编译时的模板参数计算

Question

I'm trying to deduce the greater of two template arguments at compile time. Both template arguments are of type size_t.

I have a templated type, SomeType, which takes a size_t as it's template argument. I then have a function that takes two SomeType parameters with different template size_t's and i want the return type to be a SomeType with its templated size_t to be the greater of the two input size_t sizes.

template <size_t d> struct SomeType {...}

template<size_t d1, size_t d2>
SomeType<the_larger_of_d1_and_d2> Func(SomeType<d1> A, SomeType<d2> B)
{
    ...
}

Is this possible?

Answer 1

You can compute the type directly, no need for SFINAE:

template<size_t d1, size_t d2>
SomeType<(d1 > d2 ? d1 : d2)> Func(SomeType<d1> A, SomeType<d2> B)
{
    …
}

Answer 2

The solution by @KonradRudolph is correct of course. But if you want to delve further in to template metaprogramming, it would pay-off very quickly to learn Boost.MPL . It provides a whole battery of convenience functions. Eg your question can be solved like

#include <iostream>
#include <boost/mpl/int.hpp>
#include <boost/mpl/max.hpp>

template<size_t d> 
struct SomeType
: 
    boost::mpl::int_<d> 
{};

template<size_t d1, size_t d2>
typename boost::mpl::max<SomeType<d1>, SomeType<d2> >::type
Func(SomeType<d1> A, SomeType<d2> B) 
{
    return typename boost::mpl::max<SomeType<d1>, SomeType<d2> >::type();
}

int main()
{
    SomeType<2> st2;
    SomeType<3> st3;
    boost::mpl::max<SomeType<2>, SomeType<3> >::type res = Func(st2, st3);
    std::cout << res.value;
}  

Live Example .

Some notes:

• letting SomeType inherit from boost::mpl::int_ endows it with a type and value , as well as some convenient tags. This makes it very easy to re-use other metafunctions from Boost.MPL
• the boost::mpl::max does the same ternary trick behind the scenes. It is more readible IMO, and if you ever want to change to another condition it's easy to do so.
• there is a bit of a learning curve for Boost.MPL, but the tutorial at the linked documentation should get you started.

Answer 3

If you can make use of the c++11 standard, you can use SFINAE standard support:

template<size_t one, size_t two>
struct larger {
    static constexpr typename std::enable_if<(one > two), size_t>::type value() {
        return one;
    }

    static constexpr typename std::enable_if<(two >= one, size_t>::type value() {
        return two;
    }
};

Then

template<size_t d1, size_t d2>
SomeType<larger<d1, d2>::value()> Func(SomeType<d1> A, SomeType<d2> B)
{
    ...
}

Answer 4

As I'm constantly have to look up myself (my older code) about that question again and again, I've decided to make a GIT gist , and compilation sample that allows (at least me) to quickly access some 'template' code (pun intended), to play with the meta-programmed conditional type selection stuff (also working for the 'old' c++03 standard):

Selector declaration:

template<typename FalseType, typename TrueType, bool condition>
struct ConditionalTypeSelector {
    typedef void ResultType;
};

---

Selector specialization(s):

template<typename FalseType, typename TrueType>
struct ConditionalTypeSelector<FalseType,TrueType,false> {
    typedef FalseType ResultType;
};

template<typename FalseType, typename TrueType>
struct ConditionalTypeSelector<FalseType,TrueType,true> {
    typedef TrueType ResultType;
};

---

Selected types:

struct A {
    unsigned char member;
};

struct B {
    int member;
};

struct C {
    long long member;
};

---

Testing:

#include <iostream>
#include <typeinfo>

using namespace std;

int main() {
    cout << typeid
                ( ConditionalTypeSelector
                      < A,B,(sizeof(A) > sizeof(B))>::ResultType
                ).name() << endl;
    cout << typeid
                ( ConditionalTypeSelector
                      <A,B,(sizeof(B) > sizeof(A)) >::ResultType
                ).name() << endl;
    cout << typeid
                ( ConditionalTypeSelector
                      < A,C,(sizeof(A) > sizeof(C))>::ResultType
                ).name() << endl;
    cout << typeid
                ( ConditionalTypeSelector
                      < C,B,true>::ResultType
                ).name() << endl;
    cout << typeid
                ( ConditionalTypeSelector
                      < C,A,false>::ResultType
                ).name() << endl;

    return 0;
}

It's pretty easy to change this template to use eg an enum type for specialized selections, or whatever else constant condition known at compile time should be checked.