How Can I Efficiently Convert a Generic C Lambda to an std::function Using Templates?

Lambda to std::function Conversion using Templates

In C , converting a lambda function to an std::function using templates can be a challenging task. To achieve this, template type deduction is essential, but it may face limitations in certain scenarios.

Initially, attempts to convert a lambda to std::function may fail due to missing template parameters or a mismatch in candidate matching. To resolve this, explicitly specifying the template parameters, as seen in std::function<void()> f([<unclosed>]), would suffice. However, this approach is not desirable for generic code that targets lambdas of any parameter type and count.

To address this issue, a more nuanced solution is required. While template type deduction cannot directly infer the std::function template parameters from a lambda, it can still be guided by providing an additional type constraint.

Consider the following approach: wrap the lambda function in a structure identity<std::function<void(T...)>>, where T can be deduced from the lambda parameters. This structure serves as an identity type, allowing the lambda to be passed as an argument without any type conversion.

template <typename T>
struct identity
{
  typedef T type;
};

template <typename... T>
void func(typename identity<std::function<void(T...)>>::type f, T... values) {
  f(values...);
}

Now, when calling func, the template parameters of std::function can be deduced from the identity structure. This eliminates the need for explicit template parameter specification or additional argument passing.

int main() {
  func([](int x, int y, int z) { std::cout << (x*y*z) << std::endl; }, 3, 6, 8);
  return 0;
}

This approach satisfies the requirements of converting a generic lambda to std::function without explicitly specifying the template parameters and allows the currying of variadic functions by preserving the original lambda signature.

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