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.