Why Does `std::move` Work on `const` Objects in C ?

Why is std::move Allowed on const Objects?

In C 11, the code snippet below is valid:

struct Cat {
   Cat(){}
};

const Cat cat;
std::move(cat);

This behavior may seem counterintuitive, as moving a const object implies altering its state. However, it's important to understand that std::move does not actually perform any movement.

The Trick Behind std::move

When you call std::move on a const object, the compiler simply tries to move the object. If the class does not have a constructor that accepts a const lvalue reference, it will use the implicit copy constructor to create a new object instead. This ensures that the const object's state remains unchanged.

To demonstrate this, consider the following code:

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);
}

Running this code prints "COPY," indicating that the copy constructor was used instead of the move constructor. This confirms that std::move does not modify the const object.

Implications for Performance and Debugging

While moving a const object is not inherently incorrect, it can lead to performance issues if the object does have a move constructor. In such cases, a copy is created unnecessarily. Additionally, it can make it more difficult to debug issues related to moving objects.

Conclusion

std::move can be used on const objects because it is implemented as an attempt to move the object rather than an actual move. This behavior does not violate the contract of const objects, and it allows for greater flexibility in code design. However, it is important to be aware of the potential performance implications and debugging challenges that can arise from this practice.

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