delete如何处理指针常量？ [英] How does delete deal with pointer constness?
问题描述
我正在阅读此问题删除const指针，并想进一步了解删除行为。现在，根据我的理解：
I was reading this question Deleting a const pointer and wanted to know more about delete behavior. Now, as per my understanding:
删除表达式可以通过两个步骤进行操作：
delete expression works in two steps:
- 调用析构函数
- 然后释放内存（通常调用
free（）），方法是调用操作符删除。
- invoke destructor
- then releases the memory (often with a call to
free()) by calling operator delete.
操作符删除接受 void * 。作为测试程序的一部分，我重载了 operator delete ，发现 operator delete 不接受 const 指针。
operator delete accepts a void*. As part of a test program I overloaded operator delete and found that operator delete doesn't accept const pointer.
由于运算符删除不接受const指针，并且在内部进行删除，因此调用了运算符delete，删除const指针是工作吗？
Since operator delete does not accept const pointer and delete internally calls operator delete, how does Deleting a const pointer work ?
是否删除在内部使用const_cast吗？
Does delete uses const_cast internally?
推荐答案
const_cast并没有真正做任何事情–这是一种抑制编译器抱怨对象常数的方法。 delete 关键字是一个编译器结构，编译器知道在这种情况下该怎么做，并且不关心指针的常数。
const_cast doesn't really do anything – it's a way to suppress compiler moaning about const-ness of the object. delete keyword is a compiler construct, the compiler knows what to do in this case and doesn't care about const-ness of the pointer.
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