c++ - Passing vector of derived, templated class -

i'd define general function foo takes data, perhaps manipulates underlying class variables, , returns int. however, when attempt create separate function takes vector of foo objects, compiler fails deduce template parameter. following illustrates i've tried:

#include <vector>  template <typename t> class base { public:   virtual int foo(const t& x) const = 0; };  template <typename t> class derived : public base<std::vector<t> > { // specialize vector data public:   virtual int foo(const std::vector<t>& x) const { return 0;} };  template <typename t> int bar(const t& x, const std::vector< base<t> >& y) {   if(y.size() > 0)     return y[0].foo(x); }  int main(int argc, char** argv) {   std::vector<double> x;   std::vector< derived<double> > y;   bar(x, y); } 

this fails find matching function bar, notes:

main.cc:16:5: note:   template argument deduction/substitution failed: main.cc:24:11: note:   mismatched types ‘base<t>’ , ‘derived<double>’ 

and

main.cc:24:11: note:   ‘std::vector<derived<double> >’ not derived \ ‘const std::vector<base<t> >’ 

forgive me if answer lies in already-posted thread; i've read quite number seem related, don't, knowledge, address issue.

first note std::vector<base<t> > , std::vector<derived<t> > different types, if base<std::vector<t>> base of derived<t>. type conversion doesn't happen in template type deduction. t cannot deduced matching second argument y of type std::vector<derived<double>> pass bar std::vector<base<t>>.

next, suppose make y of "right" type

std::vector< base<double> > y;  

so can pass bar. in principle can deduce t matching second parameter in bar of type std::vector<base<t>> type std::vector< base<double> > of y. t deduced double, don't forget x, pass first parameter bar, has type vector<double>, x deduce t vector<double>, of course inconsistent double deduced y. type deduction fails.

here simplified example replicates issue.