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Letting the compiler determine the lambda type with auto works just fine:

#include <memory>
#include <functional>

int main(void) {
    std::unique_ptr<int> ptr(new int(1));
    auto fn = [ capture = std::move(ptr) ] () {};
}

If the lambda type is explicitly defined, however, a compilation error occurs saying there was an attempt to call unique_ptr's deleted copy constructor:

#include <memory>
#include <functional>

int main(void) {
    std::unique_ptr<int> ptr(new int(1));
    std::function<void()> fn = [ capture = std::move(ptr) ] () {};
}

Reduced output:

/media/hdd/home/vitor/Documents/parallel-tools/tests/reusable_thread.cpp:115:59: error: use of deleted function ‘std::unique_ptr<_Tp, _Dp>::unique_ptr(const std::unique_ptr<_Tp, _Dp>&) [with _Tp = int; _Dp = std::default_delete<int>]’
In file included from /usr/include/c++/7/memory:80:0,
                 from /media/hdd/home/vitor/Documents/parallel-tools/tests/reusable_thread.cpp:110:
/usr/include/c++/7/bits/unique_ptr.h:388:7: note: declared here
       unique_ptr(const unique_ptr&) = delete;
       ^~~~~~~~~~

What exactly is going on here?

rockerbacon
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    Important thing to remember: A lambda is **not** a `std::function`. it can be stored in one (if it is copyable), but it is its own distinct class type. – NathanOliver Dec 23 '19 at 19:56

0 Answers0