'auto' and 'register' variables

One of the questions that is still being asked in the interviews is about 'auto variables' and 'register variables'. I have never come across any practical code that uses them and dont even see any need for them being used but as I say, Interview questions dont reflect how good a coder would be and even if a person answers all questions correctly, how can we judge someone's debugging skills.

Anyway, 'auto' variables are the default local variables. When you define a local variable, it is by default 'auto'. There is no need to use it as its implicit. In the new C++ standard, C++0x, 'auto' serves another purpose and may break any C++ code using auto when compiled using C++0x.

The 'register' keyword was initially used to help the compiler optimise the performance for the variables that were read/written very often. Since the compilers have evolved, nowadays they choose to ignore the requests as they can optimize the performance better than us humans.

The following is from StackOverflow discussion:

Here's an excerpt from Section 6.7.1 (footnote 101) of the C99 standard (pdf):

The implementation may treat any register declaration simply as an auto declaration. However,whether or not addressable storage is actually used, the address of any part of an object declared with storage-class specifier register cannot be computed, either explicitly (by use of the unary & operator as discussed in 6.5.3.2) or implicitly (by converting an array name to a pointer as discussed in 6.3.2.1). Thus, the only operator that can be applied to an array declared with storage-class specifier register is sizeof.

And from Section 7.1.1, Paragraph 3 of the C++ standard (pdf):

A register specifier has the same semantics as an auto specifier together with a hint to the implementation that the object so declared will be heavily used. [Note: the hint can be ignored and in most implementations it will be ignored if the address of the object is taken. —end note]

So, this seems like another case of C and C++ having "identical" features that behave the way you'd expect them most of the time, but diverge and cause confusion other times. In this situation, I think the way C does it makes sense since it

• is semantically closest to being correct and;
• doesn't do things behind your back.

Fun tidbits about register

The C++ group (WG21) wants to deprecate register:

The register keyword serves very little function, offering no more than a hint that a note says is typically ignored. It should be deprecated in this version of the standard, freeing the reserved name up for use in a future standard, much like auto has been re-used this time around for being similarly useless.

Notes from the March, 2009 meeting:

The consensus of the CWG was in favor of deprecating register.

Look what the C99 group (WG14) said about register (pdf) at a meeting:

General agreement to deprecate the “auto” keyword. Should we ask WG21 to go back to the previous use of “register” (no address)? No, this will not fly with WG21.

Sample program as follows:

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

using namespace std;

int main()
{
 int i = 123;
 auto int j = 456;
 register int k = 789;

 cout<<"Address of i = " << &i <<", Value = " << i << endl;
 cout<<"Address of j = " << &j <<", Value = " << j << endl;
 cout<<"Address of k = " << &k <<", Value = " << k << endl;

 return 0;
}

Output as follows:

Note that the address of i, j and k are very close implying that the compiler chose to ignore the 'register' keyword.

An Challenging Interview question with a difference

The following program is provided:

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

using namespace std;

//This class can be designed as you wish
class SR
{
public:
SR(int i) {}
};

int main()
{
int sum = 0;
for (int i = 1; i < 100; i++)
{
  SR ii(i);
  while (i--)
    sum += i;
}
cout<<"Expected value of sum = 161700" << endl;
cout<<"Returned value of sum = " << sum << endl;

return 0;
}

As already mentioned above, the expected output of 161700 is provided. The class 'SR' could be written as one wishes. How to do write the class 'SR' to get the desired output?

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Give it a try before looking at the solution.

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The modified program with the correct 'SR' class as follows:

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

using namespace std;

//This class can be designed as you wish
class SR
{
public:
 SR(int& i):ref(i)
 {
   var = i;
 }
 ~SR()
 {
   ref = var;
 }
private:
 int var;
 int &ref;
};

int main()
{
 int sum = 0;
 for (int i = 1; i < 100; i++)
 {
   SR ii(i);
   while (i--)
     sum += i;
 }
 cout<<"Expected value of sum = 161700" << endl;
 cout<<"Returned value of sum = " << sum << endl;

 return 0;
}

Source: Modified from here. Another similar question is available here.

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:

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.

Swap two variables without using third and in one line

Couple of weeks back I was interviewing a fresh graduate. Even though they are taught programming, I am not sure if they take it seriously and learn or practice it well. One of the questions I asked was to swap 2 numbers without using a temp variable.

Looking back now, I think it may be a bigger challenge to ask to swap numbers without using a temp variable and in one line. Below are my three different approaches but I would advise you to try it yourself before looking at the answer.

//Program to swap 2 numbers without using 3rd variable and in one line
//Program tested on Microsoft Visual Studio 2008 - Zahid Ghadialy
#include<iostream>

using namespace std;

void approach1(int& a, int& b)
{
  cout<<"\nApproach 1"<<endl;
  a^=b^=a^=b;
}

void approach2(int& a, int& b)
{
  cout<<"\nApproach 2"<<endl;
  //b=(a+b)-(a=b); - This should work but doesnt, why?
  a =((a = a + b) - (b = a - b)); 
}

void approach3(int& a, int& b)
{
  cout<<"\nApproach 3"<<endl;
  a = ((a = a * b) / (b = a / b));
}



int main()
{
  int a = 13, b = 29;
  cout<<"\nOriginal"<<endl;
  cout<<"a = "<<a<<", b = "<<b<<endl;

  approach1(a, b);
  cout<<"a = "<<a<<", b = "<<b<<endl;

  a = 13, b = 29;
  approach2(a, b);
  cout<<"a = "<<a<<", b = "<<b<<endl;

  a = 13, b = 29;
  approach3(a, b);
  cout<<"a = "<<a<<", b = "<<b<<endl;

  return 0;
}

The output is as follows:

Agreed that the above would be applicable only for integers.

'sizeof' a class

Continuation from last week. What is the sizeof of a class? Wrote a simple program as follows:

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

using namespace std;

class A
{
  int a;
};

class B
{
  int b;
  int someFunc(void) {return b;}
};

class C
{
  void someFunc1(void) {};
  void someFunc2(void) {};
};

class D : public A
{
  void someFunc(void) {};
};


int main()
{
  cout<<"Size of class A = "<<sizeof(A)<<endl;
  cout<<"Size of class B = "<<sizeof(B)<<endl;
  cout<<"Size of class C = "<<sizeof(C)<<endl;
  cout<<"Size of class D = "<<sizeof(D)<<endl;
}

The output is as follows:

After finishing this, I found this interesting article here.

Checking the 'sizeof' doubts

I noticed sometime back that one of my programs behaved unexpectedly in certain scenarios which was traced to an incorrect use of sizeof. As a result, I made a small program to make sure that my understanding of sizeof is correct. Here is the program:

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

using namespace std;

int main()
{
 char a[]="";
 cout<<"Size of a[] = "<<sizeof(a)<<endl;

 char b[]=" ";
 cout<<"Size of b[ ] = "<<sizeof(b)<<endl;

 char c[10];
 cout<<"Size of c[10] = "<<sizeof(c)<<endl;

 char* d;
 cout<<"Size of d = "<<sizeof(d)<<endl;

 char* e = c;
 cout<<"Size of e = "<<sizeof(e)<<endl;

 char f[] = "123456";
 cout<<"Size of f = "<<sizeof(f)<<endl;

 char g[] = {'1','2','3','4','5','6'};
 cout<<"Size of g = "<<sizeof(g)<<endl;

 return 0;
}

The output is as follows:

A lot of interviewers love the sizeof questions.

Have you come across any interesting sizeof problems? Please feel free to add them in comments.

String Shift and Print

After my last blog post, someone pointed out a similar problem here. The problem goes:

Write a program that generates from the string "abcdefghijklmnopqrstuvwxyz{" the following:

a

bcb

cdedc

defgfed

efghihgfe

fghijkjihgf

ghijklmlkjihg

hijklmnonmlkjih

ijklmnopqponmlkji

jklmnopqrsrqponmlkj

klmnopqrstutsrqponmlk

lmnopqrstuvwvutsrqponml

mnopqrstuvwxyxwvutsrqponm

nopqrstuvwxyz{zyxwvutsrqpon

Well there is no time limit this time and we dont even need to use google. I think this is slightly trickier than the last one.

Here is my solution:

//Program tested on Microsoft Visual Studio 2008 - Zahid Ghadialy

#include<iostream>
#include<string>

using namespace std;

int main()
{
  string str = "abcdefghijklmnopqrstuvwxyz{";
  unsigned strlen = str.length();
  string tempStr;
  int loop = 0;
  while(tempStr.length() < strlen)
  {
    int start1 = loop;
    int count1 = loop + 1;
    tempStr = str.substr(start1,count1);

    count1--;
    for(; count1 > 0; count1--)
      tempStr.append(str.substr(start1+count1-1, 1));

    cout<<tempStr;
    cout<<endl;
    loop++;
  }
  return 0;
}

ASCII Print

Few weeks back, someone I know got asked the following question in an interview:

Write a C++ program in 15 minutes to print the following in the output:

a

bcb

cdedc

defgfed

efghihgfe

fghijkjihgf

ghijklmlkjihg

hijklmnonmlkjih

ijklmnopqponmlkji

jklmnopqrsrqponmlkj

klmnopqrstutsrqponmlk

lmnopqrstuvwvutsrqponml

mnopqrstuvwxyxwvutsrqponm

nopqrstuvwxyz{zyxwvutsrqpon

You can do search on the web for any information but all your searches would be logged and used to assess your answer.

Well my obvious guess would be to take help of ASCII table. My only search was for Ascii table in google. I did try and make it in less than 15 mins. My solution is below but do try it on your own before looking at my solution.

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

using namespace std;

int main()
{
  char c=(char)97; //'a' in ASCII
  char loop = 0;
  while(c <= 'n')
  {
    char x = 0;
    for(; x <=loop; x++)
    {
      cout<<char(c+x);
    }
    for(char y = 1; y < x; y++)
    {
      cout<<char(c+x-y-1);
    }
    cout<<endl;
    c++;
    loop++;
  }

  return 0;
}

Difference between Deep copy and Shallow copy in C++

So what is the difference between 'Deep copy' (sometimes referred to as 'Hard copy') and 'Shallow copy' in C++

Shallow copy: Copies the member values from one object into another.

Deep Copy: Copies the member values from one object into another. Any pointer objects are duplicated and Deep Copied.

There are some interesting discussions on Stack Overflow here and here. A related discussions here is interesting as well.

Program to demonstrate as follows:

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

using namespace std;

class MyString
{
public:
  MyString(){}
  MyString(string str)
  {
    size = str.size();
    data = new char(size+1); //+1 for '\0'
    memcpy(data, str.c_str(), size+1);
  }
  //MyString(const MyString& copy); - Use default for Shallow
  void DeepCopy(const MyString& copy) //Deep Copy
  {
    size = copy.size;
    data = new char(size+1); //+1 for '\0'
    memcpy(data, copy.data, size+1);
  }
  int   size;
  char* data;
};

int main()
{
  MyString m1("Zahid");
  cout<<"\nm1.size = "<<m1.size<<", &m1.data = "<<(void *)m1.data<<", *m1.data = "<<m1.data<<endl;

  MyString m2(m1); //Uses the default copy constructor - Shallow Copy
  cout<<"\nm2.size = "<<m2.size<<", &m2.data = "<<(void *)m2.data<<", *m2.data = "<<m2.data<<endl;

  MyString m3; //Another default constructor
  m3.DeepCopy(m1);
  cout<<"\nm3.size = "<<m3.size<<", &m3.data = "<<(void *)m3.data<<", *m3.data = "<<m3.data<<endl;

  return 0;
}

Output as follows:

Difference between 'new', 'new operator' and 'operator new' in C++

A question asked in a forum said:

Why does void* p = new (1024); gives me a compilation error while void* p = operator new (1024); works. What is the difference between new and "operator new"?

Lets go back to the beginning. Once upon a time there was this C language that used 'malloc' to allocate memory. Then when C++ came, a new way was defined of allocating memory by the use of 'new'. Its supposed to be much safter and better way but some software gurus may differ. Memory for 'new' is allocated from 'Free Store' and memory by 'malloc' is generated from 'heap'. The 'heap' and 'Free Store' may be the same area and is a compiler implementation detail.

The syntax for 'operator new' is:

void* new (std::size_t size);

All it does is allocate a memory of size specified

On the other hand, 'new' does 2 things:

1. It calls 'operator new'

2. It calls constructor for the type of object

In the above case since 1024 is not a type, it will fail in calling its constructor.

'new' is also referred to as 'keyword new' or 'new operator' to cause more confusion :)

Here is a small example to play with 'operator new' after the example you will realise that regardless of what is passed, it always allocates a 4 byte memory.

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

using namespace std;

int main()
{
 //void* p = new (1024);

 void* p = operator new (1024);

 //Lets test if we can do anything with the memory
  cout<<"p = "<<p<<endl;
 cout<<"sizeof(p) = "<<sizeof(p)<<endl;
 int *x = static_cast<int *>(p);
 cout<<"*x before = "<<*x<<endl;
 *x = 23456;
 cout<<"*x after = "<<*x<<endl;

 void* q = operator new (0);
 cout<<"\nq = "<<q<<endl;
 cout<<"sizeof(q) = "<<sizeof(q)<<endl;
 x = static_cast<int *>(q);
 cout<<"*x before = "<<*x<<endl;
 *x = 65432;
 cout<<"*x after = "<<*x<<endl;

 void* z = operator new ('a');
 cout<<"\nz = "<<z<<endl;
 cout<<"sizeof(z) = "<<sizeof(z)<<endl;
 x = static_cast<int *>(z);
 cout<<"*x before = "<<*x<<endl;
 *x = 11111;
 cout<<"*x after = "<<*x<<endl;

 return 0;
}

The output is as follows:

My knowledge in this area is quite limited so please feel free to improve on my explanation or correct it.

Challenging problem to test your 'for loop' basics

While scouting through the web I came across this fun puzzle:

In the following code:

#include<iostream>

using namespace std;

int main()
{
 int i, n = 20;
 for (i=0; i<n; i--)
 {
   cout << "x" << endl;
 }

 return 0;
}

by changing only ONE character in the above code, meaning you cannot change 20 to 31, because you will have changed two characters, you can change 20 to 21, because you only changed the 0, do the following:

find 3 ways to make the above code print x 20 times (by changing only one character).

Give it a try, if you can do it then look at the answer below.

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The first one should be simple:

i-- becomes n--

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I guess this is the second one you got

i<n becomes i+n

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The third one which is tricky, In fact there are two similar solutions is:

i<n becomes, -i<n,
i<n becomes, ~i<n

remember BODMAS

Class initialisation and constructors

There is often lots of confusion with regards to class initialisation. People may forget to take care while using default constructors and copy constructors. Here is a simple example that tries to explain lots of concepts.

//Program tested on Microsoft Visual Studio 2008 - Zahid Ghadialy
//This program gives an example of constructors and initialisations

#include<iostream>

using namespace std;

//Example of a class
class someClass1
{
public:
  //implicit default constructor
  int x;
  int *y;
};

//Example of a class with constructor
class someClass2
{
public:
  someClass2(const someClass2& xyz)
  {
    cout<<"** copy constructor called"<<endl;
    x = xyz.x;
    y = new int(*xyz.y);
  }
  //default constructor will have to be explicitly defined
  someClass2() {};
  //overloading operator '='
  const someClass2& operator = (const someClass2& xyz)
  {
    cout<<"** operator '=' called"<<endl;
    x = xyz.x;
    y = new int(*xyz.y);
    return *this;
  }
  int x;
  int *y;
};

int main()
{
  someClass1 a; //Default initialisation
  a.x = 1234;
  a.y = new int(6789);

  cout<<"someClass1: a.x = "<<a.x<<"       a.y( " <<a.y<< " ) = "<<*(a.y)<<endl;

  someClass1 b = a; //Copy Initialisation

  cout<<"someClass1: b.x = "<<b.x<<"       b.y( " <<b.y<< " ) = "<<*(b.y)<<endl;

  someClass1 c(a); //Direct Initialisation

  cout<<"someClass1: c.x = "<<c.x<<"       c.y( " <<c.y<< " ) = "<<*(c.y)<<endl;

  //Calling default constructor
  someClass2 aa;
  aa.x = 2468;
  aa.y = new int(3579);

  cout<<"someClass2: aa.x = "<<aa.x<<"      aa.y( " <<aa.y<< " ) = "<<*(aa.y)<<endl;

  //calling copy constructor
  someClass2 bb = aa; //Copy Initialisation - note copy constructor will be called

  cout<<"someClass2: bb.x = "<<bb.x<<"      bb.y( " <<bb.y<< " ) = "<<*(bb.y)<<endl;

  //calling copy constructor
  someClass2 cc(aa); //Direct Initialisation - note copy constructor called in this case as well

  cout<<"someClass2: cc.x = "<<cc.x<<"      cc.y( " <<cc.y<< " ) = "<<*(cc.y)<<endl;

  someClass2 dd;
  //calling operator =
  dd = aa;

  cout<<"someClass2: dd.x = "<<dd.x<<"      dd.y( " <<dd.y<< " ) = "<<*(dd.y)<<endl;

  return 0;
}

The output is as follows:

There are certain things worth noting in the above example:

• In case of someClass1, since only the default constructor is used, the same pointer is used for y in all the cases. This can cause serious problems in the code if one of the classes delete the memory pointed by y for class someClass1. This problem is not present in someClass2
• If any constructor is defined, it becomes necessary to define the default constructor. You can make sure that nobody uses default constructor in case of someClass2 by making it private (I havent done that because I wanted to show operator =)
• As you can see in case of variable cc, copy constructor would always be called instead of operator =.
• operator = would only be called in case of assignment. This is a common mistake.
• In the code above, cc(aa) is better than using bb = aa even though the results are the same. The main advantage being that it will avoid confusion with operator '=' if its overloaded for a novice and the constructor can be overloaded to take more than one input in future.