问：运营商新/删除的地址 [英] Q: address of operator new/delete
问题描述
是否有可能获得运营商new / delete和运营商
new [] / delete []的地址？否则我将不得不去购买一个非便携式和/或
相当不干净的解决方案。
感谢任何提示！
- -
jb
（如果你想通过电子邮件回复，用x替换y）
Hi,
is it possible to get the address of operator new/delete and operator
new[]/delete[] portably? Otherwise I would have to go for a non-portable and
rather unclean solution.
Thanks for any hints!
--
jb
(replace y with x if you want to reply by e-mail)
推荐答案
" Jakob Bieling" <是ne ***** @ gmy.net>在留言新闻中写道：bj ************* @ news.t-online.com ...
"Jakob Bieling" <ne*****@gmy.net> wrote in message news:bj*************@news.t-online.com...
是不是可以获得运营商新/删除和运营商
new [] / delete []的地址吗？否则我将不得不寻求一种非便携式而且非常不干净的解决方案。
Hi,
is it possible to get the address of operator new/delete and operator
new[]/delete[] portably? Otherwise I would have to go for a non-portable and
rather unclean solution.
#include< new>
using namespace std;
int main（）{
void *（* ona）（size_t）=& operator new [];
}
#include <new>
using namespace std;
int main() {
void* (*ona)(size_t) = &operator new[];
}
2003年9月4日星期四17:33:18 +0200，Jakob Bieling < ne ***** @ gmy.net>
写道：
On Thu, 4 Sep 2003 17:33:18 +0200, "Jakob Bieling" <ne*****@gmy.net>
wrote:
是否有可能得到运营商新/删除和运营商的地址
新[] /删除[]可移植？否则我将不得不去寻找一个非便携式且相当不干净的解决方案。
Hi,
is it possible to get the address of operator new/delete and operator
new[]/delete[] portably? Otherwise I would have to go for a non-portable and
rather unclean solution.
AFAIK，这是好的和便携的：
#include< cstddef>
#include< new>
struct A
{
void * operator new（size_t）
{
throw std :: bad_alloc（）;
}
};
int main（）
{
void *（* ptr1）（std： ：size_t）= :: operator new;
void *（* ptr2）（std :: size_t）=& A :: operator new;
}
Tom
AFAIK, this is fine and portable:
#include <cstddef>
#include <new>
struct A
{
void* operator new(size_t)
{
throw std::bad_alloc();
}
};
int main()
{
void* (*ptr1)(std::size_t) = ::operator new;
void* (*ptr2)(std::size_t) = &A::operator new;
}
Tom
" Ron Natalie" < ro*@sensor.com>在留言中写道
新闻：3f *********************** @ news.newshosting.co m ...
"Ron Natalie" <ro*@sensor.com> wrote in message
news:3f***********************@news.newshosting.co m...
Jakob Bieling <是ne ***** @ gmy.net>在消息中写道
"Jakob Bieling" <ne*****@gmy.net> wrote in message
新闻：bj ************* @ news.t-online.com ...
news:bj*************@news.t-online.com...
是否有可能获得operator new / delete和
operator new [] / delete []的地址？否则我将不得不寻找一个非便携的
而不是不干净的解决方案。
Hi,
is it possible to get the address of operator new/delete and operator new[]/delete[] portably? Otherwise I would have to go for a non-portable and rather unclean solution.
#include< new>
使用命名空间std;
int main（）{
void *（* ona）（size_t）=& operator new [];
}
#include <new>
using namespace std;
int main() {
void* (*ona)(size_t) = &operator new[];
}
哦，我明白了。我试图将它直接转换为void *并且我不知道
知道哪个超载了我想要的运营商地址。使用static_cast它现在可以使用
。
谢谢！
-
jb
（如果你想通过电子邮件回复，用x替换y）
Oh I see. I tried to convert it to a void* directly and I prolly did not
know which overloaded operator''s address I wanted. With a static_cast it
works now.
Thanks!
--
jb
(replace y with x if you want to reply by e-mail)
这篇关于问：运营商新/删除的地址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！
