Can I Create an `std::function` from a Move-Capturing Lambda Expression?

Creating an std::function from a Move-Capturing Lambda Expression

It is possible to construct an std::function from a move-capturing lambda expression; however, certain restrictions apply.

Template Constructor

An std::function can be constructed from a move-capturing lambda using the following template constructor:

template<class f>
function(F f);</class>

Requirements

For this constructor to work, the following requirements must be met:

• The lambda expression's capture must be copy-constructible.
• The lambda expression must be Callable with the specified argument types and return type.
• The lambda's capture's copy constructor and destructor must not throw exceptions.

Move-Only Types

It is not possible to construct an std::function from a lambda that move-captures a move-only type. This is because std::function's copy constructor and assignment operator are defined in terms of a constructor that requires copy-constructible types.

Example

Consider the following code snippet:

auto pi = std::make_unique<int>(0);

// Move-capturing lambda
auto foo = [q = std::move(pi)] {
    *q = 5;
    std::cout <p>Attempting to create an std::function from this lambda using any of the following approaches will result in a compilation error:</p>
<pre class="brush:php;toolbar:false">std::function<void> bar = foo;
std::function<void> bar{foo};
std::function<void> bar{std::move(foo)};
std::function<void> bar = std::move(foo);
std::function<void> bar{std::forward<:function>>(foo)};
std::function<void> bar = std::forward<:function>>(foo);</:function></void></:function></void></void></void></void></void>

This is because pi is a move-only type, and std::function's copy constructor requires copy-constructible types.

Therefore, if you wish to use a move-capturing lambda with std::function, ensure that its capture is copy-constructible.

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