Caveat: It is not necessary to put the implementation in the header file, see the alternative solution at the end of this answer.

Anyway, the reason your code is failing is that, when instantiating a template, the compiler creates a new class with the given template argument. For example:

template<typename T>
struct Foo
{
    T bar;
    void doSomething(T param) {/* do stuff using T */}
};

// somewhere in a .cpp
Foo<int> f; 

When reading this line, the compiler will create a new class (let's call it FooInt), which is equivalent to the following:

struct FooInt
{
    int bar;
    void doSomething(int param) {/* do stuff using int */}
};

Consequently, the compiler needs to have access to the implementation of the methods, to instantiate them with the template argument (in this case int). If these implementations were not in the header, they wouldn't be accessible, and therefore the compiler wouldn't be able to instantiate the template.

A common solution to this is to write the template declaration in a header file, then implement the class in an implementation file (for example .tpp), and include this implementation file at the end of the header.

Foo.h

template <typename T>
struct Foo
{
    void doSomething(T param);
};

#include "Foo.tpp"

Foo.tpp

template <typename T>
void Foo<T>::doSomething(T param)
{
    //implementation
}

This way, implementation is still separated from declaration, but is accessible to the compiler.

Alternative solution

Another solution is to keep the implementation separated, and explicitly instantiate all the template instances you'll need:

Foo.h

// no implementation
template <typename T> struct Foo { ... };

Foo.cpp

// implementation of Foo's methods

// explicit instantiations
template class Foo<int>;
template class Foo<float>;
// You will only be able to use Foo with int or float

If my explanation isn't clear enough, you can have a look at the C++ Super-FAQ on this subject.

The compiler is allowed to make one implicit conversion to resolve the parameters to a function. This means that the compiler can use constructors callable with a single parameter to convert from one type to another in order to get the right type for a parameter.

Here's an example with converting constructors that shows how it works:

struct Foo {
    // Single parameter constructor, can be used as an implicit conversion.
    // Such a constructor is called "converting constructor".
    Foo(int x) {}
};
struct Faz {
    // Also a converting constructor.
    Faz(Foo foo) {}
};

// The parameter is of type Foo, not of type int, so it looks like
// we have to pass a Foo.
void bar(Foo foo);

int main() {
    // However, the converting constructor allows us to pass an int.
    bar(42);
    // Also allowed thanks to the converting constructor.
    Foo foo = 42;
    // Error! This would require two conversions (int -> Foo -> Faz).
    Faz faz = 42;
}

Prefixing the explicit keyword to the constructor prevents the compiler from using that constructor for implicit conversions. Adding it to the above class will create a compiler error at the function call bar(42). It is now necessary to call for conversion explicitly with bar(Foo(42))

The reason you might want to do this is to avoid accidental construction that can hide bugs.
Contrived example:

• You have a MyString class with a constructor that constructs a string of the given size. You have a function print(const MyString&) (as well as an overload print (char *string)), and you call print(3) (when you actually intended to call print("3")). You expect it to print "3", but it prints an empty string of length 3 instead.