Can You Move From a Const Object in C ?

Understanding std::move on const Objects: Unveiling the Performance Pitfall

In C , the std::move function is used to transfer ownership of an object. However, a common question arises: why is it possible to call std::move on a const object? Isn't this counterintuitive and potentially dangerous?

The trick lies in the fact that std::move(const_object) doesn't perform an actual move. Instead, it instructs the compiler to attempt a move. If the class doesn't have a constructor that accepts a const reference to an rvalue (move constructor), the compiler will default to the implicit copy constructor.

Consider the following example:

struct Cat {
   Cat(){}
   Cat(const Cat&amp;) {std::cout << "COPY";}
   Cat(Cat&amp;&amp;) {std::cout << "MOVE";}
};

int main() {
    const Cat cat;
    Cat cat2 = std::move(cat); // prints "COPY"
}

As you can see, even though cat is a const object, the compiler safely constructs cat2 as a copy. This is because there's no move constructor available for const objects in this example.

The potential pitfall highlighted by Scott Meyers in "Effective Modern C " is a performance issue rather than a stability issue. In essence, attempting to move from a const object when a move constructor isn't available will result in a less efficient copy.

While it might seem logical to restrict std::move from operating on const objects, it's not a straightforward solution due to the following reasons:

• It would not prevent all cases of unintended copying since it's still possible for non-const objects to lack a move constructor.
• It would interfere with generic template code that intentionally uses the fallback copy constructor.
• It would restrict cases where a move constructor from a const object might be intentionally desired.

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