C++ variadic function template in std::functional - c++

Question

basically I want to integrate a multi-dimensional integral with this recursion.
But the problem itself is a general one. It's not specific for integration.
#include "math.h"
#include <iostream>
#include <functional>
double f(double x,double y,double z){
return x+y+z+1;
}
//Base
double redDim(std::function<double(double)> &f){
return f(0); //a silly integrator for testing
}
// Recursion
template<typename Tfirst=double, typename... Trest>
auto redDim(std::function<double(Tfirst first,Trest... rest)> &f){
return redDim([=](Trest... R){return redDim([=](double x){return f(x,R...);});});
}
int main(){
std::cout<<redDim(f)<<std::endl;
return 0;
}
The problem is, compiler says:
c:\C++\templateTutorial\templateTut.cpp: In function 'int main()':
c:\C++\templateTutorial\templateTut.cpp:24:19: error: no matching function for call to 'redDim(double (&)(double, double, double))'
cout<<redDim(f)<<endl;
^
c:\C++\templateTutorial\templateTut.cpp:12:8: note: candidate: 'double redDim(std::function<double(double)>&)'
double redDim(std::function<double(double)> &f){
^~~~~~
c:\C++\templateTutorial\templateTut.cpp:12:8: note: no known conversion for argument 1 from 'double(double, double, double)' to 'std::function<double(double)>&'
c:\C++\templateTutorial\templateTut.cpp:17:6: note: candidate: 'template<class Tfirst, class ... Trest> auto redDim(std::function<double(Tfirst, Trest ...)>&)'
auto redDim(std::function<double(Tfirst first,Trest... rest)> &f){
^~~~~~
c:\C++\templateTutorial\templateTut.cpp:17:6: note: template argument deduction/substitution failed:
c:\C++\templateTutorial\templateTut.cpp:24:19: note: mismatched types 'std::function<double(Tfirst, Trest ...)>' and 'double(double, double, double)'
cout<<redDim(f)<<endl;
^
The terminal process terminated with exit code: 1
So why is the type of f not matching the requirements of redDim() ?
Thus I can't even test, if my method works.
I hope you could help me!

Answer 1

A function pointer is not a std::function.
Template argument deduction does not do any type conversions, other than a few to-base cases.
Add a redim( double(*f)(Args...) ) template that calls the std function one.
template<class...Args>
auto redDim( double(*f)(Args...) {
return redDim( std::function<double(Args...)>{ f } );
}
and it should work.
This can deduce the signature of the function pointer. It then explicitly converts to a std::function, which then matches your other redim functions.
You do have to explicitly convert your other lambdas to std::functions:
// Recursion
template<typename Tfirst, typename... Trest>
auto redDim(std::function<double(Tfirst first,Trest... rest)> f){
return redDim(
std::function<double(Trest...)>{
[=](Trest... R){
return redDim(
std::function<double(double)>{
[=](double x){return f(x,R...);}
}
);
}
}
);
}
Live example.
Also change the argument for the last redDim to std::function<double(double)> (not a reference). Alternatively, a const&.

Answer 2

Based on Yakk - Adam Nevraumont's answer, the following code should compile:
#include "math.h"
#include <iostream>
#include <functional>
double f(double x,double y,double z){
return x+y+z+1;
}
//Base
double redDim(const std::function<double(double)> &f){
return f(0); //a silly integrator for testing
}
// Recursion
template<typename Tfirst=double, typename... Trest>
auto redDim(const std::function<double(Tfirst first,Trest... rest)> &f) {
return redDim(
std::function<double(Trest...)>{
[=](Trest... R)->double{return redDim([=](double x){return f(x,R...);});}});
}
template<class...Args>
auto redDim( double(*f)(Args...)) {
return redDim( std::function<double(Args...)>{ f } );
}
int main(){
std::cout<<redDim(f)<<std::endl;
return 0;
}
As Yakk already pointed out, std::function and a function pointer don't have the same type. Please note that I've also changed the argument type of redDim to const std::function<...> &). You could also pass rvalue references using the && syntax.