Do built-in types have move semantics?

Question

Consider this code:

#include <iostream> using namespace std;  void Func(int&& i) {     ++i; }  int main() {     int num = 1234;     cout << "Before: " << num << endl;     Func(std::move(num));     cout << "After: " << num << endl; } 

Its output is:

Before: 1234 After: 1235 

Clearly, i is being modified inside Func, as it is bound to parameter i after being "converted" to an r-value reference by std::move.

Well, my point:

Moving an object means transferring ownership of resources from one object into another. However, built-in types holds no resources because they themselves are the resources. It makes no sense to transfer the resources they hold. As shown by the example, num's value is modified. Its resource, its self, is the one being modified.

Do built-in types have move semantics?

Also, Do built-in type objects after it is moved (if it is) a well-defined behavior?

Answer 1

And so, is the one shown by the example a well-defined behavior?

Yes, the behaviour shown in the example is the only behaviour allowed by the standard. That is because std::move doesn't move. The things that move are move constructors and move assignment operators.

All std::move does is change an lvalue into an xvalue, so that it can bind to rvalue references. It does not invoke any constructor or anything else. Changing the value category happens at the type level. Nothing happens at runtime.

Rvalue references are still references: they refer to the original object. The function increments the original integer through the reference given.

If a function takes an argument by reference, no copies nor moves happen: the original object is bound to the reference.

If a function takes an argument by value, then we might have a move.

However, fundamental types don't have move constructors. Moves degrade to copies in that case.