Does the following code has undefined behavior due to r-value usage

Question

According to the post r-value causes a warning without the use of std::move, I'm wondering if the following code has undefined behavior:

struct Animal
{
  virtual void foo() const
  {
    std::cout << "error" << std::endl;
  }
};

struct Dog : public Animal
{
  void foo() const
  {
    std::cout << "wuff" << std::endl;
  }
};

struct A
{
  A( const Animal& a) : _a( a ) {}

  void foo()
  {
    _a.foo();
  }


  const Animal& _a;
};

int main()
{
  A a( Dog{} );
  a.foo();

}

On my machine, the output is (as expected) wuff.

score 4 · Answer 1 · answered Nov 19 '16 at 16:16

4

Indeed, a.a_ is a dangling reference in the statement a.foo();, and so the evaluation of the expression _a.foo has undefined behaviour. The lifetime of the temporary object Dog{} ends at the end of the full-expression (and is not extended).

answered Nov 19 '16 at 16:16

Kerrek SB

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Do you know any fix or workaround? – abraham_hilbert Nov 19 '16 at 19:37
@abraham_hilbert: Take the value by const pointer: `A(const Animal* p) : a_(*p) {}`. That makes it *much* harder to accidentally pass the address of prvalue in. – Kerrek SB Nov 20 '16 at 00:00

Does the following code has undefined behavior due to r-value usage

1 Answers1