部分元類減法不能編譯

Question

我想寫在編譯時做基本操作的struct Fraction。請注意，這不具有任何實際用途 - 我只是將其作爲練習。

我開始了與此：

namespace internal 
{ 
    // Euclid algorithm 
    template <int A, int B> 
    struct gcd 
    { 
     static int const value = gcd<B, A % B>::value; 
    }; 

    // Specialization to terminate recursion 
    template <int A> 
    struct gcd<A, 0> 
    { 
     static int const value = A; 
    }; 
} 

template <int Numerator, int Denominator> 
struct Fraction 
{ 
    // Numerator and denominator simplified 
    static int const numerator = Numerator/internal::gcd<Numerator, Denominator>::value; 
    static int const denominator = Denominator/internal::gcd<Numerator, Denominator>::value; 

    // Add another fraction 
    template <class Other> struct add 
    { 
     typedef Fraction< 
      Numerator * Other::denominator + Other::numerator * Denominator, 
      Denominator * Other::denominator 
     > type; 
    }; 
}; 

這編譯和工作原理：Fraction<1,2>::add< Fraction<1,3> >::type會Fraction<5,6>。現在，我嘗試添加減法：

template <class Other> 
struct sub 
{ 
    typedef typename Fraction<Numerator, Denominator>::add< 
     Fraction<-Other::numerator, Other::denominator> 
    >::type type; 
}; 

但我得到一個編譯器錯誤，我不明白：

Error: "typename Fraction<Numerator, Denominator>::add" uses "template<int Numerator, int Denominator> template <class Other> struct Fraction::add" which is not a type 

有人能向我解釋什麼，編譯器說，爲什麼我不能做我想要的東西？順便說一下，我使用g++ 4.4.6。

Answer 1

使用模板關鍵字。

template <class Other> 
struct sub 
{ 
    typedef typename Fraction<Numerator, Denominator>::template add< 
     Fraction<-Other::numerator, -Other::denominator> 
    >::type type; 
}; 

http://liveworkspace.org/code/26f6314be690d14d1fc2df4755ad99f6

閱讀本Where and why do I have to put the "template" and "typename" keywords?爲更好的解釋。