ANSWERS TO QUIZ 1
	      ---------------------


Question 1
----------


The difference is that one of the variable is defined type "extern". extern
exposes variables in one implementation file to a different implementation file and hence can be considered as a global declaration of the variable.


Question 2
----------

a. valid
b. valid
c. invalid : correct declaration : string Namespace;
d. invalid : correct declaration : string catch_22;
e. invalid : correct declaration : char one_or_2;
f. valid


Question 3
---------

pi is a null pointer and in foobar you are trying to dereference a null pointer which is an invalid operation.

simple fix :

int foobar(int *pi)
{
   if(pi != 0)
    {
	*pi = 1024;
	return (*pi);
    }
   return -1; // you can return anything you want

}



Question 4
------------

You are returning the address of a stack allocated variable (b). After function foo returns, the memory for b is deallocated from the stack, hence you are actually returning the address of something which does not exist in memory.



Question 5
----------
One of the several right answers

int *sort(int *array, int n)
{

 for(int i=0;i<n;i++)
 {
   for(int j=i+1;j<n;j++)
	{
	   if(array[i] > array[j])
	     {
	       swap(&array[i],&array[j]);
	     }
	}
 }

 return array;
}

void swap(int *a, int *b)
{
  int temp = *a;
  *a = *b;
  *b = temp;
}


Question 6
----------

One of the many right answers.

#include <iostream>


using namespace std;
int main()
{
  const int MAX = 255;
  char line[MAX];
  int a_cnt= 0, e_cnt = 0, i_cnt = 0, o_cnt = 0, u_cnt = 0; 

  memset(line,0,MAX);

  cin.getline(line, MAX);

  for(int i=0;i<MAX;i++)
  {
	char temp = tolower(line[i]);
	switch(temp)
	 {
	   case 'a': 
		a_cnt++;
		break;

	   case 'e': 
		e_cnt++;
		break;
	   
	   case 'i': 
		i_cnt++;
		break;

	   case 'o':
		o_cnt++;
		break;

	 case 'u': 
		u_cnt++;
		break;
	 }
  }

 cout<<" "<<a_cnt<<" "<<e_cnt<<" "<<i_cnt<<" "<<o_cnt<<" "<<u_cnt<<endl;
}