c++ - 不调用放置释放函数

Question

我编写的以下代码必须同时调用放置释放和分配函数。

#include <iostream>
using namespace std;

struct A
{
    void * operator new [] (size_t t, int, int)
    {
        cout << "3 parameters allocation" << endl;
        return ::operator new[](t);
    }

    void operator delete [] (void *p, int, int)
    {
        cout << "3 parameters deallocation" << endl;
        return ::operator delete[](p);
    }
};

int main() 
{
    A *a = new (5,5) A[10]; //operator new [] (size_t, int, int) invoked
    delete [] a; //Error. Overload resolution is failed.
}

demo

5.3.4/22 说 N3797:

A declaration of a placement deallocation function matches the declaration of a placement allocation function if it has the same number of parameters and, after parameter transformations (8.3.5), all parameter types except the first are identical. If the lookup finds a single matching deallocation function, that function will be called; otherwise, no deallocation function will be called.

它不是在C++11中实现的还是我的误解？

Answer 1

如果提供了特定于类的释放函数，则没有以范围解析运算符为前缀的delete-expression 是非良构的，除非具有一个或两个参数的特定于类的释放函数是可用:

10 - If the type is complete and if deallocation function lookup finds both a usual deallocation function with only a pointer parameter and a usual deallocation function with both a pointer parameter and a size parameter, then the selected deallocation function shall be the one with two parameters. Otherwise, the selected deallocation function shall be the function with one parameter.

所以你需要:

• 提供 void operator delete[](void*);,
• 提供void operator delete[](void*, size_t);，或者
• 编写::delete[]一个。