Tuesday Coding Tip 24 — Getting template type from an existing object

Question

Tuesday coding tips are super short posts about various tidbits mainly from C++, but also from other programming languages I use. You can also follow the #TuesdayCodingTips hashtag on Mastodon and Linkedin.

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Nice APIs expose their template types as member aliases so that anybody can get them easily. For other types, you might be able to deduce the type by calling decltype on a return value of some of their functions.

Sometimes, none of that is possible. But there is still a way to get to it. You need to write a template structure that only matches on the type you care about and alias the template parameters so you can access them later, a technique known as ‘type trait’.

#include <vector>

// Base template, only for the compiler
template<class T>
struct UnderlyingVectorType {};

// Overload for vector, note that templates don't have to match
template<class T, class A>
struct UnderlyingVectorType<std::vector<T, A>> {
    using StorageType = T;
    using AllocatorType = A;
};

// Nice types already expose their template types
template<class T>
class NiceType {
public:
    using StorageType = T;
};

int main() {
    int a;
    
    // error: no type named 'type' in 'UnderlyingVectorType<int>'
    // because it resolves to the base template
    // UnderlyingVectorType<decltype(a)>::storageType b;

    std::vector<float> vec;
    using VecStorageType = UnderlyingVectorType<decltype(vec)>::StorageType;
    VecStorageType vec2;
    static_assert(std::is_same_v<VecStorageType, float>);

    NiceType<double> nice;
    decltype(nice)::StorageType myDouble = 0.0;
    static_assert(std::is_same_v<decltype(nice)::StorageType, double>);
}