Re: "lifetime of temporary bound to reference..."

James Kanze <>
Tue, 27 Jan 2009 00:53:00 -0800 (PST)
On Jan 26, 7:25 pm, Victor Bazarov <> wrote:

Igor R. wrote:

Does the rule in the subj applies to the following:

First, very importantly, the rule in the subject doesn't exist.
Whether a temporary is bound to a reference has no effect on its
lifetime. The only time the lifetime of a temporary is extended
is if a reference is initialized with a rvalue. (The
difference, of course, is whether a temporary is bound to a
reference is a transitive relationship; whether a temporary was
initialized with an rvalue isn't.)

std::pair<int, int> getPair()
  return std::pair<int, int>(1, 2);

int main()
  // is this legal?
  const std::pair<int, int> &p1 = getPair();

Yes, this is legal. The temporary returned by 'getPair' will
have the same lifetime as the 'p1' reference (i.e. until
'main' returns).

Formally, it might be a copy of the temporary (which is also a
temporary) which has its lifetime extended (but I think this
possibility will be removed from C++0x).

  // and this?
  std::pair<int, int> &p2 = getPair();

No, this is not legal. A temporary can only be bound to a
reference to const.

No! A reference can only be initialized with an rvalue if it is
const and non volatile. But it's quite possible to arrange for
a temporary to bind to a non-const reference (const_cast, member
functions, etc.).

This point is in some ways related to my initial comment, above:
in order to bind a temporary to a non-const reference, you must
somehow arrange for the expression to be an lvalue. And since
the initialization expression of the temporary is not an rvalue,
the lifetime of the temporary will NOT be extended.

This is probably best explained with a simple example:

    struct S
        S& me() { return *this ; }
    } ;

    S& rs = S().me() ;

The reference rs is clearly bound to a temporary, even though
the reference is non-const. On the other hand, the reference
was not initialized with an rvalue expression, so the lifetime
of the temporary will NOT be extended beyond the end of the full
expression. (Technically, the temporary object S() will be
destructed before the reference rs is initialized. In practice,
there's no way a conforming program can tell, however.)

James Kanze (GABI Software)
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