lambda - CUDA Thrust shortcut math functions -

is there way automatically wrap cuda math function in functor 1 can apply thrust::transform without having write functor manually? functionality (i gather) std::function provides?

thrust::placeholders doesn't seem math functions. std::function doesn't seem available.

example code:

#include <thrust/transform.h> #include <thrust/device_vector.h> #include <iostream> #include <functional> #include <math.h>  struct myfunc{     __device__      double operator()(double x,double y){     return hypot(x,y);     } };  int main(){      double x0[10] = {3.,0.,1.,2.,3.,4.,5.,6.,7.,8.};     double y0[10] = {4.,0.,1.,2.,3.,4.,5.,6.,7.,8.};      thrust::device_vector<double> x(x0,x0+10);     thrust::device_vector<double> y(y0,y0+10);     thrust::device_vector<double> r(10);      (int i=0;i<10;i++) std::cout << x0[i] <<" ";    std::cout<<std::endl;     (int i=0;i<10;i++) std::cout << y0[i] <<" ";    std::cout<<std::endl;      // works:     thrust::transform(x.begin(),x.end(),y.begin(),r.begin(), myfunc());      // doesn't compile:     using namespace thrust::placeholders;     thrust::transform(x.begin(),x.end(),y.begin(),r.begin(), hypot(_1,_2));      // nor this:     thrust::transform(x.begin(),x.end(),y.begin(),r.begin(), std::function<double(double,double)>(hypot));       (int i=0;i<10;i++) std::cout << r[i] <<" ";    std::cout<<std::endl; } 

converting comment answer:

as @jaredhoberock stated, there no automatic way achieve want. there syntactic / typing overhead.

one way reduce overhead of writing separate functor (as did my_func) use lambdas. since cuda 7.5 there experimental device lambda feature allows folllowing:

auto h = []__device__(double x, double y){return hypot(x,y);}; thrust::transform(x.begin(),x.end(),y.begin(),r.begin(), h); 

you need add following nvcc compiler switch compile this:

nvcc --expt-extended-lambda ... 

---

another approach convert function functor using following wrapper:

template<typename sig, sig& s> struct wrapper;  template<typename r, typename... t, r(&function)(t...)> struct wrapper<r(t...), function> {     __device__     r operator() (t&... a)     {         return function(a...);     } }; 

you use this:

 thrust::transform(x.begin(),x.end(),y.begin(),r.begin(), wrapper<double(double,double), hypot>());