展示位置新运算符是否禁用默认的新运算符？

Question

我有这样的代码段

#include <cstring>
#define DISABLE_DEFAULT_NEW                                                    \
  void* operator new(size_t aSize, void* aPtr);                                \
  void operator delete(void* aPtr);

class A
{
public:
  DISABLE_DEFAULT_NEW
  A() {}
};

int
main()
{

  A* a = new A();
  return 0;
}

编译时，会看到类似这样的错误消息

disable_new.cpp:17:10: error: no matching function for call to 'operator new'
  A* a = new A();
         ^
disable_new.cpp:9:3: note: candidate function not viable: requires 2 arguments, but 1 was provided
  DISABLE_DEFAULT_NEW
  ^
disable_new.cpp:3:9: note: expanded from macro 'DISABLE_DEFAULT_NEW'
  void* operator new(size_t aSize, void* aPtr);                                \
        ^
1 error generated.

我的问题是默认的新运算符在哪里？我希望new的默认语法仍然有效，规范中是否提到了行为问题？

Answer 1

他们只是隐藏