关于effective-cpp 条款52的一些疑问
正常的new语句会有两步操作,一是分配内存,二是调用相应类的构造函数构造对象。
现在假设在第二步,即构造函数调用时抛异常
// mem_placement.h
#ifndef __MEM_PLACEMENTNEW_H__
#define __MEM_PLACEMENTNEW_H__
#include <iostream>
using namespace std;
class Apple
{
public:
Apple();
static void * operator new(std::size_t size, std::ostream &logStream) throw(std::bad_alloc);
static void operator delete(void *pMemory, std::size_t size) throw();
// 如果定义了placement new,必须定义额外参数一致的placement delete,才能防止内存泄漏。
static void operator delete(void *pMemory, std::ostream &logStream) throw();
};
#endif
// mem_placementnew.cpp
#include "mem_placemantnew.h"
Apple::Apple()
{
#if 0
int *pn = new int[1000000000000L];
delete [] pn;
#endif
throw std::bad_alloc();
}
void * Apple::operator new (size_t size, ostream &logStream) throw(std::bad_alloc)
{
logStream << "request for memory!\n";
if (size != sizeof(Apple))
{
return ::operator new (size);
}
while (true)
{
// 尝试分配内存
void *pMem = malloc(size);
if (pMem)
{
return pMem;
}
// 分配失败, 调用失败处理函数
std::new_handler globalHandler = std::set_new_handler(0);
std::set_new_handler(globalHandler);
if (globalHandler)
{
(*globalHandler)();
}
else
{
throw std::bad_alloc();
}
}
}
void Apple::operator delete (void *pMemory, size_t size) throw()
{
cout << "custom delete called: " << size << endl;
// C++保证删除null指针永远安全,首先需要实现这一保证
if (0 == pMemory)
{
return;
}
if (size != sizeof(Apple))
{
::operator delete (pMemory);
}
// 归还、释放内存
::free(pMemory);
}
void Apple::operator delete (void *pMemory, std::ostream &logStream) throw()
{
cout << "custom placement delete called\n";
logStream << "release memory\n";
// C++保证删除null指针永远安全,首先需要实现这一保证
if (0 == pMemory)
{
return;
}
// 归还、释放内存
::free(pMemory);
}
书中阐述到自定义的placement new必须有相同额外参数的placement delete才会在构造抛异常时,由运行系统自动调用对应的placement delete释放已分配的内存。
现在的问题是我应该怎么测试异常抛出之后确实调用了void Apple::operator delete (void *pMemory, std::ostream &logStream) throw(); 这个函数呢?因为抛异常,程序直接就terminate掉了。看不见有调进去。有兄弟知道怎么调试这个来验证书本内容吗?