class person
  {
    private:
      char fname[20];
      char lname[20];
      int age;
    public:
  };

1.  Write a member function called close_age for the class
    person (see above), which has another person as 
    parameter.  Your function should return true if the two
    persons are with ten years of each other.  Otherwise false.

bool person::close_age(person p)
{
  if (p.age - age <= 10 && age - p.age <= 10)
    return true;
  else
    return false;
}

2.  Write a nonmember function called some_close which takes
    a person, an array of persons and the size of the array 
    as parameters.  The function returns true if that person 
    is within ten years of somebody in the array.  Otherwise false.

bool some_close(person p, person A[], int size)
{
  int i;

  for (i=0; i<size; i++)
  {  
    if (A[i].close_age(p))
      return true;
  }

  return false;
}

3.  Write a nonmember function called all_close which takes
    two arrays of persons and the sizes of the arrays as 
    parameters.  The function returns true if everybody in 
    the first array is within ten years of somebody in the 
    second array.

bool all_close(person A[], person B[], int size_A, int size_B)
{
  int i;

  for (i=0; i<size_A; i++)
  {  
    if (!some_close(A[i],B,size_B))
      return false;
  }

  return true;
}

4.  Write a member function of person called older which takes 
    an integer as parameter.  The function returns true if the
    person's age is greater than the number passed as parameter.

bool person::older(int x)
{
  if (age > x)
    return true;
  else
    return false;
} 

5.  Write a nonmember function called young_adults which takes
    two arrays of persons and the size of the first array as
    parameters.  The function should copy all the persons in the first 
    array between the ages of 18 and 21 (inclusive) into the second 
    array, and return the number of persons now in the second array.
    (Assume the second array starts out empty.)

int young_adults(person A[], person B[], int size)
{
  int i,j=0;

  for (i=0; i<size; i++)
  {
    if (A[i].older(17) && !A[i].older(21))
    {
      B[j] = A[i];
      j++
    }
  }
  
  return i;
}

6.  Write a function called mess_up that takes a two strings as  
    parameters.  The function should make the second string be
    a messed up version of the first string, such that the first
    letter is moved to the end, and then "ay" is added to the end.
    In other words, if the first string is "pig", then the second
    string will become "igpay".

void mess_up(char s[], char t[])
{
  int i;

  for (i=1; s[i]!='\0'; i++)
    t[i-1] = s[i];
  
  t[i-1] = s[0];
  t[i] = 'a';
  t[i+1] = 'y';
  t[i+2] = '\0'; 
}

7.  Write a function called pig_latin which takes two strings as
    parameters.  The function make the second string be a pig
    latin version of the first string.  If a word begins with
    a consonant, then the pig latin version is to move the first
    letter from the beginning to the end and add "ay".  If a
    word begins with a vowel, then the pig latin version is just
    to add "way" at the end.  So the pig latin version of "pig"
    is "igpay".  And the p latin version of "ink" is "inkway".
    (Vowels are 'a', 'e', 'i', 'o' and 'u'.  Don't worry about
     capital letters.)

void pig_latin(char s[], char t[])
{
  int i;

  if (s[0] != 'a' && s[0] != 'e' && s[0] != 'i'
                  && s[0] != 'o' && s[0] != 'u')
    mess_up(s,t);
  else
  {
    for (i=0; s[i] != '\0'; i++)     
      t[i] = s[i];
  
    t[i] = 'w';
    t[i+1] = 'a';
    t[i+2] = 'y';
    t[i+3] = '\0';
  }
}

8.  Write a function that takes an input files stream and an
    output file stream as parameters.  Both files are assumed 
    to be already open.  The file associated with the input
    file stream contains only words (no punctuation, no numbers,
    no capital letters).  The function should make the output
    file be a pig latin version of the input file.

void pig_latin_file(ifstream &in, ofstream &out)
{
  char s[20];
  char t[20];

  in >> s;
  while (!in.eof())
  {
    pig_latin(s,t);
    out << t;
    in >> s;
  }
}

9.  Give the output of the following program:

int main()
{
  int A[10];
  int i, sum=0;
 
  for (i=0; i<10; i++)
    A[i] = i;
  
  for (i=0; i<10; i++)
    {
      if (A[i] % 2 != 0)
        sum += A[i];
    }

  cout << "The sum is " << sum << "\n";
  return 0;
}

The sum is 25

10.  Give the output of the following program:

int main()
{
  int A[10];
  int i, sum=0;
 
  for (i=0; i<10; i++)
    A[i] = i;
  
  for (i=1; i<10; i++)
    A[i] += A[i-1];

  for (i=0; A[i] < 12; i++)
    cout << A[i] << " ";

  cout << endl;
  return 0;
}

0 1 3 6 10