资源获取即初始化RAII

//class Resource {
//public:
//    Resource(parms p): r(allocate(p)) { }
//    ~Resource() { release(r); }
//    // also need to define copy and assignment
//private:
//    resource_type *r;           // resource managed by this type
//    resource_type *allocate(parms p);     // allocate this resource
//    void release(resource_type*);         // free this resource
//};

//void fcn()
//{
//    Resource res(args);   // allocates resource_type
//    // code that might throw an exception
//    // if exception occurs, destructor for res is run automatically
//    // ...
//}  // res goes out of scope and is destroyed automatically

#include <exception>
#include <iostream>
#include <unistd.h>
#include <stdexcept>

class Resource {
public:
// 不用allocate这个函数时，可以使用这个构造函数也是一样的
//Resource(int const & number) : r(new int [number]())
//{
//    std::cout << "allocate 4k bytes" << std::endl;
//}
Resource(int const & number) : r(allocate(number)) { }
~Resource() { release(r); }
private:
// I as a programmer, believe this function may throw exception
int* allocate(const int & number) const
{
std::cout << "allocate 4k bytes " << std::endl;
return new int[number]();           // 小括号表示初始化为0，在linux下这句话也是编不过的，必须把小括号拿掉
}
// I as a programmer, believe that this function will not throw exception
void release(int* pRes) const throw()
{
std::cout << "free 4k bytes " << std::endl;
delete [] pRes;
// delete pRes;
}
int* r;
};

void fcn() throw()
{
usleep(1000);
Resource const res(1024);
}

void fcn2() throw(std::logic_error) // 程序员认为这个函数可能抛出std::exception类型或者其派生类类型的异常
{
usleep(1000);
Resource const res(1000);
// throw std::exception("a! yi chang le!!!");
throw std::logic_error("a! yi chang le!!!");
}

int main()
{
//while(true) // 用来验证我的RAII是不是好用，一是注释析构中的release，然后取消注释，通过任务管理器能看的很清楚
//{
//    fcn();
//}

// 把上面注释起来，下面用来看出异常时，能否有效

while(true)
{
try
{
fcn2();
}
catch(std::exception const & e)
{
std::cout << e.what() << std::endl;
}
}
return 0;
}