A generic sort program with 'functors' and 'templates'

Picked up this question from here and made a program out of it. It may be a good idea to quickly brush functors and templates if required.

The program sorts the input Vector provided regardless of the type.

//Program tested on Microsoft Visual Studio 2008 - Zahid Ghadialy
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

class Int
{
public:
  Int() {x_ = 0;}
  Int(const int &x) {x_ = x;}
  int getID (void) {return x_;}
  int x_;
};

class Str
{
public:
  Str() {x_ = "";}
  Str(const string &x) {x_ = x;}
  string getID (void) {return x_;}
  string x_;
};

template <typename Object> class Comparator {
  public:
    bool operator()(const Object &o1, const Object &o2) const 
    {
      return (const_cast<Object&>(o1).getID() < const_cast<Object&>(o2).getID());
    }

    bool operator()(const Object *o1, const Object *o2) const {
      return (o1->getID() < o2->getID());
    }
};

template <typename VecObject> void Display(VecObject v)
{
  VecObject::iterator it;
  for(it = v.begin(); it != v.end(); ++it)
  {
    cout<<it->getID()<<", ";
  }
  cout<<endl;
}

int main()
{
  vector<Int> objects1;
  objects1.push_back(Int(3));
  objects1.push_back(Int());
  objects1.push_back(Int(77));

  //print the output
  cout<<"objects1 before sort = ";
  Display(objects1);
  std::sort(objects1.begin(), objects1.end(), Comparator<Int> ());
  cout<<"objects1 after sort  = ";
  Display(objects1);

  std::vector<Str> objects2;
  objects2.push_back(Str("Hello Hello"));
  objects2.push_back(Str("apple?"));
  objects2.push_back(Str());
  objects2.push_back(Str("1 Jump"));

  //print the output
  cout<<"objects2 before sort = ";
  Display(objects2);
  std::sort(objects2.begin(), objects2.end(), Comparator<Str> ());
  cout<<"objects2 after sort  = ";
  Display(objects2);

  return 0;
}

The output is as follows:

Debugging Mutex and Locks

When there are multiple Mutex's in the program, it may be required to find if a particular thread is locked far longer than necessary. This can cause problems and the output may not be what is expected.

To get round this, I took an old example from here and modified it to help me print some debugging info.

I modified the printSomething() in Singleton.h to add a sleep as follows:

  void printSomething(char *name, int count)
 {
   Lock guard(mutex_);
   Sleep(10);
   Lock guard2(mutex_);
   std::cout << name << " loop " << count << std::endl;
 }

and I modified the Mutex.h as follows:

//Example from http://www.relisoft.com/Win32/active.html
#if !defined _MUTEX_H_
#define _MUTEX_H_

class Mutex
{
   friend class Lock;
public:
   Mutex () { InitializeCriticalSection (& _critSection); }
   ~Mutex () { DeleteCriticalSection (& _critSection); }
private:
   void Acquire ()
   {
     DWORD start = GetTickCount();
     EnterCriticalSection (& _critSection);
     DWORD elapsed = GetTickCount() - start;
     if (elapsed > 15)
     {
       //Debugging Info
        std::cout<<"Debugging Info: Waited at mutex for "<<elapsed<<std::endl;
     }
   }
   void Release ()
   {
       LeaveCriticalSection (& _critSection);
   }

   CRITICAL_SECTION _critSection;
};

#endif 

The output is as follows:

The small problem with the above approach is that if a Thread is deadlocked, we may not get the debug output as we would have to kill the process.

Add two unsigned integers without using '+'

I found this very interesting discussion (and the program) to add two numbers without the use of the arithmetic operator '+'. The obvious guess would be to use the Bitwise operators.

The logic behind the addition operation using the bitwise operators is as follows:

integer1 =  3  = 0011b
integer2 =  5  = 0101b

    first operation        second operation      third operation
        0011                0011                 shift by one
        0101                0101                   
       ______              ______
 XOR    0110           AND  0001  ------>        <<1   0010

    Again...  
                   first operation   second operation   third operation
previous XOR        0110                0110            shift by one
previous <<1        0010                0010
                   ______              ______
               XOR  0100            AND 0010  ------>  <<1   0100

    Again...  
                   first operation  second operation    third operation
previous XOR        0100                0100            shift by one
previous <<1        0100                0100
                   ______              ______
                XOR 0000           AND  0100  ------>  <<1   1000

    Again...  
                   first operation   second operation   
previous XOR        0000               0000            
previous <<1        1000               1000
                    ______            ______
                XOR 1000          AND  0000  
When AND iguals 0 the result of XOR is iqual to the sum of the two integers

The program is as follows:

//Program to add two numbers by using boolean operators
//Ref: http://www.daniweb.com/forums/thread84950.html
//Program tested on Microsoft Visual Studio 2008 - Zahid Ghadialy
#include<iostream>

using namespace std;

unsigned long add(unsigned integer1, unsigned integer2)
{
  unsigned long xor, and, temp;

  and = integer1 & integer2; /* Obtain the carry bits */
  xor = integer1 ^ integer2; /* resulting bits */
    
  while(and != 0 ) /* stop when carry bits are gone */
  {
    and <<= 1; /* shifting the carry bits one space */
    temp = xor ^ and; /* hold the new xor result bits*/
    and &= xor; /* clear the previous carry bits and assign the new carry bits */
    xor = temp; /* resulting bits */
  }
  return xor; /* final result */
}

int main()
{
  int num1 = 13, num2 = 27;

  cout << num1 << " + " << num2 << " = " <<add(num1, num2) << endl;

  return 0;
}

The output is:

13 + 27 = 40

See the original discussion here.

Consequences of Ignoring Compiler Warnings

Though this is a fictional program that I have picked up from here, I have seen similar problems in real life.

Lets say my program (below) was expected to return this

but instead returned:

Program as follows:

//Program tested on Microsoft Visual Studio 2008 - Zahid Ghadialy
#include<iostream>

using namespace std;

class Bclass
{
public:
 Bclass() {}
 virtual void func()
 {
   cout<<"In Bclass::func()"<<endl;
 }
};

class Dclass : public Bclass
{
public:
 Dclass() {}
 virtual void func()
 {
   cout<<"In Dclass::func()"<<endl;
   Bclass:func();
 }
};

int main()
{
 Dclass d;
 d.func();
 //...
  return 0;
}

The clue of the problem was given by the compiler that generated the following warning:

main.cpp(23) : warning C4102: 'Bclass' : unreferenced label

pointing to the line

Bclass:func();

The problem lies in the fact that due to a single : rather than :: the line above is being treated as label and since its an unreferenced label it goes back to the start of the same function.

You can read the complete discussion here. Make sure to check the warnings the next time your program behaves unexpectedly.