Can I reuse an rvalue reference parameter to return an rvalue reference?

Question

Consider the following code:

struct MyString
{
  // some ctors

  MyString& operator+=( const MyString& other ); // implemented correctly
};

MyString operator+( const MyString& lhs, const MyString& rhs )
{
  MyString nrv( lhs );
  nrv += rhs;
  return nrv;
}

MyString&& operator+( MyString&& lhs, const MyString& rhs )
{
  lhs += rhs;
  return std::move( lhs ); // return the rvalue reference we received as a parameter!
}

This works for the following use-case

MyString a, b, c; // initialized properly
MyString result = a + b + c;

But it creates a dangling reference for

const MyString& result = a + b + c;

Now, I understand why that is and how to fix it (returning an ravlue instead of an rvalue reference) but I consider it a usage error if someone writes the above as the code looks like it is asking for trouble. Is there any "canonical" real-world example where the above operator returning a rvalue reference is a problem? What is a convincing reason why I should always return an rvalue from operators?

This issue was also brought up in Defect report 1138.

Jesse Good
– Jesse Good

2013-04-25 23:11:59 +00:00
Commented Apr 25, 2013 at 23:11 — Jesse Good
– Jesse Good, Commented Apr 25, 2013 at 23:11

Daniel Frey · Accepted Answer · 2013-04-26 19:37:56Z

The example you are looking for is a range-based for statement:

MyString a, b, c;
for( MyCharacter mc : a + b + c ) { ... }

In this case the result of a + b + c is bound to a reference, but the nested temporary (generated by a + b and returned as an rvalue reference by (a + b) + c) is destroyed before the range-based for loop is executed.

The standard defines range-based for loops in

6.5.4 The range-based for statement [stmt.ranged]

¹ For a range-based for statement of the form

for (for-range-declaration:expression)statement

let range-init be equivalent to the expression surrounded by parentheses

( expression )

and for a range-based for statement of the form

for (for-range-declaration:braced-init-list)statement

let range-init be equivalent to the braced-init-list. In each case, a range-based for statement is equivalent to
{
   auto && __range = range-init;
   for ( auto __begin = begin-expr,
              __end = end-expr;
         __begin != __end;
         ++__begin ) {
      for-range-declaration = *__begin;
      statement
   }
}

Note that auto && __range = range-init; would extend the lifetime of a temporary returned from range-init, but it does not extend the lifetime of nested temporaries inside of range-init.

Kerrek SB · Accepted Answer · 2013-04-25 23:05:57Z

5

Instead of asking for trouble, you should trust the string's own move constructor:

MyString operator+(MyString lhs, MyString rhs)
{
    lhs += std::move(rhs);
    return std::move(lhs);
}

Now both MyString x = a + b; and MyString y = MyString("a") + MyString("b"); work efficiently.

answered Apr 25, 2013 at 23:05

Kerrek SB

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4 Comments

Daniel Frey Over a year ago

For x = a + b, an extra copy of b is created which is not needed. But as I said, I know how to fix it, the question was why returning an rvalue reference is a problem.

Kerrek SB Over a year ago

@DanielFrey: Yeah, I guess, one of the two could be a reference... Returning a reference is only OK if you can guarantee that the reference refers to something that exists, which is hard and non-obvious for the user.

Kerrek SB Over a year ago

@DanielFrey: Since appending at the front is presumably not worth optimizing, I'd probably change the second parameter only to a const MyString &, as you have in your original code.

Captain Obvlious Over a year ago

@DanielFrey Passing by value instead of by reference was presented at Going Native 2012 - STL11 Magic && Secrets starting at the 35:10 mark. It's worth watching if you haven't seen it.

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