Re: Overload resolution and copy constructors that take non-const references

James Kanze <>
Fri, 9 May 2008 03:28:52 -0700 (PDT)
On May 7, 11:44 am, Anthony Williams <> wrote:

Should the following compile, and what should it print?

I'm not sure, but it's a very delicate point, and it wouldn't
surprise me if a lot of compilers get it wrong.

#include <memory>
#include <iostream>

void foo(std::auto_ptr<int> x)


struct dummy
    dummy(std::auto_ptr<int> const&)

void foo(dummy x)
    std::cout<<"dummy const ref"<<std::endl;

int main()
    std::auto_ptr<int> const x(new int);

MSVC 9.0 and g++ 4.3 disagree.

MSVC compiles it and prints "dummy const ref", which is what I expected.

g++ refuses to compile it, saying that:

test_copy.cpp: In function 'int main()':
test_copy.cpp:23: error: passing 'const std::auto_ptr<int>' as 'this' argu=

ment o

f 'std::auto_ptr<_Tp>::operator std::auto_ptr_ref<_Tp1>() [with _Tp1 = i=

nt, _Tp

= int]' discards qualifiers
test_copy.cpp:23: error: initializing argument 1 of 'void foo(std::auto_=



implying that it chose the first overload of foo.

Both functions are found during lookup, and are part of the
initial set of candidate functions. The question is whether the
first is viable. It meets the first condition (it can be called
with a single argument). The exact wording it:

    Second, for F to be a viable function, there shall exist
    for each argument an implicit conversion sequence
    ( that converts that argument to the
    corresponding parameter of F. If the parameter has
    reference type, the implicit conversion sequence
    includes the operation of binding the reference, and the
    fact that a reference to non-const cannot be bound to an
    rvalue can affect the viability of the function (see

The question hinges on whether there is "an implicit conversion
sequence" which converts an std::auto_ptr<int> const to an
std::auto_ptr<int>. And I think that the standard is far from
clear about this. In particular, =A713.3.3.1 says:

    When the parameter type is not a reference, the implicit
    conversion sequence models a copy-initialization of the
    parameter from the argument expression. The implicit
    conversion sequence is the one required to convert the
    argument expression to an rvalue of the type of the
    parameter. [Note: when the parameter has a class type,
    this is a conceptual conversion defined for the purposes
    of clause 13; the actual initialization is defined in
    terms of constructors and is not a conversion. ] Any
    difference in top-level cv-qualification is subsumed by
    the initialization itself and does not constitute a
    conversion. [Example: a parameter of type A can be
    initialized from an argument of type const A. The
    implicit conversion sequence for that case is the
    identity sequence; it contains no =BFconversion=BF from
    const A to A.] When the parameter has a class type and
    the argument expression has the same type, the implicit
    conversion sequence is an identity conversion. When the
    parameter has a class type and the argument expression
    has a derived class type, the implicit conversion
    sequence is a derived-to-base Conversion from the
    derived class to the base class. [Note: there is no
    such standard conversion; this derived-to-base
    Conversion exists only in the description of implicit
    conversion sequences.] A derived-to-base Conversion has
    Conversion rank (

Note that there is actually a contradiction in this text for the
case which interests us: "the implicit conversion sequence
models a copy-initialization of the parameter from the argument
expression", which bans the initialization of an std::auto_ptr<>
from an std::auto_ptr<> const, and "Any difference in top-level
cv-qualification is subsumed by the initialization itself and
does not constitute a conversion", which explicitly says that
the case in question does *not* constitute a conversion, i.e.
that it is treated as an exact match.

Of course, once the first function has crossed the line and is
considered viable, it's obviously the "best viable function"; if
it is to be eliminated, it can only be because it isn't viable.
Globally, I think that the intent is probably that the function
shouldn't be viable, since the argument cannot be used to
initialize a variable of the type of the argument. But the
wording in the paragraph I just cited seems contradictory enough
that it can be interpreted either way.

James Kanze (GABI Software)
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