如何在容器和类型均为参数的情况下创建模板函数？

Question

This might be a trivial question but is driving me crazy. I want to define a single function foo() that would work with different containers like: vector<int> , vector<double> , set<int> and set<double> .

I have tried to define foo like this:

template<typename CONT, typename T>
   int foo(CONT<T>){
      //evaluate x
      return (int) x ;
   }

This kind of definition doesn't work, but I can't understand why.

How can I achieve something similar?

Answer 1

The way to specify both the container class template and its instantiation is to use template template parameters:

template <template <typename...> class Cont, typename T>
int foo(Cont<T>) {
    ...
}

Note that Cont is using a variable number of arguments because otherwise it wouldn't cover the unknown number of defaulted template arguments the standard containers have.

Answer 2

Consider this:

template< class ContainerT >
int foo( ContainerT const& c ) {
}

Then ContainerT can be any thing, including std::vector<int> , std::vector<std::string> or even std::map<std::string, int> . So you don't need to add a new template parameter and if you need to know the type just use value_type of your container:

typedef typename ContainerT::value_type container_type; // Or T in your foo