Solution:

#include<iostream.h>
void main() { 
   cout << "Hello World!" ; 
}

    f = 32 + (9/5) c


Solution:

#include<iostream.h>
void main() { 
   double c ;
   cin >> c ;
   cout << ( 32.0 + (9.0/5.0) * c ) << endl ;
}




Here is a C++ program which prints the letters a through z
each on a single line:
   #include<iostream.h>
   void main() 
   {
      char c ;
      for (  c='a'; c<='z'; c++ )
      {
         cout << c << endl ;
      }
   }

Write a C++ program which prints  all the two letter
combinations aa, ab, ac, etc., through zz each on a single
line.


Solution:
#include<iostream.h>

void main() {

   char c ;
   for (  c='a'; c<='z'; c++ ) {

      char d ;
      for (  d='a'; d<='z'; d++ )  {
         cout << c << d << endl ;
      }
   }
}




Write a C++ function called rvrStr
which takes a string argument and reverses the order of the 
letters in the string. Example:
#include<iostream.h>
#include<strings.h>

void rvrStr(char * s ) {
   // you write the code here
}

void main() {
   char * s =  "able was i" ;
   rvrStr(s) ;
   cout << s << endl ;
}

prints out i saw elba.


Solution:
This was presented in class as program
SereverSenil.C
and available with the usual class materials.


Here is another possibility:
void rvrStr(char * s ) {
   char * p ;
   p = s + strlen(s) - 1 ;
   while (p>s) {
      char t ;
      t = *s ;
      *s = *p ;
      *p = t ;
      p-- ;
      s++ ;
   }
}




Write a C++ function which takes three arguments:
an integer value x, an integer array A[]
and the array size n, and returns either:

   
 An integer i for which A[i]==x,
   
 or -1 in the case that the value x is not
         anywhere in the array A[].

The function arguments and return type are:
   int findInt( int x, int A[], int n ) {

      // body of function, you do it.

   }



Solution:
int findInt( int x, int A[], int n )
{
   int i ;
   for ( i=0; i<n; i++ ) {
      if ( x==A[i] ) return i ;
   }
   return -1 ;
}




Write a C++ function 
int sumLL(N * n)  which takes a pointer to
a linked list of integers and returns the sum of all the integers
in the linked list.
The linked list is defined using the structure:
   struct N {
      int theInt ;
      N * next ;
   }

Example. if N * p points to the three element list containing
integers 3, 4 and -1 it returns 6.


Solution:
int sumLL(N * n)
{
   int sum = 0 ;
   while ( n!=NULL ) {
      sum += n->theInt ;
      n = n->next ;
   }
   return sum ;
}





Write a C++ function:
  N * bePos(N * l) {
  // your write the function
  }

which takes a linked list of integers, using the definition
of a linked list given in the previous problem, and
returns the linked list remaining after removing all the
nodes whose theInt values are negative.


Solution:
N * bePos(N * l) 
{
   // assume dummy header 
   Node * p = l ;
   while ( p->next!=NULL ) {
      if ( p->next->theInt < 0 ) {
         // remove
         p->next = p->next->next ;
      }
      else {
         // keep
         p = p->next ;
      }
   return l ;
}