blocks|key|2583076|text|这在任何情况下都不起作用，如果我们假设工作意味着拥有定义良好的行为，而不是看起来像是工作(即不崩溃)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2583077|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

This doesn't work in any case, if we assume work means having well-defined behavior rather than appearing to work (i.e. not crashing)

blocks|key|4198321|text|不，这是未定义的行为，不管项目是否有一个简单的析构函数。如果析构函数是微不足道的，它可能看起来“工作”，而实际上它是内存泄漏，等等。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4198322|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

No, this is undefined behavior regardless of whether or not the item has a trivial destructor. If the destructor is trivial it may appear to "work" when in fact it's leaking memory, etc.

blocks|key|2583128|text|我要说的是，这介于已定义的实现和未定义的实现之间。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2583129|2583130|+5.3.5/2：“在第一种选择(delete对象)中，delete操作数的值可以是...指向由前一个new-expression创建的非数组对象的指针...|blockquote|2583131|2583132|2583133|指针的值在按照您的方式使用时不会更改，因此，如果使用sizeof(char*)+==+sizeof(int*)，这应该会按预期工作。该特定比较的结果是由实现定义的，如果假设为假，则行为是未定义的。|offset|length|style|CODE|2583134|所以它真的不是特别安全。|2583135|entityMap^0|0|0|0|0|0|Q|T|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|S|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|T|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|U|8|@]|9|@]|A|$]]|$1|G|3|-4|5|6|7|V|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|W|8|@$J|X|K|Y|L|M]]|9|@]|A|$]]|$1|N|3|O|5|6|7|Z|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|Q|$]]

I'm going to say this is somewhere between implementation defined and undefined. 

<blockquote>
 5.3.5/2: "In the first alternative (delete object), the value of the
 operand of delete may be ... a pointer to a
 non-array object created by a previous new-expression ... .
</blockquote>

The value of the pointer does not change when used the way you did so, so this should work as expected, provided <code>sizeof(char*) == sizeof(int*)</code>. The result of that particular comparison is implementation defined, and if the assumption is false then the behavior is undefined.

So it really really isn't particularly safe.

blocks|key|524833|text|很容易看出这是一个危险的错误。这两种类型的内存分配和释放方式可能完全不同且不兼容。这包括填充、垃圾收集、记账、特定于类的内存操作等。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|524834|#include+<cstddef>
#include+<cstdlib>
#include+<iostream>

using+namespace+std;

class+A
{

++++public:

++++++++void*+operator+new+(size_t+size)
++++++++{
++++++++++++cout+<<+"A::operator+new+(size_t)"+<<+endl;
++++++++++++return+malloc(size);
++++++++}

++++++++void*+operator+new+[]+(size_t+size)
++++++++{
++++++++++++cout+<<+"A::operator+new+[]+(size_t)"+<<+endl;
++++++++++++return+malloc(size);
++++++++}

++++++++void+operator+delete+(void*+ptr)
++++++++{
++++++++++++cout+<<+"A::operator+delete+(void*)"+<<+endl;
++++++++++++free(ptr);
++++++++}

++++++++void+operator+delete+[]+(void*+ptr)
++++++++{
++++++++++++cout+<<+"A::operator+delete+[]+(void*)"+<<+endl;
++++++++++++free(ptr);
++++++++}

};

class+B
{

++++public:

++++++++void*+operator+new+(size_t+size)
++++++++{
++++++++++++cout+<<+"B::operator+new+(size_t)+with+some+B-specific+stuff"+<<+endl;
++++++++++++return+malloc(size);
++++++++}

++++++++void*+operator+new+[]+(size_t+size)
++++++++{
++++++++++++cout+<<+"B::operator+new+[]+(size_t)+with+some+B-specific+stuff"+<<+endl;
++++++++++++return+malloc(size);
++++++++}

++++++++void+operator+delete+(void*+ptr)
++++++++{
++++++++++++cout+<<+"B::operator+delete+(void*)+with+some+B-specific+stuff"+<<+endl;
++++++++++++free(ptr);
++++++++}

++++++++void+operator+delete+[]+(void*+ptr)
++++++++{
++++++++++++cout+<<+"B::operator+delete+[]+(void*)+with+some+B-specific+stuff"+<<+endl;
++++++++++++free(ptr);
++++++++}

};


int+main+(int,+char**)
{

++++union{
++++++++A*+a;
++++++++B*+b;
++++}+u;

++++u.a+=+new+A();
++++delete+u.b;

++++u.a+=+new+A[5];
++++delete+[]+u.b;

}|code-block|syntax|javascript|524835|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

It is easy to see this is a dangerous error. The two types might have completely different and incompatible ways of memory allocation and deallocation. This includes padding, garbage collection, bookkeeping, class-specific memory manipulation, etc. Just don't do it.

<pre><code>#include &lt;cstddef&gt;
#include &lt;cstdlib&gt;
#include &lt;iostream&gt;

using namespace std;

class A
{

 public:

 void* operator new (size_t size)
 {
 cout &lt;&lt; "A::operator new (size_t)" &lt;&lt; endl;
 return malloc(size);
 }

 void* operator new [] (size_t size)
 {
 cout &lt;&lt; "A::operator new [] (size_t)" &lt;&lt; endl;
 return malloc(size);
 }

 void operator delete (void* ptr)
 {
 cout &lt;&lt; "A::operator delete (void*)" &lt;&lt; endl;
 free(ptr);
 }

 void operator delete [] (void* ptr)
 {
 cout &lt;&lt; "A::operator delete [] (void*)" &lt;&lt; endl;
 free(ptr);
 }

};

class B
{

 public:

 void* operator new (size_t size)
 {
 cout &lt;&lt; "B::operator new (size_t) with some B-specific stuff" &lt;&lt; endl;
 return malloc(size);
 }

 void* operator new [] (size_t size)
 {
 cout &lt;&lt; "B::operator new [] (size_t) with some B-specific stuff" &lt;&lt; endl;
 return malloc(size);
 }

 void operator delete (void* ptr)
 {
 cout &lt;&lt; "B::operator delete (void*) with some B-specific stuff" &lt;&lt; endl;
 free(ptr);
 }

 void operator delete [] (void* ptr)
 {
 cout &lt;&lt; "B::operator delete [] (void*) with some B-specific stuff" &lt;&lt; endl;
 free(ptr);
 }

};


int main (int, char**)
{

 union{
 A* a;
 B* b;
 } u;

 u.a = new A();
 delete u.b;

 u.a = new A[5];
 delete [] u.b;

}
</code></pre>

Doing this:

<pre><code> union{
 int * integer;
 char * character;
 } u;
 u.integer = new int;
 delete u.character;

 u.integer = new int[5];
 delete [] u.character;
</code></pre>

I assume this wouldn't work if any of these types have non trivial destructors, but is this ok?

Deleting an aliased pointer

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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执行此操作： union{     int * integer;     char * character; } u; u.integer = new int; delete u.character; u.integer = new int[5]; delete [] u.character;我假设如果这些类型中的任何一个都有非平凡的析构函数，那么这将不会起作用，但是这可以吗？

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