C++ 模板歧義實例化

Question

我正在嘗試使用模板進行一些插值，但出現“模棱兩可的模板實例化”錯誤。 這是代碼

// interpolation rules
enum InterRule {trap, rect, trapSum};

// Rectangle rule
template <int n, int k, InterRule rule, class Expr> struct Integration {
    static double integrate(double a, double b){
        return (b-a)/n * Expr::eval(a + (k-1)*(b-a)/n) + Integration<n, k - 1, rule, Expr>::integrate(a,b);
    }
};

// termination case
template <int n, InterRule rule, class Expr> struct Integration<n,0,rule,Expr> {
    static double integrate(double a, double b){
        return 0;
    }
};

// Trapezoidal rule
template <int n, int k, class Expr> struct Integration<n, k, trap, Expr> {
    static double integrate(double a, double b){
        return (b-a)/n * (Expr::eval(a)/2 + Integration<n,k-1,trapSum,Expr>::integrate(a,b) + Expr::eval(b)/2);
    }
};

// Trapezoidal sum
template <int n, int k, class Expr> struct Integration<n, k, trapSum, Expr> {
    static double integrate(double a, double b){
        return Expr::eval(a + k*(b-a)/n) + Integration<n,k-1,trapSum,Expr>::integrate(a,b);
    }
};

基本上，我正在嘗試實施梯形規則，以便靜態展開。 但是，編譯器似乎對使用“終止案例”還是“梯形和”感到困惑。 我做錯了什么，有解決方法嗎？ 如果k==0無論InterRule規則的類型如何，我都想強制它使用“終止案例”。

編輯附加代碼以使其運行：

// the types of expressions (+,-,*, etc.)
enum ExprType { mul, divide, add, sub, constant};

// constant
template <ExprType eType, class Left, class Right, int coeff, int power> struct Expr {
    static double eval(double x){
        return coeff * std::pow(x, power);
    }
};


int main()
{

    double a = 1;
    double b = 2;

    // Expr defines the function f(x) = x
    Integration<50, 50, trap, Expr<constant,int,int,1,1>> inte2;
    std::cout << inte2.integrate(a,b) << std::endl;

    return 0;
}

Answer 1

您可以嘗試消除添加額外模板參數的歧義。

也許像

// .................................................VVVVVVVVVVV
template <int n, int k, InterRule rule, class Expr, bool = true>
struct Integration {
   // ...
};

顯式使用false的基本情況

template <int n, InterRule rule, class Expr>
struct Integration<n, 0, rule, Expr, false> { // <--- false !
    static double integrate(double a, double b){
        return 0;
    }
};

並在遞歸調用中添加正確的參數

    return (b-a)/n * Expr::eval(a + (k-1)*(b-a)/n)
       + Integration<n, k - 1, rule, Expr, (k>1)>::integrate(a,b);
    // ....................................^^^^^

下面是一個完整的編譯示例

#include <cmath>
#include <iostream>

// interpolation rules
enum InterRule {trap, rect, trapSum};

// Rectangle rule
template <int n, int k, InterRule rule, class Expr, bool = true>
struct Integration {
    static double integrate(double a, double b){
        return (b-a)/n * Expr::eval(a + (k-1)*(b-a)/n)
           + Integration<n, k - 1, rule, Expr, (k>1)>::integrate(a,b);
    }
};

// termination case
template <int n, InterRule rule, class Expr>
struct Integration<n, 0, rule, Expr, false> {
    static double integrate(double a, double b){
        return 0;
    }
};

// Trapezoidal rule
template <int n, int k, class Expr>
struct Integration<n, k, trap, Expr> {
    static double integrate(double a, double b){
        return (b-a)/n * (Expr::eval(a)/2
           + Integration<n,k-1,trapSum,Expr,(k>1)>::integrate(a,b)
           + Expr::eval(b)/2);
    }
};

// Trapezoidal sum
template <int n, int k, class Expr>
struct Integration<n, k, trapSum, Expr> {
    static double integrate(double a, double b){
        return Expr::eval(a + k*(b-a)/n)
           + Integration<n,k-1,trapSum,Expr,(k>1)>::integrate(a,b);
    }
};

// the types of expressions (+,-,*, etc.)
enum ExprType { mul, divide, add, sub, constant};

// constant
template <ExprType eType, class Left, class Right, int coeff, int power> struct Expr {
    static double eval(double x){
        return coeff * std::pow(x, power);
    }
};


int main ()
 {

    double a = 1;
    double b = 2;

    // Expr defines the function f(x) = x
    Integration<50, 50, trap, Expr<constant,int,int,1,1>> inte2;
    std::cout << inte2.integrate(a,b) << std::endl;
 }