How to apply Template Method Pattern for functions with different signatures? - c++

Question

How does one provide a unified interface to sets of functions, that are used in the same way? To illustrate, please look at the set of given library functions:
/* existing library functions */
/* the signatures are different: some return int, some float */
/* set of input related functions */
int getInputValue() { return 42; }
size_t getInputSize() { return 1; }
/* set of output related functions */
int getOutputValue() { return 21; }
size_t getOutputSize() { return 1; }
/* set of parameter related functions */
float getParameterValue() { return 3.14; }
size_t getParameterSize() { return 1; }
and assume they are used in the same way:
if (getSize() > 0) {
T value = getValue()
A) What is a good way to provide getSize() and getValue()?
I first though that Template Method Pattern is what I want, but I couldn't apply it, because in contrast to the Worker in the Template Method Pattern, my functions have different signatures.
So what I did instead:
/* I want to provide a uniform interface */
/* the specific part of inputs, outputs and parameters is in the traits */
struct input_traits {
typedef int value_type;
static int getValue() { return getInputValue(); }
static size_t getSize() { return getInputSize(); }
};
struct output_traits {
typedef int value_type;
static int getValue() { return getOutputValue(); }
static size_t getSize() { return getOutputSize(); }
};
struct parameter_traits {
typedef float value_type;
static float getValue() { return getParameterValue(); }
static size_t getSize() { return getParameterSize(); }
};
/* the common part (they are used in the same way) is in the Helper */
template<typename traits>
class CommonUsage {
public:
void use()
{
if (traits::getSize() > 0) {
typename traits::value_type value = traits::getValue();
}
}
};
int main()
{
CommonUsage<input_traits>().use();
CommonUsage<output_traits>().use();
CommonUsage<parameter_traits>().use();
}
B) Is this a good approach?

Answer 1

A. If i understood your question correctly you should use an abstract class.
Look at the next code, it essentially does the same as your code.
This is how i would do it.
#include <iostream>
template <typename value_type>
class Traits {
public:
virtual value_type getValue() const = 0;
virtual size_t getSize() const = 0;
virtual ~Traits() { }
};
class input_traits: public Traits <int>{
public:
virtual int getValue() const {
return 42;
}
virtual size_t getSize() const {
return 1;
}
};
class parameter_traits: public Traits <double>{
public:
virtual double getValue() const {
return 3.14;
}
virtual size_t getSize() const {
return 1;
}
};
class CommonUsage {
public:
template <typename value_type>
void use(const Traits<value_type>& traitsObject) {
if (traitsObject.getSize() > 0) {
std::cout << traitsObject.getValue();
}
}
};
int main() {
CommonUsage().use(parameter_traits());
return 0;
}

Answer 2

As an alternative to user's answer (this time using template specialization) is:
template <class T>
struct traits {
T getValue() const { throw std::runtime_exception("..."); }
size_t getSize() const { return 0; }
};
template <>
struct traits<int> {
int getValue() const { return 42; }
size_t getSize() const { return 1; }
};
template <>
struct traits<float> {
int getValue() const { return 3.145; }
size_t getSize() const { return 1; }
};
// do template aliasing
using input_traits = traits<int>;
using parameter_traits = traits<float>;
struct CommonUsage {
template <typename T>
static void use(const traits<T> &traits) {
if (traits.getSize() > 0)
std::cout << traits.getValue() << std::endl;
}
};
int main(int arg, char **argv) {
CommonUsage::use(input_traits());
CommonUsage::use(parameter_traits());
}
There are advantages/disadvantages to both approaches. If you use template specialization, you don't pay the overhead of virtual methods.