这个是UB ... [英] This HAS to be UB...
问题描述
请记住，我是一名C程序员;好吧，无论如何这里是C ++
计划......
__________________________________________________ ____________________
#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）;
返回mem;
}
static void deallocate（void * const mem，std :: size_t size）
throw（） {
std :: printf（" custom_allocator :: deallocate（％p，％lu）\ n"，
（void *）mem，（unsigned long）尺寸）;
::操作员删除（mem）;
}
};
模板< typename T> ;
struct allocator_base {
static void * operator new（std :: size_t size）
th row（std :: bad_alloc（））{
返回custom_allocator :: allocate（size）;
}
static void * operator new []（std :: size_t size）
throw（std :: bad_alloc（））{
返回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 =新buf2;
删除b;
b =新buf2 [5];
删除[] b;
返回0;
}
__________________________________________________ ____________________
在GCC上我得到以下输出：
custom_allocator :: allocate（00246C50,2234）
custom_allocator :: deallocate（00246C50,2234） ）
custom_allocator :: allocate（00247760,11178）
custom_allocator :: deallocate（00247760,11174）
在MSVC 8上，我得到：
custom_allocator :: allocate（00362850,2234）
custom_allocator :: deallocate（00362850,2234）
custom_allocator :: allocate（00366B68,11170）
custom_allocator :: deallocate（00366B68,2234）
它们是否适合UB？ WTF正在进行中？海湾合作委员会似乎是准确的，最低价格为...... b $！DAMN！
谢谢大家的时间。
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.
推荐答案
" Chris M. Thomasson" < no@spam.invalidwrote in message
news：2y ******************* @ newsfe06.iad ...
"Chris M. Thomasson" <no@spam.invalidwrote in message
news:2y*******************@newsfe06.iad...
请记住我是C程序员;好吧，无论如何这里是C ++
计划......
__________________________________________________ ____________________
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];
};
template<std::size_t T_size>
class buf {
char mem[T_size];
};
我将虚拟dtor添加到buf1，输出没有变化。
I add virtual dtor to buf1, and no change in output.
>
class buf2：public buf< 1234> ;,public allocator_base< buf2 {
char mem2 [1000];
};
>
class buf2 : public buf<1234>, public allocator_base<buf2{
char mem2[1000];
};
>
int main（）{
buf2 * b =新buf2;
删除b;
b =新buf2 [5];
删除[] b;
返回0;
}
__________________________________________________ ____________________
>
int main() {
buf2* b = new buf2;
delete b;
b = new buf2[5];
delete [] b;
return 0;
}
__________________________________________________ ____________________
[...]
[...]
Chris M. Thomasson写道：
Chris M. Thomasson wrote:
请记住我是C程序员;好吧，无论如何这里是C ++
程序...
[..]
在GCC上我得到以下输出：
custom_allocator :: allocate（00246C50,2234）
custom_allocator :: deallocate（00246C50,2234）
custom_allocator ::分配（00247760,11178）
custom_allocator :: deallocate（00247760,11178）
在MSVC 8上我得到：
custom_allocator :: allocate（00362850,2234）
custom_allocator :: deallocate（00362850,2234）
custom_allocator :: allocate（00366B68,11170）
custom_allocator :: deallocate（00366B68,2234）
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给出相同的输出，BTW。
MSVC 9 gives the same output, BTW.
它们是否都归功于UB？ WTF正在进行中？ GCC似乎很准确
至少...... DAMN！
Are they both right due to UB? WTF is going on? GCC seems to be accurate
at least... DAMN!
嗯，运算符delete []的默认实现*不*
具有size论点。实际上有两个允许的声明
运算符delete []：
void operator delete []（void * ptr）throw（）;
和
void operator delete []（void * ptr，const std :: nothrow&）throw（）;
我不知道还能告诉你什么。
V
-
请在通过电子邮件回复时删除资本''A'
我没有回复最热门的回复，请不要问
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
" Victor Bazarov" < v。******** @ comAcast.netwrote in message
news：gc ********** @ news.datemas.de ...
"Victor Bazarov" <v.********@comAcast.netwrote in message
news:gc**********@news.datemas.de...
Chris M. Thomasson写道：
Chris M. Thomasson wrote:
>请记住我是C程序员;好吧，无论如何这里是C ++
程序......
[...]
在GCC上我得到以下输出：
custom_allocator :: allocate（00246C50,2234）
custom_allocator :: deallocate（00246C50,2234）
custom_allocator :: allocate（00247760,11178）
custom_allocator :: deallocate（00247760,11178）
在MSVC 8上我得到：
custom_allocator :: allocate（00362850,2234）
custom_allocator :: deallocate（00362850,2234）
custom_allocator :: allocate（00366B68,11170）
custom_allocator :: deallocate（00366B68,2234）
>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给出相同的输出，BTW。
MSVC 9 gives the same output, BTW.
> UB是否正确？ WTF正在进行中？ GCC似乎准确至少...... DAMN！
>Are they both right due to UB? WTF is going on? GCC seems to be accurate
at least... DAMN!
嗯，运算符delete []的默认实现确实*不*
size论点。实际上有两个允许声明的
运算符delete []：
void operator delete []（void * ptr）throw（）;
和
void operator delete []（void * ptr，const std :: nothrow&）throw（）;
我不知道还能告诉你什么。
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.
这必须是GCC扩展吗？这很奇怪，好吧，也许不那么奇怪
因为它必须只是100％UB。那好吧。我最初以为我可以利用它获得的优惠。不是!!!
; ^ /
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!!!
;^/
这篇关于这个是UB ...的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
