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comp.lang.c++

This HAS to be UB...

Chris M. Thomasson

10/2/2008 7:53:00 PM

Keep in mind that I am a C programmer; well, anyway here is the C++
program...
______________________________________________________________________
#include <cstdio>
#include <cstdlib>
#include <new>


struct custom_allocator {
static void* allocate(std::size_t size)
throw(std::bad_alloc()) {
void* const mem = ::operator new(size);
std::printf("custom_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("custom_allocator::deallocate(%p, %lu)\n",
(void*)mem, (unsigned long)size);
::operator delete(mem);
}
};


template<typename T>
struct allocator_base {
static void* operator new(std::size_t size)
throw(std::bad_alloc()) {
return custom_allocator::allocate(size);
}

static void* operator new[](std::size_t size)
throw(std::bad_alloc()) {
return custom_allocator::allocate(size);
}

static void operator delete(void* mem)
throw() {
if (mem) {
custom_allocator::deallocate(mem, sizeof(T));
}
}

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


template<std::size_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_allocator::allocate(00246C50, 2234)
custom_allocator::deallocate(00246C50, 2234)
custom_allocator::allocate(00247760, 11174)
custom_allocator::deallocate(00247760, 11174)




On MSVC 8 I get:

custom_allocator::allocate(00362850, 2234)
custom_allocator::deallocate(00362850, 2234)
custom_allocator::allocate(00366B68, 11170)
custom_allocator::deallocate(00366B68, 2234)





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




thank you all for your time.


12 Answers

Chris M. Thomasson

10/2/2008 7:56:00 PM

0

"Chris M. Thomasson" <no@spam.invalid> wrote in message
news:2y9Fk.39817$hX5.39354@newsfe06.iad...
> Keep in mind that I am a C programmer; well, anyway here is the C++
> program...
> ______________________________________________________________________
[...]

> template<std::size_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

10/2/2008 8:10:00 PM

0

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_allocator::allocate(00246C50, 2234)
> custom_allocator::deallocate(00246C50, 2234)
> custom_allocator::allocate(00247760, 11174)
> custom_allocator::deallocate(00247760, 11174)
>
>
>
>
> On MSVC 8 I get:
>
> custom_allocator::allocate(00362850, 2234)
> custom_allocator::deallocate(00362850, 2234)
> custom_allocator::allocate(00366B68, 11170)
> custom_allocator::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.

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

10/2/2008 8:26:00 PM

0

"Victor Bazarov" <v.Abazarov@comAcast.net> wrote 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_allocator::allocate(00246C50, 2234)
>> custom_allocator::deallocate(00246C50, 2234)
>> custom_allocator::allocate(00247760, 11174)
>> custom_allocator::deallocate(00247760, 11174)
>>
>>
>>
>>
>> On MSVC 8 I get:
>>
>> custom_allocator::allocate(00362850, 2234)
>> custom_allocator::deallocate(00362850, 2234)
>> custom_allocator::allocate(00366B68, 11170)
>> custom_allocator::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.h4g

10/2/2008 11:18:00 PM

0

In article <2y9Fk.39817$hX5.39354@newsfe06.iad>, "Chris M. Thomasson"
<no@spam.invalid> wrote:

> Keep in mind that I am a C programmer; well, anyway here is the C++
> program...
> ______________________________________________________________________
> #include <cstdio>
> #include <cstdlib>
> #include <new>
>
> struct custom_allocator {
> static void* allocate(std::size_t size)
> throw(std::bad_alloc()) {
***^^***

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

How did this even compile?

Chris M. Thomasson

10/3/2008 4:19:00 AM

0

"blargg" <blargg.h4g@gishpuppy.com> wrote in message
news:blargg.h4g-0210081818220001@192.168.1.4...
> In article <2y9Fk.39817$hX5.39354@newsfe06.iad>, "Chris M. Thomasson"
> <no@spam.invalid> wrote:
>
>> Keep in mind that I am a C programmer; well, anyway here is the C++
>> program...
>> ______________________________________________________________________
>> #include <cstdio>
>> #include <cstdlib>
>> #include <new>
>>
>> struct custom_allocator {
>> static void* allocate(std::size_t size)
>> throw(std::bad_alloc()) {
> ***^^***
>
>> void* const mem = ::operator new(size);
>> std::printf("custom_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

10/3/2008 8:44:00 AM

0

On Oct 2, 9:52 pm, "Chris M. Thomasson" <n...@spam.invalid> wrote:
> 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_allocator {
> static void* allocate(std::size_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("custom_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("custom_allocator::deallocate(%p, %lu)\n",
> (void*)mem, (unsigned long)size);
> ::operator delete(mem);
> }
> };

> template<typename 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_allocator::allocate(size);
> }

> static void* operator new[](std::size_t size)
> throw(std::bad_alloc()) {
> return custom_allocator::allocate(size);
> }

> 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_allocator::deallocate(mem, sizeof(T));
> }
> }
> static void operator delete [](void* mem, std::size_t size)
> throw() {
> if (mem) {
> custom_allocator::deallocate(mem, size);
> }
> }
> };

> template<std::size_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_allocator::allocate(00246C50, 2234)
> custom_allocator::deallocate(00246C50, 2234)
> custom_allocator::allocate(00247760, 11174)
> custom_allocator::deallocate(00247760, 11174)

> On MSVC 8 I get:

> custom_allocator::allocate(00362850, 2234)
> custom_allocator::deallocate(00362850, 2234)
> custom_allocator::allocate(00366B68, 11170)
> custom_allocator::deallocate(00366B68, 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.kanze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

Hendrik Schober

10/3/2008 11:47:00 PM

0

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

10/3/2008 11:54:00 PM

0

Chris M. Thomasson wrote:
> [...]
> On MSVC 8 I get:
>
> custom_allocator::allocate(00362850, 2234)
> custom_allocator::deallocate(00362850, 2234)
> custom_allocator::allocate(00366B68, 11170)
> custom_allocator::deallocate(00366B68, 2234)

That code can be simplified further. Using VC9, this

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

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

void deallocate(void* const mem, std::size_t size) throw() {
std::printf("deallocate(%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

Chris M. Thomasson

10/4/2008 5:13:00 AM

0

>
"James Kanze" <james.kanze@gmail.com> wrote in message
news:7878ab49-834f-4bbc-b687-efdd8f31f1f3@z66g2000hsc.googlegroups.com...
> On Oct 2, 9:52 pm, "Chris M. Thomasson" <n...@spam.invalid> wrote:
> > 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_allocator {
> > static void* allocate(std::size_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("custom_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("custom_allocator::deallocate(%p, %lu)\n",
> > (void*)mem, (unsigned long)size);
> > ::operator delete(mem);
> > }
> > };

> > template<typename 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_allocator::allocate(size);
> > }

> > static void* operator new[](std::size_t size)
> > throw(std::bad_alloc()) {
> > return custom_allocator::allocate(size);
> > }

> > 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_allocator::allocate(00246C50, 2234)
> > custom_allocator::deallocate(00246C50, 2234)
> > custom_allocator::allocate(00247760, 11174)
> > custom_allocator::deallocate(00247760, 11174)

> > On MSVC 8 I get:

> > custom_allocator::allocate(00362850, 2234)
> > custom_allocator::deallocate(00362850, 2234)
> > custom_allocator::allocate(00366B68, 11170)
> > custom_allocator::deallocate(00366B68, 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_allocator {
static void* allocate(std::size_t size)
throw(std::bad_alloc) {
void* const mem = std::malloc(size);
if (! mem) {
throw std::bad_alloc();
}
std::printf("custom_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("custom_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_allocator::allocate(size);
}

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

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

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




template<std::size_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_allocator::allocate(00246C50, 1028)
custom_allocator::deallocate(00246C50, 1028)
custom_allocator::allocate(002472A8, 2240)
custom_allocator::deallocate(002472A8, 2240)
custom_allocator::allocate(002472A8, 11204)
custom_allocator::deallocate(002472A8, 11204)




On every version of MSVC, I get:

custom_allocator::allocate(00365B28, 1028)
custom_allocator::deallocate(00365B28, 1028)
custom_allocator::allocate(00362850, 2240)
custom_allocator::deallocate(00362850, 2240)
custom_allocator::allocate(00366FA8, 11204)
custom_allocator::deallocate(00366FA8, 2240)



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

Chris M. Thomasson

10/4/2008 5:22:00 AM

0


"Victor Bazarov" <v.Abazarov@comAcast.net> wrote 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_allocator {
static void* allocate(std::size_t size)
throw(std::bad_alloc) {
void* const mem = std::malloc(size);
if (! mem) {
throw std::bad_alloc();
}
std::printf("custom_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("custom_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_allocator::allocate(size);
}

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

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

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




template<std::size_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?