STL container as template parameter in function, error in call

Question

Cant understand what is wrogn with code, second function definition or call of this function in main? I think, but not sure, problem in call, cause without calling code compiles well. Compiler gcc

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

template<class T>
void show_element(T ob)
{
    cout << ob << " ";
}

template<template<class> class S, class T>
void show_sequence(S<T> sequence)
{
    for_each(sequence.begin(), sequence.end(), show_element<T>);    
}

int main(int argc, char const *argv[])
{
    std::vector<int> v(20, 0);

    //here the problem
    show_sequence<std::vector<int>, int>(v);

    return 0;
}

Answer 1

std::vector isn't a template of one parameter, it takes an allocator type as well. You can use it as vector<T> simply because the second parameter has a default (std::allocator<T>).

As it's written, your template function cannot accept any standard container, since off the top of my head, none take just a single type parameter.

An approach that would work, and not require you to know how many template parameters a container requires, is to accept a container type (not template), and glean the value type from the container type.

template<class Seq>
void show_sequence(Seq const& sequence)
{
    typedef typename Seq::value_type T;
    for_each(sequence.begin(), sequence.end(), show_element<T>);    
}

All standard containers have a value_type member, so this will work with any of them. Furthermore, it will work with any container that takes its cue from the standard library.

Answer 2

The problem is that std::vector is a template but std::vector<int> is a type.

When you are giving the second one to the function, you are giving one type and not a template.

So, you can rewrite your function as :

template<class S>
void show_sequence(S sequence)

Moreover, vector does not take only one template paramete but two (see StoryTeller answer)