Why Don't Built-In Types Exhibit Move Semantics?

Do Built-In Types Exhibit Move Semantics?

The provided code demonstrates the unexpected behavior where the built-in type integer num is modified within the Func function, even though it is passed as an r-value reference using std::move. This behavior contradicts the common understanding that std::move transfers ownership and resources.

Essence of Move Semantics

Move semantics refers to the transfer of ownership and resources from one object to another. In this context, the resources are the data values or allocated memory belonging to an object.

Applicability to Built-In Types

Built-in types, such as integers, themselves represent the resources they hold. Therefore, it makes little sense to transfer ownership of their resources. Instead, they simply pass the reference to the object, allowing modifications to be reflected in the original object.

The Role of std::move

std::move does not actually move an object. Its purpose is to convert l-value references (references to objects with defined locations) into x-value references (references to objects that have expired). This conversion allows the reference to bind to r-value references within the receiving function.

Post-Move Behavior

After invoking std::move on a built-in type object, the reference passed to the function still refers to the original object. Any modifications made to the object are reflected in the original.

Conclusion

Unlike user-defined types, which can define custom move semantics, built-in types have no such behavior. std::move does not invoke any move constructors or assignment operators for built-in types. Instead, it simply casts them to r-value references for binding purposes. This behavior is well-defined by the C standard and explains the unexpected modification of num in the given code example.

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