我正在使用 c++11 (g++ v4.7.2)
我已经为“Base”类重载了 operator new 和 operator delete。显然他们不应该在调用 new/delete 时调用构造函数/析构函数,因为我没有在重载的 new/delete 中实现 ctor/dtor 调用。但是输出却与此相反
//Output of below program
Base new opnd.cpp 87
Base ctor
10
Base dtor
Base delete
为什么要为重载的 operator new/delete 调用 ctor/dtor?
#include <iostream>
using namespace std;
#define NEW new(__FILE__, __LINE__)
#define DELETE delete
class Base
{
public:
Base():m_i(10){ cout << "Base ctor" << endl; }
virtual ~Base(){ cout << "Base dtor" << endl; }
void* operator new(size_t size, const char* file, int line) throw(std::bad_alloc);
void operator delete(void *rawMem, size_t size);
int geti(){ return m_i; }
private:
int m_i;
};
void* Base::operator new(size_t size, const char* file, int line) throw(std::bad_alloc)
{
void *p;
cout << "Base new " << file << " " << line << endl;
//Handle 0 byte requests
if(size == 0)
size = 1;
if(size != sizeof(Base))
{
return ::operator new(size); // To handle new requests for derived classes
}
while(true)
{
p = malloc(size);
if(p)
return p;
new_handler globalHandler = set_new_handler(0);
set_new_handler(globalHandler);
if(globalHandler) (*globalHandler)();
else throw std::bad_alloc();
}
}
void Base::operator delete(void *rawMem, size_t size)
{
cout << "Base delete" << endl;
if(rawMem == 0)
return;
if(size != sizeof(Base))
{
::operator delete(rawMem); //To handle delete requests for derived classes
return;
}
free(rawMem);
}
int main()
{
Base *b = NEW Base;
cout << b->geti() << endl;
DELETE b;
return 0;
}
最佳答案
operator new
函数(无论是全局函数还是类特定函数)并不是 new
表达式的完整实现。这只是分配功能。当您在代码中编写 new T
时,会发生以下情况:
选择并调用名为
operator new
的适当分配函数来为对象获取空间。T
的构造函数在从点 1 获得的空间上被调用。
这意味着无法通过编写自己的 operator new
来绕过构造函数调用 - 调用构造函数是由语言完成的,而不是由分配函数完成的。
关于c++ - 为什么要为重载的 operator new 调用类构造函数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23385266/