删除POD的[] [英] delete[] of POD
问题描述
int * p;
p = new int [10];
void * q =（void *）p;
delete [] q;
如果p保证是POD，这是否有效？ （不要问为什么......有一个
的原因我需要这个。显然我会在函数中创建p，在另一个
函数中使用p知道什么在一个函数中输入p是和删除p为
简单不知道p是什么）
读取MSDN似乎
"全局运算符删除函数，如果声明，则采用单个参数
$ b $类型为void *，其中包含指向要释放的对象的指针。
返回类型为void（operator delete无法返回值）。存在两种形式
用于类成员操作员删除功能：
所以应该没问题但是这是MSDN指南： - ）
--- bye
int *p;
p = new int[10];
void *q = (void*)p;
delete[] q;
Is this valid if p is guaranteed to be a POD? (don''t ask why... there is a
reason I need this. Clearly I will create p in function, use p in another
function that know what type p is and delete p in a function that for
simplicity won''t know what p is)
Reading the MSDN it seems that
"The global operator delete function, if declared, takes a single argument
of type void *, which contains a pointer to the object to deallocate. The
return type is void (operator delete cannot return a value). Two forms exist
for class-member operator delete functions:"
so it should be ok But this is the MSDN guide :-)
--- bye
推荐答案
MaxMax写道：
int * p;
p = new int [10];
void * q =（void *）p;
delete [] q;
这是否有效p保证是POD吗？ （不要问为什么......有一个原因我需要这个。显然我会在函数中创建p，在另一个知道什么类型p的函数中使用p并删除p用于简单的功能不知道p是什么）
阅读MSDN似乎
全局运算符删除函数，如果声明，采用类型为void *的单个参数，其中包含指向要释放的对象的指针。
返回类型为void（operator delete无法返回值）。对于类成员操作员删除功能存在两种形式：
所以它应该没问题但是这是MSDN指南:-)
--- bye
int *p;
p = new int[10];
void *q = (void*)p;
delete[] q;
Is this valid if p is guaranteed to be a POD? (don''t ask why... there is a
reason I need this. Clearly I will create p in function, use p in another
function that know what type p is and delete p in a function that for
simplicity won''t know what p is)
Reading the MSDN it seems that
"The global operator delete function, if declared, takes a single argument
of type void *, which contains a pointer to the object to deallocate. The
return type is void (operator delete cannot return a value). Two forms exist
for class-member operator delete functions:"
so it should be ok But this is the MSDN guide :-)
--- bye
operator delete []在标准中有一个void *作为参数，所以你很好。
-
问候，
Ferdi Smit（理学硕士）
电邮： Fe ******** @ cwi.nl
房间：C0.07电话： 4229
INS3可视化和3D界面
CWI荷兰阿姆斯特丹
operator delete[] has a void* as parameter in the standard, so you''re fine.
--
Regards,
Ferdi Smit (M.Sc.)
Email: Fe********@cwi.nl
Room: C0.07 Phone: 4229
INS3 Visualization and 3D Interfaces
CWI Amsterdam, The Netherlands
* Ferdi Smit：
* Ferdi Smit:
MaxMax写道：
int * p;
p = new int [10];
void * q =（void *）p;
delete [] q;
如果p保证是POD，这是否有效？ （不要问为什么......有一个原因我需要这个。显然我会在函数中创建p，在另一个知道什么类型p的函数中使用p并删除p用于简单的功能不知道p是什么）
阅读MSDN似乎
全局运算符删除函数，如果声明，采用类型为void *的单个参数，其中包含指向要释放的对象的指针。
返回类型为void（operator delete无法返回值）。对于类成员操作员删除功能存在两种形式：
所以应该没问题但是这是MSDN指南： - ）
int *p;
p = new int[10];
void *q = (void*)p;
delete[] q;
Is this valid if p is guaranteed to be a POD? (don''t ask why... there is a
reason I need this. Clearly I will create p in function, use p in another
function that know what type p is and delete p in a function that for
simplicity won''t know what p is)
Reading the MSDN it seems that
"The global operator delete function, if declared, takes a single argument
of type void *, which contains a pointer to the object to deallocate. The
return type is void (operator delete cannot return a value). Two forms exist
for class-member operator delete functions:"
so it should be ok But this is the MSDN guide :-)
operator delete []在标准中有一个void *作为参数，所以你没事。
operator delete[] has a void* as parameter in the standard, so you''re fine.
对不起，这是一个不正确的推理。 />
标准将此定义为未定义行为，如果某人不应该看到这个含义，它会在说明73中明确指出void *：
[上下文：删除数组]" ...使用
类型void *的指针无法删除对象因为..."。
-
答：因为它弄乱了人们通常阅读文字的顺序。
问：为什么这么糟糕东西？
A：热门发布。
问：usenet和电子邮件中最烦人的是什么？
Sorry, that''s an incorrect inference.
The standard defines this as Undefined Behavior, and in case somebody should
not see that implication, it states that explicitly for void* in note 73:
[context: delete array] "... an object cannot be deleted using a pointer of
type void* because ...".
--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
Alf P. Steinbach写道：
Alf P. Steinbach wrote:
operator delete []在标准中有一个void *作为参数，所以你没事。
operator delete[] has a void* as parameter in the standard, so you''re fine.
对不起，这是一个不正确的推理。
标准将此定义为未定义行为，如果某人不应该看到这种含义，它会在注释73中明确指出void *：
[context：delete array ]...使用
类型void *的指针无法删除对象因为...。
Sorry, that''s an incorrect inference.
The standard defines this as Undefined Behavior, and in case somebody should
not see that implication, it states that explicitly for void* in note 73:
[context: delete array] "... an object cannot be deleted using a pointer of
type void* because ...".
我认为指的是一个虚空的数组是不可能的（大小为0，类似的原因，为什么一个（空的）实例化对象的大小至少为
1）;因此你不能删除[]一个真正的空白*。但在max'的情况下，
基础数据实际上是另一种类型（int *）。运算符delete []
取无效*，所以无论你传递什么指针，它都会在运算符运行之前首先转换为
void *。 />
-
问候，
Ferdi Smit（理学硕士）
电子邮件： Fe********@cwi.nl
房间：C0.07电话：4229
INS3可视化和3D界面
CWI荷兰阿姆斯特丹
I think that refers to an array of void which is impossible (size 0, a
similar reason to why an (empty) instantiated object has size at least
1); and thus you can''t delete[] a true void*. But in max''s case the
underlying data is in fact of another type (int*). The operator delete[]
takes a void*, so whatever pointer you pass it, it will be cast to a
void* first before the operator runs anyway.
--
Regards,
Ferdi Smit (M.Sc.)
Email: Fe********@cwi.nl
Room: C0.07 Phone: 4229
INS3 Visualization and 3D Interfaces
CWI Amsterdam, The Netherlands
这篇关于删除POD的[]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！
