I've created a small utility function for string conversion so that I don't have to go around creating ostringstream objects all over the place
template<typename T>
inline string ToString(const T& x)
{
std::ostringstream o;
if (!(o << x))
throw BadConversion(string("ToString(") + typeid(x).name() + ")");
return o.str();
}
I want to provide some specialisations of this method for instances where there is no default overloaded << operator for stringstream (i.e. std::pair, std::set, my own classes) and I've run into difficulties with the templating. I will illustrate with the std::pair example, if I want to be able to
string str = ToString(make_pair(3, 4));
the only way I can think of to do it is to define the explicit specialisation for int
template<>
inline string ToString(const pair<int,int>& x)
{
std::ostringstream o;
if (!(o << "[" << x.first << "," << x.second << "]"))
throw BadConversion(string("ToString(pair<int,int>)"));
return o.str();
}
is there a way I can define this for the generic case?
template<>
inline string ToString(const pair<T1,T2>& x)
{
std::ostringstream o;
if (!(o << "[" << x.first << "," << x.second << "]"))
throw BadConversion(string("ToString(pair<T1,T2>)"));
return o.str();
}
The act of creating a new definition of a function, class, or member of a class from a template declaration and one or more template arguments is called template instantiation. The definition created from a template instantiation is called a specialization.
There are two types of templates in C++, function templates and class templates.
Explanation: Template specialization is used when a different and specific implementation is to be used for a specific data type. In this program, We are using integer and character.
There are three kinds of templates: function templates, class templates and, since C++14, variable templates. Since C++11, templates may be either variadic or non-variadic; in earlier versions of C++ they are always non-variadic.
Don't specialize the template, but overload it. The compiler will figure out what function template to take by ordering them according to their specialization of their function parameter types (this is called partial ordering).
template<typename T1, typename T2>
inline string ToString(const std::pair<T1, T2>& x) {
std::ostringstream o;
if (!(o << "[" << x.first << "," << x.second << "]"))
throw BadConversion(string("ToString(pair<T1,T2>)"));
return o.str();
}
In general, partial ordering will result in exactly what you expect. In more detail, consider having these two functions
template<typename T> void f(T);
template<typename T, typename U> void f(pair<T, U>);
Now, to see whether one is at least as specialized as the other one, we test the following for both function templates:
Example for the above:
X1
into T
gives us some type, call it X1
. argument deduction of X1
against pair<T, U>
won't work. So the first is not at least as specialized as the second template. Y1
and Y2
into pair<T, U>
yields pair<Y1, Y2>
. Doing argument deduction against T
of the first template works: T
will be deduced as pair<Y1, Y2>
. So the second is at least as specialized as the first. The rule is, a function template A is more specialized than the other B, if A is at least as specialized as B, but B is not at least as specialized as A. So, the second wins in our example: It's more specialized, and it will be chosen if we could in principle call both template functions.
I'm afraid, that overview was in a hurry, i only did it for type parameters and skipped some details. Look into 14.5.5.2
in the C++ Standard Spec to see the gory details. g
What about using type traits for serializing different types to the stream like this:
template<typename T>
struct Traits {
static inline bool ToStream(std::ostringstream& o, const T& x) {
return o << x;
}
};
template<typename T1, typename T2>
struct Traits<std::pair<T1, T2> > {
static inline bool ToStream(std::ostringstream& o, const std::pair<T1, T2>& x) {
return o << "[" << x.first << "," << x.second << "]";
}
};
template<typename T>
inline std::string ToString(const T& x)
{
std::ostringstream o;
if (!Traits<T>::ToStream(o, x))
return "Error";
return o.str();
}
Note: "template<>" from the specialization part is optional, the code compiles fine without it. You can further add methods in the traits for the exception message.
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