並分配一個整數值吧...

FindIntegralType<15>::type test(4000); 

我得到如下：

error: no matching function for call to ‘FindIntegralType2<15u, NextIntegralType<unsigned char> >::FindIntegralType2(int)’ 
note: candidates are: 
note: constexpr FindIntegralType2<15u, NextIntegralType<unsigned char> >::FindIntegralType2() 
note: candidate expects 0 arguments, 1 provided 
note: constexpr FindIntegralType2<15u, NextIntegralType<unsigned char> >::FindIntegralType2(const FindIntegralType2<15u, NextIntegralType<unsigned char> >&) 
note: no known conversion for argument 1 from ‘int’ to ‘const FindIntegralType2<15u, NextIntegralType<unsigned char> >&’ 

好像我的遞歸是不是'平倉」。任何人都可以將我指向正確的方向嗎？我使用GCC 4.6。

編輯：
我發現後，我錯過了前：
Automatically pick a variable type big enough to hold a specified number

指向升壓答案（他們總是有）：
boost_integer

這應該解決我的實際需求和求知慾。

Answer 1

您的問題是_type::type評估爲std::conditional<...>::type，而不是FindIntegralType2<...>::type。將其更改爲typedef typename _type::type::type type;（太多type x_X）。這應該可以解決你的問題。