模板函数作为参数传递给模板函数 - 悬而未决的功能类型错误

Question

我试图传递的接收的参考到本征MatrixBase到另一个函数模板F_2那将，用它做什么函数模板F_1：

//Function that will be argument to F_2 
template<typename derived_matrix,typename derived_float> 
void F_1(const MatrixBase<derived_matrix>& x,derived_float t,MatrixBase<derived_matrix>& store){ 
    store = 2*x; 
} 

template<typename derivative_function,typename const_derived_matrix,typename derived_matrix,typename derived_float> 
void F_2(derivative_function derivatives,const_derived_matrix x0,derived_float t0,derived_float dt,derived_matrix& output){ 
    //Do something with F_1(= derivatives) 
    derived_matrix k1; 
    derived_matrix k2; 
    derived_matrix k3; 
    derived_matrix k4; 
    derivatives(x0,t0,k1); 
    derivatives(x0+dt*k1/2,t0+dt/2,k2); 
    derivatives(x0+dt*k2/2,t0+dt/2,k3); 
    derivatives(x0+dt*k3/2,t0+dt,k4); 
    output = x0+dt*(k1+2*k2+2*k3+k4)/6; 
} 

但我从F_2函数调用收到错误“未解决重载函数类型”：

double t = 0; 
double dt = 0.1; 
Vector2d x; 
x << 1,2; 
Vector2d out_value; 
F_2(F_1,x,t,dt,out_value); //unresolved overloaded function type 

它减少到最低程度，我只能做一个普通的F_2模板函数接收F_1功能时，它（F_1 ）不是模板：

#include <iostream> 
#include <functional> 
using namespace std; 

template<typename f,typename derived_float> 
void F_2(f wrapped,derived_float x1,derived_float& x2){ 
    wrapped(x1,x2); 
} 

void F_1_works(double& x,double& store_output){ 
    store_output = 2*x; 
} 

template<typename T> 
void F_1_broken(double& x,T& store_output){ 
    store_output = 2*x; 
} 

int main(){ 
    double out= 0; 
    double x = 25; 
    F_2(F_1_works,x,out); //compiles! 
    F_2(F_1_broken,x,out); //Error: 
         //  no matching function for call to ‘F_2(<unresolved overloaded function type>, double&, double&)’ 
         //  couldn't deduce template parameter ‘f’ 
    cout<<out<<endl; 
    return 0; 
} 

1. 我怎样才能让编译器推断的“包装”在F_2（模板参数f）的类型？
2. 您有什么建议可以与Eigen做不同吗？

Answer 1

试试这个：

F_2(F_1_broken<double>,x,out); 

编译器有没有办法在这里演绎的类型。 所以你必须帮助他:-)