Re: Why shall a non-const reference not be bound to a bit-field?

From:

"Alf P. Steinbach" <alfps@start.no>

Newsgroups:

comp.lang.c++

Date:

Fri, 07 Jul 2006 00:23:30 +0200

Message-ID:

<4h5gr7F1q601bU1@individual.net>

* Lighter:

On page 163 of In the C++ standard document(9.6 Bit-fields), I find
three rules on bit-fields:

Rule 1, "A bit-field shall not be a static member."

Rule 2, "A non-const reference shall not be bount to a bit-field"

Rule 3, "Note: if the initializer for a reference of type const T& is
an lvalue that refers to a bit-field, the reference is bound to a
temporary initialized to hold the value of the bit-field; the reference
is not bound to the bit-field directly."

Visual Studio 2005, however, can correctly compile and run the
following code fragment:

typedef int BIT;

struct BITSET
{
    BIT a : 1;
    BIT b : 1;
    BIT c : 1;
    BIT d : 5;
};

class Test
{
public:
    static BITSET m; // Violation of the Rule 1
};

BITSET Test::m = {0};

int main()
{
    BITSET a = {1, 0, 1};
    BITSET& b = a; // Violation of the Rule 2

                const BITSET& c = a;
    a.a = 0; // After this statement, c.a is also set to 0. Violation of
the rule 3
}

Maybe someone will say: "They are just that Microsoft doesn't abide by
the C++ standard", but what I want know is why the C++ standard
committee made such restrictions. I cannot find enough motivation for
the C++ standard committee to do like this. I think VS 2005 did right
to break these rules.

The above code does not demonstrate any rule breaking.

A static bit-field would be like

struct Foo { static int a : 1; };

And so on -- please post your validation results with the code changed
to use actual bitfields, not structs.

If you know the whys, please tell me. Thanks in advance.

For the why's of the standard, please ask in [comp.std.c++].

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