This HAS to be UB...

**Chris M. Thomasson** · Oct 2 '08, 07:55 PM

Re: This HAS to be UB...

"Chris M. Thomasson" <no@spam.invali dwrote in message
news:2y9Fk.3981 7$hX5.39354@new sfe06.iad...

Keep in mind that I am a C programmer; well, anyway here is the C++
program...
_______________ _______________ _______________ _______________ __________

[...]

template<std::s ize_t T_size>
class buf {
char mem[T_size];
};

I add virtual dtor to buf1, and no change in output.

>
>
class buf2 : public buf<1234>, public allocator_base< buf2{
char mem2[1000];
};

>
>
int main() {
buf2* b = new buf2;
delete b;
>
b = new buf2[5];
delete [] b;
>
return 0;
}
_______________ _______________ _______________ _______________ __________

[...]

**Victor Bazarov** · Oct 2 '08, 08:15 PM

Re: This HAS to be UB...

Chris M. Thomasson wrote:

Keep in mind that I am a C programmer; well, anyway here is the C++
program...
[..]
>
On GCC I get the following output:
>
custom_allocato r::allocate(002 46C50, 2234)
custom_allocato r::deallocate(0 0246C50, 2234)
custom_allocato r::allocate(002 47760, 11174)
custom_allocato r::deallocate(0 0247760, 11174)
>
>
>
>
On MSVC 8 I get:
>
custom_allocato r::allocate(003 62850, 2234)
custom_allocato r::deallocate(0 0362850, 2234)
custom_allocato r::allocate(003 66B68, 11170)
custom_allocato r::deallocate(0 0366B68, 2234)

MSVC 9 gives the same output, BTW.

Are they both right due to UB? WTF is going on? GCC seems to be accurate
at least... DAMN!

Well, the default implementation of the operator delete[] does *not*
have the "size" argument. In fact there are two allowed declarations of
the operator delete[]:

void operator delete[](void* ptr) throw();

and

void operator delete[](void* ptr, const std::nothrow&) throw();

I'm not sure what else to tell you.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

**Chris M. Thomasson** · Oct 2 '08, 08:25 PM

Re: This HAS to be UB...

"Victor Bazarov" <v.Abazarov@com Acast.netwrote in message
news:gc39rg$q2m $1@news.datemas .de...

Chris M. Thomasson wrote:

>Keep in mind that I am a C programmer; well, anyway here is the C++
>program...
>[..]
>>
>On GCC I get the following output:
>>
>custom_allocat or::allocate(00 246C50, 2234)
>custom_allocat or::deallocate( 00246C50, 2234)
>custom_allocat or::allocate(00 247760, 11174)
>custom_allocat or::deallocate( 00247760, 11174)
>>
>>
>>
>>
>On MSVC 8 I get:
>>
>custom_allocat or::allocate(00 362850, 2234)
>custom_allocat or::deallocate( 00362850, 2234)
>custom_allocat or::allocate(00 366B68, 11170)
>custom_allocat or::deallocate( 00366B68, 2234)

>
MSVC 9 gives the same output, BTW.
>

>Are they both right due to UB? WTF is going on? GCC seems to be accurate
>at least... DAMN!

>
Well, the default implementation of the operator delete[] does *not* have
the "size" argument. In fact there are two allowed declarations of the
operator delete[]:
>
void operator delete[](void* ptr) throw();
>
and
>
void operator delete[](void* ptr, const std::nothrow&) throw();
>
I'm not sure what else to tell you.

This has to be GCC extension? This is weird, well, perhaps not so weird
because it simply MUST be 100% UB. Oh well. I initially thought I could take
advantage of it; NOT!!!

;^/

**blargg** · Oct 2 '08, 11:25 PM

Re: This HAS to be UB...

In article <2y9Fk.39817$hX 5.39354@newsfe0 6.iad>, "Chris M. Thomasson"
<no@spam.invali dwrote:

Keep in mind that I am a C programmer; well, anyway here is the C++
program...
_______________ _______________ _______________ _______________ __________
#include <cstdio>
#include <cstdlib>
#include <new>
>
struct custom_allocato r {
static void* allocate(std::s ize_t size)
throw(std::bad_ alloc()) {

***^^***

void* const mem = ::operator new(size);
std::printf("cu stom_allocator: :allocate(%p, %lu)\n",
(void*)mem, (unsigned long)size);
return mem;
}

[...]

How did this even compile?

**Chris M. Thomasson** · Oct 3 '08, 04:15 AM

Re: This HAS to be UB...

"blargg" <blargg.h4g@gis hpuppy.comwrote in message
news:blargg.h4g-021008181822000 1@192.168.1.4.. .

In article <2y9Fk.39817$hX 5.39354@newsfe0 6.iad>, "Chris M. Thomasson"
<no@spam.invali dwrote:
>

>Keep in mind that I am a C programmer; well, anyway here is the C++
>program...
>______________ _______________ _______________ _______________ ___________
>#include <cstdio>
>#include <cstdlib>
>#include <new>
>>
>struct custom_allocato r {
> static void* allocate(std::s ize_t size)
> throw(std::bad_ alloc()) {

***^^***
>

> void* const mem = ::operator new(size);
> std::printf("cu stom_allocator: :allocate(%p, %lu)\n",
> (void*)mem, (unsigned long)size);
> return mem;
> }

[...]
>
How did this even compile?

I don't know! It did! Well, blame MSVC 8+ and GCC! ARGH... Well, if it
didn;t compile I would have NOT asked the contrived question indeed!

:^|

**James Kanze** · Oct 3 '08, 09:25 AM

Re: This HAS to be UB...

On Oct 2, 9:52 pm, "Chris M. Thomasson" <n...@spam.inva lidwrote:

Keep in mind that I am a C programmer; well, anyway here is
the C++ program...

It looks to me like you're attacking some fairly tricky stuff.
You'd probably be better of starting with something simpler if
you're still learning C++. However...

_______________ _______________ _______________ _______________ __________
#include <cstdio>
#include <cstdlib>
#include <new>

struct custom_allocato r {
static void* allocate(std::s ize_t size)
throw(std::bad_ alloc()) {

That should doubtlessly be:
throw( std::bad_alloc )
What you've said is that the only exception type which will
escape from your function is a pointer to a function returning
an std::bad_alloc and taking no arguments. I really don't think
you meant to say that you're going to throw pointers to
functions.

In practice, exception specifications are not really that
useful, except when they're empty. (It's very important in
certain cases to know that a function cannot throw any
exceptions, but it's rarely useful to know that it can't throw
certain types of exceptions.)

void* const mem = ::operator new(size);
std::printf("cu stom_allocator: :allocate(%p, %lu)\n",
(void*)mem, (unsigned long)size);
return mem;
}

static void deallocate(void * const mem, std::size_t size)
throw() {
std::printf("cu stom_allocator: :deallocate(%p, %lu)\n",
(void*)mem, (unsigned long)size);
::operator delete(mem);
}
};

template<typena me T>
struct allocator_base {
static void* operator new(std::size_t size)

The static isn't really necessary: allocation and deallocation
member functions (operator new and operator delete) are always
static, whether you declare them so or not. (On the other hand,
it doesn't hurt.)

throw(std::bad_ alloc()) {
return custom_allocato r::allocate(siz e);
}

static void* operator new[](std::size_t size)
throw(std::bad_ alloc()) {
return custom_allocato r::allocate(siz e);
}

static void operator delete(void* mem)

Just curious: since you require the size in delete[], why don't
you require it here? Derivation can mean that the size isn't a
constant, e.g.:

class Base : public allocator_base< Base >
{
// ...
} ;

class Derived : public Base
{
// ...
} ;

Base* p = new Derived ;
// ...
delete p ;

(This supposes, of course, that Base has a virtual destructor.)

throw() {
if (mem) {
custom_allocato r::deallocate(m em, sizeof(T));
}
}
static void operator delete [](void* mem, std::size_t size)
throw() {
if (mem) {
custom_allocato r::deallocate(m em, size);
}
}
};

template<std::s ize_t T_size>
class buf {
char mem[T_size];
};

class buf2 : public buf<1234>, public allocator_base< buf2{
char mem2[1000];
};

int main() {
buf2* b = new buf2;
delete b;

b = new buf2[5];
delete [] b;
return 0;
}

_______________ _______________ _______________ _______________ __________

On GCC I get the following output:

custom_allocato r::allocate(002 46C50, 2234)
custom_allocato r::deallocate(0 0246C50, 2234)
custom_allocato r::allocate(002 47760, 11174)
custom_allocato r::deallocate(0 0247760, 11174)

On MSVC 8 I get:

custom_allocato r::allocate(003 62850, 2234)
custom_allocato r::deallocate(0 0362850, 2234)
custom_allocato r::allocate(003 66B68, 11170)
custom_allocato r::deallocate(0 0366B68, 2234)

Are they both right due to UB? WTF is going on? GCC seems to
be accurate at least... DAMN!

Well, there's no undefined behavior. You're program seems
perfectly legal and well defined to me. It looks like a bug in
VC++, see §12.5/5:

When a delete-expression is executed, the selected
deallocation function shall be called with the address
of the block of storage to be reclaimed as its first
argument and (if the two-parameter style is used) the
size of the block as its second argument.

And I can't think of any way of interpreting "the size of the
block" to mean anything other than the size requested in the
call to operator new.

--
James Kanze (GABI Software) email:james.kan ze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientier ter Datenverarbeitu ng
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

**Chris M. Thomasson** · Oct 4 '08, 05:15 AM

Re: This HAS to be UB...

>
"James Kanze" <james.kanze@gm ail.comwrote in message
news:7878ab49-834f-4bbc-b687-efdd8f31f1f3@z6 6g2000hsc.googl egroups.com...

On Oct 2, 9:52 pm, "Chris M. Thomasson" <n...@spam.inva lidwrote:

Keep in mind that I am a C programmer; well, anyway here is
the C++ program...

It looks to me like you're attacking some fairly tricky stuff.
You'd probably be better of starting with something simpler if
you're still learning C++. However...

I was exploring the feature in C++ delete operator in which the size of the
allocation is returned along with the pointer to allocated memory. One could
create heavily optimized custom memory allocator using that important piece
of information.

_______________ _______________ _______________ _______________ __________
#include <cstdio>
#include <cstdlib>
#include <new>

struct custom_allocato r {
static void* allocate(std::s ize_t size)
throw(std::bad_ alloc()) {

That should doubtlessly be:
throw( std::bad_alloc )
What you've said is that the only exception type which will
escape from your function is a pointer to a function returning
an std::bad_alloc and taking no arguments. I really don't think
you meant to say that you're going to throw pointers to
functions.

That was definitely a typo/error on my part.

In practice, exception specifications are not really that
useful, except when they're empty. (It's very important in
certain cases to know that a function cannot throw any
exceptions, but it's rarely useful to know that it can't throw
certain types of exceptions.)

I thought it would be prudent to give the overloaded operator new an
exception specification of `std::bad_alloc '. Also, I wanted to give an empty
specification to the overload of operator delete. As to how useful it is...
Well, I don't quite know.

void* const mem = ::operator new(size);
std::printf("cu stom_allocator: :allocate(%p, %lu)\n",
(void*)mem, (unsigned long)size);
return mem;
}

static void deallocate(void * const mem, std::size_t size)
throw() {
std::printf("cu stom_allocator: :deallocate(%p, %lu)\n",
(void*)mem, (unsigned long)size);
::operator delete(mem);
}
};

template<typena me T>
struct allocator_base {
static void* operator new(std::size_t size)

The static isn't really necessary: allocation and deallocation
member functions (operator new and operator delete) are always
static, whether you declare them so or not. (On the other hand,
it doesn't hurt.)

Its a habit of mine. Also, using printf in C++ is another habit.

throw(std::bad_ alloc()) {
return custom_allocato r::allocate(siz e);
}

static void* operator new[](std::size_t size)
throw(std::bad_ alloc()) {
return custom_allocato r::allocate(siz e);
}

static void operator delete(void* mem)

Just curious: since you require the size in delete[], why don't
you require it here? Derivation can mean that the size isn't a
constant, e.g.:
>
class Base : public allocator_base< Base >
{
// ...
} ;
>
class Derived : public Base
{
// ...
} ;
>
Base* p = new Derived ;
// ...
delete p ;

(This supposes, of course, that Base has a virtual destructor.)

[...]

_______________ _______________ _______________ _______________ __________

On GCC I get the following output:

custom_allocato r::allocate(002 46C50, 2234)
custom_allocato r::deallocate(0 0246C50, 2234)
custom_allocato r::allocate(002 47760, 11174)
custom_allocato r::deallocate(0 0247760, 11174)

On MSVC 8 I get:

custom_allocato r::allocate(003 62850, 2234)
custom_allocato r::deallocate(0 0362850, 2234)
custom_allocato r::allocate(003 66B68, 11170)
custom_allocato r::deallocate(0 0366B68, 2234)

Are they both right due to UB? WTF is going on? GCC seems to
be accurate at least... DAMN!

Well, there's no undefined behavior. You're program seems
perfectly legal and well defined to me. It looks like a bug in
VC++, see §12.5/5:

It definitely looks like a bug is MSVC++. I get erroneous behavior on
versions 6 through 9.

When a delete-expression is executed, the selected
deallocation function shall be called with the address
of the block of storage to be reclaimed as its first
argument and (if the two-parameter style is used) the
size of the block as its second argument.

And I can't think of any way of interpreting "the size of the
block" to mean anything other than the size requested in the
call to operator new.

I thought that MSVC was crapping out because `allocator_base ' was a
template. So I created another little test which hopefully has all the bugs
fixed:
_______________ _______________ _______________ _______________ ______________
#include <cstdio>
#include <cstdlib>
#include <new>

struct custom_allocato r {
static void* allocate(std::s ize_t size)
throw(std::bad_ alloc) {
void* const mem = std::malloc(siz e);
if (! mem) {
throw std::bad_alloc( );
}
std::printf("cu stom_allocator: :allocate(%p, %lu)\n",
(void*)mem, (unsigned long)size);
return mem;
}

static void deallocate(void * const mem, std::size_t size)
throw() {
if (mem) {
std::printf("cu stom_allocator: :deallocate(%p, %lu)\n",
(void*)mem, (unsigned long)size);
std::free(mem);
}
}
};

struct allocator_base {
void* operator new(std::size_t size)
throw(std::bad_ alloc) {
return custom_allocato r::allocate(siz e);
}

void* operator new [](std::size_t size)
throw(std::bad_ alloc) {
return custom_allocato r::allocate(siz e);
}

void operator delete(void* mem, std::size_t size)
throw() {
custom_allocato r::deallocate(m em, size);
}

void operator delete [](void* mem, std::size_t size)
throw() {
custom_allocato r::deallocate(m em, size);
}
};

template<std::s ize_t T_size>
class buf : public allocator_base {
char mem[T_size];
public:
virtual ~buf() throw() {}
};

class buf2 : public buf<1234{
char mem2[1000];
};

int main() {
buf<1024>* b1 = new buf<1024>;
delete b1;

buf2* b2 = new buf2;
delete b2;

b2 = new buf2[5];
delete [] b2;

return 0;
}

_______________ _______________ _______________ _______________ ______________

On every version of GCC I have, I get the following output on a 32-bit
machine:

custom_allocato r::allocate(002 46C50, 1028)
custom_allocato r::deallocate(0 0246C50, 1028)
custom_allocato r::allocate(002 472A8, 2240)
custom_allocato r::deallocate(0 02472A8, 2240)
custom_allocato r::allocate(002 472A8, 11204)
custom_allocato r::deallocate(0 02472A8, 11204)

On every version of MSVC, I get:

custom_allocato r::allocate(003 65B28, 1028)
custom_allocato r::deallocate(0 0365B28, 1028)
custom_allocato r::allocate(003 62850, 2240)
custom_allocato r::deallocate(0 0362850, 2240)
custom_allocato r::allocate(003 66FA8, 11204)
custom_allocato r::deallocate(0 0366FA8, 2240)

Well, MSVC has a fairly nasty bug indeed. Anyway, what do you think James?

**Chris M. Thomasson** · Oct 4 '08, 05:25 AM

Re: This HAS to be UB...

"Victor Bazarov" <v.Abazarov@com Acast.netwrote in message
news:gc39rg$q2m $1@news.datemas .de...

Chris M. Thomasson wrote:

>Keep in mind that I am a C programmer; well, anyway here is the C++
>program...

[...]

>
Well, the default implementation of the operator delete[] does *not* have
the "size" argument. In fact there are two allowed declarations of the
operator delete[]:
>
void operator delete[](void* ptr) throw();
>
and
>
void operator delete[](void* ptr, const std::nothrow&) throw();
>
I'm not sure what else to tell you.

I think that

void operator delete [](void*, std::size_t) throw();

is a valid declaration. I mean, even Comeau compiles the following program
without any warnings:
_______________ _______________ _______________ _______________ ________
#include <cstdio>
#include <cstdlib>
#include <new>

struct custom_allocato r {
static void* allocate(std::s ize_t size)
throw(std::bad_ alloc) {
void* const mem = std::malloc(siz e);
if (! mem) {
throw std::bad_alloc( );
}
std::printf("cu stom_allocator: :allocate(%p, %lu)\n",
(void*)mem, (unsigned long)size);
return mem;
}

static void deallocate(void * const mem, std::size_t size)
throw() {
if (mem) {
std::printf("cu stom_allocator: :deallocate(%p, %lu)\n",
(void*)mem, (unsigned long)size);
std::free(mem);
}
}
};

struct allocator_base {
void* operator new(std::size_t size)
throw(std::bad_ alloc) {
return custom_allocato r::allocate(siz e);
}

void* operator new [](std::size_t size)
throw(std::bad_ alloc) {
return custom_allocato r::allocate(siz e);
}

void operator delete(void* mem, std::size_t size)
throw() {
custom_allocato r::deallocate(m em, size);
}

void operator delete [](void* mem, std::size_t size)
throw() {
custom_allocato r::deallocate(m em, size);
}
};

template<std::s ize_t T_size>
class buf : public allocator_base {
char mem[T_size];
public:
virtual ~buf() throw() {}
};

class buf2 : public buf<1234{
char mem2[1000];
};

int main() {
buf<1024>* b1 = new buf<1024>;
delete b1;

buf2* b2 = new buf2;
delete b2;

b2 = new buf2[5];
delete [] b2;

return 0;
}
_______________ _______________ _______________ _______________ ________

Humm... Is Comeau screwing up and compiling non-compliant code without so
much as a warning?

**Hendrik Schober** · Oct 4 '08, 10:55 AM

Re: This HAS to be UB...

Victor Bazarov wrote:

>[...]

>
Well, the default implementation of the operator delete[] does *not*
have the "size" argument. In fact there are two allowed declarations of
the operator delete[]:
>
void operator delete[](void* ptr) throw();
>
and
>
void operator delete[](void* ptr, const std::nothrow&) throw();

While that's true for global 'operator delete', my reading of
3.7.3.2/2 seems to indicate that class-specific versions can
indeed have a second argument of type 'std::size_t'. ICBWT.

I'm not sure what else to tell you.
>
V

Schobi

**Hendrik Schober** · Oct 4 '08, 10:55 AM

Re: This HAS to be UB...

Chris M. Thomasson wrote:

[...]
On MSVC 8 I get:
>
custom_allocato r::allocate(003 62850, 2234)
custom_allocato r::deallocate(0 0362850, 2234)
custom_allocato r::allocate(003 66B68, 11170)
custom_allocato r::deallocate(0 0366B68, 2234)

That code can be simplified further. Using VC9, this

#include <cstdio>
#include <cstdlib>
#include <new>

void* allocate(std::s ize_t size) throw(std::bad_ alloc) {
void* const mem = ::operator new(size);
std::printf("al locate(%p, %lu)\n", (void*)mem, (unsigned long)size);
return mem;
}

void deallocate(void * const mem, std::size_t size) throw() {
std::printf("de allocate(%p, %lu)\n", (void*)mem, (unsigned long)size);
if (mem) ::operator delete(mem);
}

struct buf2 {
char mem[1024];
void* operator new(std::size_t size) throw(std::bad_ alloc) {
return allocate(size);
}

void* operator new[](std::size_t size) throw(std::bad_ alloc) {
return allocate(size);
}

void operator delete(void* mem, std::size_t size) throw() {
deallocate(mem, size);
}

void operator delete [](void* mem, std::size_t size) throw() {
deallocate(mem, size);
}
};

int main() {
buf2* b = new buf2;
delete b;

b = new buf2[5];
delete [] b;

return 0;
}

shows the same behavior for me.

Debugging shows that VC9 doesn't call 'operator new[]' for 'new buf2[5]'
(it calls 'operator new' instead), but calls 'operator delete[]' for
'delete[] b'.
Either I'm missing something really obvious, or that's a plain bug.

[...]

Schobi

**James Kanze** · Oct 6 '08, 09:15 AM

Re: This HAS to be UB...

On Oct 4, 1:46 am, Hendrik Schober <spamt...@gmx.d ewrote:

Victor Bazarov wrote:

[...]

Well, the default implementation of the operator delete[]
does *not* have the "size" argument. In fact there are two
allowed declarations of the operator delete[]:

void operator delete[](void* ptr) throw();

and

void operator delete[](void* ptr, const std::nothrow&) throw();

While that's true for global 'operator delete', my reading of
3.7.3.2/2 seems to indicate that class-specific versions can
indeed have a second argument of type 'std::size_t'. ICBWT.

Actually, the only requirement for defining an operator delete
function is that the first argument have type void*. You can
declare and define delete functions with any other arguments you
want. What is relevant here is the fact that *IF* there is no
operator delete[]( void* ) in the class, but there is a operator
delete[]( void*, std::size_t ), the latter will be used as the
"usual" deallocator, and not only for placement delete.

And at global scope, the standard defines three operator
delete[]: (void*), (void*, std::nothrow const&) and (void*,
void*).

--
James Kanze (GABI Software) email:james.kan ze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientier ter Datenverarbeitu ng
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

**James Kanze** · Oct 6 '08, 09:25 AM

Re: This HAS to be UB...

On Oct 4, 7:13 am, "Chris M. Thomasson" <n...@spam.inva lidwrote:

"James Kanze" <james.ka...@gm ail.comwrote in message

news:7878ab49-834f-4bbc-b687-efdd8f31f1f3@z6 6g2000hsc.googl egroups.com...

On Oct 2, 9:52 pm, "Chris M. Thomasson" <n...@spam.inva lidwrote:

Keep in mind that I am a C programmer; well, anyway here is
the C++ program...

It looks to me like you're attacking some fairly tricky stuff.
You'd probably be better of starting with something simpler if
you're still learning C++. However...

I was exploring the feature in C++ delete operator in which
the size of the allocation is returned along with the pointer
to allocated memory. One could create heavily optimized custom
memory allocator using that important piece of information.

Certainly. It could be, in certain cases. But you're using a
fairly advanced feature of C++, one that many experienced C++
programmers aren't too familiar with. *IF* you're basically a C
programmer, and not too familiar with C++, you should probably
gain that familiarity first.

[...]

In practice, exception specifications are not really that
useful, except when they're empty. (It's very important in
certain cases to know that a function cannot throw any
exceptions, but it's rarely useful to know that it can't
throw certain types of exceptions.)

I thought it would be prudent to give the overloaded operator
new an exception specification of `std::bad_alloc '. Also, I
wanted to give an empty specification to the overload of
operator delete. As to how useful it is... Well, I don't
quite know.

Given that the standard does use that exception specifier for
the global operator new, it's probably a good idea to follow
suite. But I don't think that it's really that useful, and in
general, I wouldn't bother unless the specifier were empty.
(Note that the contract of the operator new function is slightly
different if it has an empty exception specifier. Unless it has
an empty specifier, operator new() and operator new[] may not
return a null pointer; if the version chosen has an empty
specifier, they can.)

void* const mem = ::operator new(size);
std::printf("cu stom_allocator: :allocate(%p, %lu)\n",
(void*)mem, (unsigned long)size);
return mem;
}
static void deallocate(void * const mem, std::size_t size)
throw() {
std::printf("cu stom_allocator: :deallocate(%p, %lu)\n",
(void*)mem, (unsigned long)size);
::operator delete(mem);
}
};
template<typena me T>
struct allocator_base {
static void* operator new(std::size_t size)

The static isn't really necessary: allocation and
deallocation member functions (operator new and operator
delete) are always static, whether you declare them so or
not. (On the other hand, it doesn't hurt.)

Its a habit of mine. Also, using printf in C++ is another habit.

The static doesn't hurt. Using printf is a very bad habit,
however.

throw(std::bad_ alloc()) {
return custom_allocato r::allocate(siz e);
}
static void* operator new[](std::size_t size)
throw(std::bad_ alloc()) {
return custom_allocato r::allocate(siz e);
}
static void operator delete(void* mem)

Just curious: since you require the size in delete[], why don't
you require it here? Derivation can mean that the size isn't a
constant, e.g.:

class Base : public allocator_base< Base >
{
// ...
} ;

class Derived : public Base
{
// ...
} ;

Base* p = new Derived ;
// ...
delete p ;
(This supposes, of course, that Base has a virtual destructor.)

[...]

_______________ _______________ _______________ _______________ __________
On GCC I get the following output:
custom_allocato r::allocate(002 46C50, 2234)
custom_allocato r::deallocate(0 0246C50, 2234)
custom_allocato r::allocate(002 47760, 11174)
custom_allocato r::deallocate(0 0247760, 11174)
On MSVC 8 I get:
custom_allocato r::allocate(003 62850, 2234)
custom_allocato r::deallocate(0 0362850, 2234)
custom_allocato r::allocate(003 66B68, 11170)
custom_allocato r::deallocate(0 0366B68, 2234)
Are they both right due to UB? WTF is going on? GCC seems to
be accurate at least... DAMN!

Well, there's no undefined behavior. You're program seems
perfectly legal and well defined to me. It looks like a bug in
VC++, see §12.5/5:

It definitely looks like a bug is MSVC++. I get erroneous
behavior on versions 6 through 9.

It's quite possible. This is such a rarely used feature, I
doubt that it gets much testing. (In practice, I don't think
I've ever used new[]/delete[] in over 15 years of C++.)

When a delete-expression is executed, the selected
deallocation function shall be called with the address
of the block of storage to be reclaimed as its first
argument and (if the two-parameter style is used) the
size of the block as its second argument.
And I can't think of any way of interpreting "the size of the
block" to mean anything other than the size requested in the
call to operator new.

I thought that MSVC was crapping out because `allocator_base '
was a template. So I created another little test which
hopefully has all the bugs fixed:
_______________ _______________ _______________ _______________ ______________

[...[

_______________ _______________ _______________ _______________ ______________
>
On every version of GCC I have, I get the following output on
a 32-bit machine:

custom_allocato r::allocate(002 46C50, 1028)
custom_allocato r::deallocate(0 0246C50, 1028)
custom_allocato r::allocate(002 472A8, 2240)
custom_allocato r::deallocate(0 02472A8, 2240)
custom_allocato r::allocate(002 472A8, 11204)
custom_allocato r::deallocate(0 02472A8, 11204)

On every version of MSVC, I get:

custom_allocato r::allocate(003 65B28, 1028)
custom_allocato r::deallocate(0 0365B28, 1028)
custom_allocato r::allocate(003 62850, 2240)
custom_allocato r::deallocate(0 0362850, 2240)
custom_allocato r::allocate(003 66FA8, 11204)
custom_allocato r::deallocate(0 0366FA8, 2240)

Well, MSVC has a fairly nasty bug indeed. Anyway, what do you
think James?

It looks like an error in the compiler, but it's certainly a
minor one, since it concerns a feature that has no real use in
practice.

--
James Kanze (GABI Software) email:james.kan ze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientier ter Datenverarbeitu ng
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

This HAS to be UB...

This HAS to be UB...

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