Problem 2
---------
Write a class called person with a data field for first name,
a data field for last name, and a data field indicating if the
person is married or not.

Write a member function called get_married which takes another 
person as input.  The function should give this other person 
the same last name as the person used to call the function, 
and change the marital status of both persons.

As usual the class should contain the function prototype.

Answer 2
--------
class person {
    private:
	char fname[20];
	char lname[20];
	bool married;
    private:
	void get_married(person p_other);
};

void person::get_married(person p_other)
{
    strcpy(p_other.lname,lname);
    p_other.married = true;
    married = true;
}


Problem 3
---------
Write a function called same_nums which takes two input streams
as input.  The function assumes that those input streams are
each already associated to a file of integers.  The function 
should return true if the files are the same (meaning that they
contain the same integers in the same order).  Otherwise,
return false.

Answer 3
--------
bool same_nums(istream &in1, istream &in2)
{
    int num1.num2;

    in1 >> num1;
    in2 >> num2;

    while(!in1.eof() && !in2.eof())
      {
	if (num1 != num2)
	    return false;
	in1 >> num1;
	in2 >> num2;
      }

    if (in1.eof() && in2.eof())
   	return true;
    else
 	return false;
}

Problem 4
---------
Write a function called c_follows_b that takes a string as input 
and returns the number of times that b is immediately followed
by c in the string.

Example
-------
Suppose the string is:	abcbabbcabcbc
Then the function will return 4.

Answer 4
--------
int c_follows_b(char s[])
{
    int count = 0;
    int i;

    for (i=1; s[i] != '\0'; i++)
	if (s[i-1] == 'b' && s[i] == 'c')
	    count++;

    return count;
}