Re: [Boost-users] [serialize] struct save_array_type in oserializer.hpp

From: Ovanes Markarian (om_boost_at_[hidden])
Date: 2007-04-12 10:10:10

Well,

the standart states in [expr.mul] §5.6 section 4:
The binary / operator yields the quotient, and the binary % operator yields the remainder from the
division of the first expression by the second. If the second operand of / or % is zero the
behavior is undefined; otherwise (a/b)*b + a%b is equal to a. If both operands are nonnegative
then the remainder is nonnegative; if not, the sign of the remainder is implementation-defined...

and later in [expr.add] §5.7 secion 6:
When two pointers to elements of the same array object are subtracted, the result is the
difference of the subscripts of the two array elements. The type of the result is an
implementation-defined signed integral type; this type shall be the same type that is defined as
ptrdiff_t in the <cstddef> header (18.1)...

In your second example you have no pointer arithmetics (subtraction, which you had in the first
example). You have only cast to the pointer type, which is in fact unsigned and the produced
warning is correct.

Using ptrdiff_t instead of int results in no warning at MSVC 8.0, where ptrdiff_t is defined as:

typedef _W64 int    ptrdiff_t;

Which is signed ;)

The following test code in MSVC:

ptrdiff_t result = sizeof(int)/(reinterpret_cast<int*>(&arr[0])-reinterpret_cast<int*>(&arr[1]));

std::cout << result << '\n';
std::cout << typeid(reinterpret_cast<int*>(&arr[0])-reinterpret_cast<int*>(&arr[1])).name() <<'\n';
std::cout <<
typeid(sizeof(int)/(reinterpret_cast<int*>(&arr[0])-reinterpret_cast<int*>(&arr[1]))).name()
<<'\n';

Produces the output:
0
int
unsigned int

The first output is implementation defined (size_t/negative number) but you get no warning.

With Kind Regards,

Ovanes Markarian

According to the standart IMO this line is correct.

On Thu, April 12, 2007 15:08, Oleg V. Zhylin wrote:
> At least for VC7.1 compiler
>
>   int a, b;
>
>   cout<<typeid(sizeof(a)/(a - b)).name()<<endl;
>
>  prints "unsingned int" and issues no warnings about int -> size_t
> conversion. Thus I still wonder if usage of int type in
>
>> >         int count = sizeof(t) / (
>> >             static_cast<const char *>(static_cast<const void
>> *>(&t[1]))
>
> dictated by consideration for some other C++ compiler. VC7.1 does issue
> a warning
>
> warning C4267: 'initializing' : conversion from 'size_t' to 'int',
> possible loss of data
>
> for that code. I would suggest to consider either change type of
> variable count from int to size_t or disable C4267 for this piece of
> code for MS compilers.
>
> --- Ovanes Markarian <om_boost_at_[hidden]> wrote:
>
>>
>> On Wed, April 11, 2007 17:51, Oleg V. Zhylin wrote:
>> > Hi,
>> >
>> >   Is there a reason to use type int in the code at line 487 in
>> > oserializer.hpp?
>> >
>> >         int count = sizeof(t) / (
>> >             static_cast<const char *>(static_cast<const void
>> *>(&t[1]))
>> >
>> >             - static_cast<const char *>(static_cast<const void
>> > *>(&t[0]))
>> >
>> >   C++ standard draft states in section 5.3.3 [expr.sizeof]
>> >
>> > The result is a constant of an implementation-defined
>> > type which is the same type as that which is named
>> > size_t in the standard header <cstddef>(18.1).
>>
>>
>> I assume otherwise you will get a warning, that a sign can be lost
>> through the assignment. The
>> problem here is that expression (type* - type*) results not in an
>> unsigned type (size_t) but a
>> signed difference type. If you divide unsigned value through a signed
>> value you get a signed
>> value. And then you would (according to your suggestion) assign it to
>> an unsigned value =>
>> warning.
>>
>>
>>
>> With Kind Regards,
>>
>> Ovanes Markarian
>>
>> _______________________________________________
>> Boost-users mailing list
>> Boost-users_at_[hidden]
>> http://lists.boost.org/mailman/listinfo.cgi/boost-users
>>
>
>
> WBR Oleg V. Zhylin   ovz_at_[hidden]
>
>
>
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