Question

In

template <typename T>
T const & foo(T const & dflt) 
  { return /* ... */ ? /* ... */ : dflt; }

int x = foo(5);

Does the reference to temporary “survive” until it’s assigned to x?

(I know it survives the duration of the foo() call, but the assignment makes me hesitate)

The code in question has some legitimate uses for relying on const references here, i.e. I would need two separate functions

T const & foo_cr(T  const & dflt);
T const & foo_v(T dflt);

which I want to avoid (just relying on different signatures wiht the same name seems a bit risky to me.)

Answer 1

Does the reference to temporary “survive” until it’s assigned to x?

Generally, yes. If foo returns dflt, that reference (to a temporary 5) is valid until the end of the full-expression it was created in, that means, until after the initialization.

C++03, 12.2/3:

When an implementation introduces a temporary object of a class that
has a non-trivial constructor (12.1), it shall ensure that a
constructor is called for the temporary object. Similarly, the
destructor shall be called for a temporary with a non-trivial
destructor (12.4). Temporary objects are destroyed as the last step in
evaluating the full-expression (1.9) that (lexically) contains the
point where they were created. This is true even if that evaluation
ends in throwing an exception.

More comes in 12.2/4 and 12.2/5, especially the exceptions of this rule.

I can’t see what happens when it returns the ... you left out, but I guess you weren’t asking about this.