Let's say I have something like this bit of (duplicated) code that I'd like to refactor using a template:
#include <iostream>
#include <algorithm>
#include <set>
struct IntFoo {
auto f(int arg, std::set<int> s = {1, 2, 3}) {
return std::find(s.begin(), s.end(), arg) != s.end();
}
};
struct FloatFoo {
auto f(float arg, std::set<float> s = {4.0f, 5.0f, 6.0f}) {
return std::find(s.begin(), s.end(), arg) != s.end();
}
};
int main() {
std::cout << IntFoo().f(3) << std::endl;
std::cout << FloatFoo().f(4.0f) << std::endl;
}
As you can see, beyond the variance in type there is also the change in the default arguments given to f()
's second parameter.
Best I could come up with was this:
#include <iostream>
#include <algorithm>
#include <set>
template<typename T, typename Def>
struct Foo {
auto f(T arg, std::set<T> s = Def::defaults){
return std::find(s.begin(), s.end(), arg) != s.end();
}
};
struct FooIntDefaults {
static constexpr std::initializer_list<int> defaults{1, 2, 3};
};
struct FooFloatDefaults {
static constexpr std::initializer_list<float> defaults{4.0f, 5.0f, 6.0f};
};
using IntFoo = Foo<int, FooIntDefaults>;
using FloatFoo = Foo<float, FooFloatDefaults>;
This works, but is a bit verbose. I don't fancy these helper structs much.
Ideally I'd like to pass the default arguments in the using
line somehow. Is there some better way?
CodePudding user response:
You can use parameter pack for specifying default arguments, e.g.
template<typename T, T... defaults>
struct Foo {
auto f(T arg, std::set<T> s = {defaults...}){
return std::find(s.begin(), s.end(), arg) != s.end();
}
};
using IntFoo = Foo<int, 1, 2, 3>; // specify default arguments when defining type
using FloatFoo = Foo<float, 4.0f, 5.0f, 6.0f>; // specify default arguments when defining type
BTW: Note that float
can't be used as non-type template parameter before C 20.