How Can I Convert Lambdas to `std::function` in C Without Explicitly Specifying Template Parameters?

Converting Lambdas to std::function with Templates

In C , converting a lambda to an std::function without explicitly specifying template parameters presents a challenge. Let's explore the problem and its solution.

Understanding the Issue

When defining an std::function without specifying template parameters, the compiler cannot deduce the type of the lambda argument. This is because the lambda's type is dependent on the types of its parameters, which are unknown to the compiler at the time of template instantiation.

The Solution: Identity Type

To circumvent this issue, we can wrap the lambda in an identity type, which ignores dependent types during template type deduction. Here's an example:

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...);
}

This approach allows us to pass a lambda to the func() function without explicitly specifying the template parameters.

Limitations

This solution, however, has limitations:

• It requires passing additional arguments to the function, which may not be desirable in all scenarios.
• It cannot handle the case where the lambda captures variables from the surrounding scope.

Alternative Approach

If the lambda doesn't capture variables and you're willing to explicitly specify the template parameters, you can use the following approach:

std::function<void()> f = [](int x, int y, int z) { std::cout << (x*y*z) << std::endl; };

This method is simpler and works well when the template parameters are known at compile time.

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