c++ - 全局分配函数和 const void*

Question

根据 C++11，下面的代码是否构成“未定义行为”(由于使用了 const_cast，请参见下面的引用)？

const void* p = operator new(123);
operator delete(const_cast<void*>(p));

来自 C++11 标准 (3.7.4.2.3) 的相关引述:

The value of the first argument supplied to a deallocation function may be a null pointer value; if so, and if the deallocation function is one supplied in the standard library, the call has no effect. Otherwise, the behavior is undefined if the value supplied to operator delete(void*) in the standard library is not one of the values returned by a previous invocation of either operator new(std::size_t) or operator new(std::size_t, const std::nothrow_t&) in the standard library

如果答案是否定的，请提供来自 C++11 标准的引述以证实这一点。

Answer 1

它不是未定义的。推理是这样的:

1. operator new 返回 void*，因此保证返回可修改(非常量)内存:[support.dynamic]

   void* operator new(std::size_t size);
   

2. const_cast 如果引用的对象不是 const，则丢弃常量是有效的:[expr.const.cast]§7，指的是 [dcl.type.cv]，特别是§3+4

   3 A pointer or reference to a cv-qualified type need not actually point or refer to a cv-qualified object, but it is treated as if it does; a const-qualified access path cannot be used to modify an object even if the object referenced is a non-const object and can be modified through some other access path.

   4 Except that any class member declared mutable (7.1.1) can be modified, any attempt to modify a const object during its lifetime (3.8) results in undefined behavior.

3. const_cast 不修改操作数的值:[expr.const.cast]§3:

   ... The result of a pointer const_cast refers to the original object.