Can an Rvalue Reference Passed to a Function Still Behave as an Rvalue?

Can an Rvalue Reference Behave Like an Lvalue Within a Function?

In programming, an rvalue reference, denoted by &&, is typically used to refer to an object that is going to be moved. However, the question arises: if an rvalue reference is passed by reference to a function, does it continue to act as an rvalue? Let's explore this scenario with a specific example.

The code snippet in the question:

void foo(string&& bar) {
  string* temp = &bar;
  cout << *temp << " @:" << temp << endl;
}

defines a function foo() that takes an rvalue reference to a string. Within the function, the code assigns the address of bar to a pointer variable temp. The question is, at this point, is bar still an rvalue or has it been transformed into an lvalue?

According to the definition provided in the answer, anything that has a name is considered an lvalue. Since bar has a name within the function, it is an lvalue, regardless of the fact that its type is an rvalue reference to a string.

However, the distinction between an rvalue and an lvalue is significant in terms of what can be bound to them. Rvalues can only be bound to rvalue references, while lvalues can bind to both lvalue and rvalue references.

In the case of foo(), since bar is specifically declared as an rvalue reference, the compiler guarantees that the value bound to it is an rvalue. This knowledge allows the function to safely treat bar as an rvalue and perform operations such as moving its resources.

In conclusion, while an rvalue reference within a function may behave like an lvalue in terms of its usage within expressions, it retains its original nature as an rvalue. The distinction between the two primarily lies in the rules governing what can be bound to these references.

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