CODEWITHSUNDEEP
c++c++11rvalue-reference
drwowe
drwowe

Reputation: 2055

rvalue refs and std::move

Can someone explain why B doesn't compile, but C does? I don't understand why std::move is required since the variable is already an rvalue ref.

struct A {
  int x;
  A(int x=0) : x(x) {}
  A(A&& a) : x(a.x) { a.x = 0; }
};

struct B : public A {
  B() {}
  B(B&& b) : A(b) {}  // compile error with g++-4.7
};

struct C : public A {
  C() {}
  C(C&& c) : A(std::move(c)) {}  // ok, but why?
};

Upvotes: 4

Views: 754

Answers (2)

Howard Hinnant
Howard Hinnant

Reputation: 218750

In the statement:

B(B&& b)

The parameter b is declared with the type: rvalue reference to B.

In the statement:

A(b)

The expression b is an lvalue of type B.

And lvalue expressions can not bind to rvalue references: specifically the rvalue reference in the statement:

A(A&& a)

This logic follows cleanly from other parts of the language. Consider this function:

void
f(B& b1, B b2, B&& b3)
{
   g(b1);
   g(b2);
   g(b3);
}

Even though the parameters of f are all declared with different types, the expressions b1, b2 and b3 are all lvalue expressions of type B, and thus would all call the same function g, no matter how g is overloaded.

In C++11 it is more important than ever to distinguish between a variable's declaration, and the expression that results from using that variable. And expressions never have reference type. Instead they have a value category of precisely one of: lvalue, xvalue, prvalue.

The statement:

A(std::move(c))

is ok, because std::move returns an rvalue reference. The expression resulting from a function call returning an rvalue reference has value category: xvalue. And together with prvalues, xvalues are considered rvalues. And the rvalue expression of type C:

std::move(c)

will bind to the rvalue reference parameter in: A(A&& a).

I find the following diagram (originally invented by Bjarne Stroustrup) very helpful:

       expression
          /  \
    glvalue  rvalue
     /  \    /  \
lvalue  xvalue  prvalue

Upvotes: 25

K-ballo
K-ballo

Reputation: 81349

Because named variables are not rvalues, even when declared &&. If it has a name then its not temporary, thus you need to use std::move.

Upvotes: 14

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