Selection between classes at runtime for instantiation

Question

I want to construct an object of a type which is selected at runtime based on an enum class value. The classes A and B have multiple constructors with various numbers of arguments which may be used at runtime. I have attempted this problem by trying to write a factory method returning a shared_ptr to the base class. Here is my ill-formed attempt:

enum class Type
{
    A,
    B // C, etc
};

struct Base{};

struct A : public Base
{
    A(const int a) { std::cout << "A::A(int)"; }
    A(double d) { std::cout << "A::A(double)"; }
    A(int a, double d) { std::cout << "A::A(int,double)"; }
};

struct B : public Base
{
    B(int i, double d) { std::cout << "B::B(double)"; }
    B(int a, double d, double dd) { std::cout << "B::B(int,double, double)"; }
};

template <typename... Args>
std::shared_ptr<Base> make_Type(Type t, Args &&...args)
{
    if (t == Type::A)
        return std::make_shared<A>(std::forward<Args>(args)...);
    else if (t == Type::B)
        return std::make_shared<B>(std::forward<Args>(args)...);
    else
        assert(false);
}

No doubt, this doesn't work as the following call triggers a compilation error as the other branch for B doesn't have any constructor with single argument:

make_Type(Type::A,1);

gcc14 returns the following:

error: no matching function for call to 'B::B(int)'

I can create a static version with if constexpr. But how to attempt a runtime version? (Possibly without using typeinfo). Any alternative approach is also welcome.

"No doubt, this doesn't work". I dunno, works fine for me. Show a minimal reproducible example, an example that actually exhibits the problem you need help with. — Igor Tandetnik
– Igor Tandetnik, Commented Sep 2, 2024 at 18:29
@IgorTandetnik: your first parameter is a compile time type, not defined at runtime. But even that works: godbolt.org/z/sP6zYPW96 — Thomas Weller
– Thomas Weller, Commented Sep 2, 2024 at 18:41
@IgorTandetnik It won't work for the case make_Type(Type::A,1) as the other branch now doesn't compile as B doesn't have a compatible constructor taking single arg. — Arnab Mahanti
– Arnab Mahanti, Commented Sep 2, 2024 at 19:11
So, include that in the code shown in the question, along with the error message you are seeing. — Igor Tandetnik
– Igor Tandetnik, Commented Sep 2, 2024 at 19:12
Your types have different sets of constructor signatures, you know what exact signature you are going to use, but you don't know the type? This doesn't look like a real-life scenario. — n. m. could be an AI
– n. m. could be an AI, Commented Sep 2, 2024 at 19:30

Igor Tandetnik · Accepted Answer · 2024-09-02 18:49:11Z

2

You might be looking for something like this:

template <typename T, typename... Args>
std::shared_ptr<T> safe_make_shared(Args&&... args) {
    if constexpr (std::is_constructible_v<T, Args&&...>) {
        return std::make_shared<T>(std::forward<Args>(args)...);
    } else {
        return nullptr;
    }
}


template <typename... Args>
std::shared_ptr<Base> make_Type(Type t, Args &&...args)
{
    switch (t) {
        case Type::A : return safe_make_shared<A>(std::forward<Args>(args)...);
        case Type::B : return safe_make_shared<B>(std::forward<Args>(args)...);
        default: assert(false); return nullptr;
    }
}

Demo

The idea is to have a safe_make_shared function that creates the object if the type is actually constructible with the given set of arguments, or returns null otherwise (or it could assert, or throw an exception as a fallback). The important part is that safe_make_shared can be instantiated with any set of parameters, but only attempts to pass the "right" set of parameters to the object's constructor.

answered Sep 2, 2024 at 18:49

Igor Tandetnik

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1 Comment

Arnab Mahanti Over a year ago

the safe_make_shared seems to have solved the issue. the catch was the is_constructible_v. Thanks. I will check in detail.

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