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In the following snippet:
std::vector<double> a(100, 4.2);
auto* a_ptr = a.data();
auto b = std::move(a);
auto* b_ptr = b.data();
std::cout << ((b_ptr == a_ptr) ? "TRUE" : "FALSE") << '\n';
does the C++ standard guarantee that b_ptr is always equal to a_ptr after std::move? Running the code on wandbox prints TRUE.
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No. They will point to the local object that's been moved from; as described above, you can't make any assumptions about that object's state after the move.
std::move is used to indicate that an object t may be "moved from", i.e. allowing the efficient transfer of resources from t to another object. In particular, std::move produces an xvalue expression that identifies its argument t . It is exactly equivalent to a static_cast to an rvalue reference type.
Move Constructor And Semantics: std::move() is a function used to convert an lvalue reference into the rvalue reference. Used to move the resources from a source object i.e. for efficient transfer of resources from one object to another. std::move() is defined in the <utility> header.
That's pretty much the only time you should write std::move , because C++ already uses move automatically when copying from an object it knows will never be used again, such as a temporary object or a local variable being returned or thrown from a function. That's it.
From cppreference.com :
After container move construction (overload (6)), references, pointers, and iterators (other than the end iterator) to other remain valid, but refer to elements that are now in *this. The current standard makes this guarantee via the blanket statement in §23.2.1[container.requirements.general]/12, and a more direct guarantee is under consideration via LWG 2321.
Pointers to elements are not invalidated, including pointers to the first element.
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