CSCI 15A                                               Jim Murphy

We have seen how to check for certain conditions
using the if, if ... else, and switch statements but
to repeat some process until we get it correct as in
giving name and PIN at the ATM machine we need
looping - ability to repeat part of our code some
number of times
For this we can use the "while loop" as:
while (condition)
{
iterative part
}
For example we could use the bank class and ATM:
#include "ouratm.h"
#include "ourbank.h"
int main( )
{
bank BofCS;
ATM myatm;
int custNum, found = FALSE;
while (!found)
{
  myatm.getNameAndPIN( );
  BofCS.findCustomer(myatm, custNum, found);
  if (!found) 
     myatm.message
        ("Not a valid name and PIN try again.");
}
It is possible that you may know ahead of time how
many times you want to execute a loop - or the end
user may know and could be asked as in: p241.cpp
// Using this while loop averaging method, the user
// must know in advance, the number of input values
// that are to be averaged
#include <iostream.h>
int main()
{
  int number;
  // Initialize sum, counter, and n:
  float sum = 0;
  int counter = 1;
  int n;
  cout << "Enter number of entries to be averaged: ";
  cin >> n;
  // While there are more numbers to be averaged
  while(counter <= n)
  {
    // Execute this block n times:
    cout << "Enter number “<<counter << “: “;  
    // <-Repeat these
    cin >> number;             // <-   three
    sum = sum + number;        // <-statements
    // counter holds the number of loop repetitions.
    // counter is incremented n times. 
    // When counter exceeds n
    // (counter <= n is false), the loop is terminated.
    counter = counter + 1;
  }
  float average = sum / float(n);
       100.0*isum/n
  cout << endl
       << "     Average: " << average << endl;
  return 0;
}
make p241
then run with:
 p241
Enter number of entries to be averaged: 4
Enter number: 60
Enter number: 65
Enter number: 70
Enter number: 75

     Average: 67.5

If the while condition is true then the iterative part
is executed once then the condition is evaluated again
This keeps up until the condition is false at which
point the iterative part is not executed
int x = 0;
while (x < 10)
  {
   cout << x << endl;
   x = x + 1;
   }
cout << x << endl;
while (*(x++)=*(y++));// ok to copy until null character
   {
    cout << “ oh my “ << endl;
    x = x + 1;
    }

LESSON 10

A counter controlled loop can be used if the number
of iterations is known at program writing as in:

src/p243.cpp
#include <iostream.h>
int main()
{
  // Iterate a loop n times
  int n = 4;
  int counter = 1;
  while(counter <= n)
  {
    // The repeated process outputs the changing 
    // value of counter:
    cout << "Iteration# " << counter << endl;
    //
    // The counter is incremented to eventually
    //  terminate the loop:
    counter = counter + 1;
  }
  cout << "After loop, counter = " << counter << endl;
  return 0;
}
p243
Iteration# 1
Iteration# 2
Iteration# 3
Iteration# 4
After loop, counter = 5

Another example adds up the first n integers where
n is set to 5 in the example from:
src/p244.cpp
// An example of a counter-controlled while loop
#include <iostream.h>
int main()
{
  // Initialize the objects used in the while loop
  int n = 5;
  int accumulator = 0;
  int counter = 1;
  // The counter controlled while loop
  while(counter <= n)
  {
    accumulator = accumulator + counter;
    counter = counter + 1;
  }
  cout << "Sum of the first " << n << " integers is " 
     << accumulator;
  return 0;
}

p244
Sum of the first 5 integers is 15pitbull 24>

Note the missing endl on the last cout  statement

Watch out for missing brackets which is allowed but
may not be what was intended as in:
int counter = 1;
while (counter <=20)
    cout << counter << endl;
counter = counter + 1; // never executed

The case study on pages 248 to 253 considers the
problem of finding the range of a large collection
of numbers - for a small set of numbers it is easy for
us to pick out the largest and smallest numbers 
present visually - for a large set of numbers we need
an algorithm

Take the first number on the list as the only contender
and consider it to be both the largest and the smallest
Certainly true if there was only this one number on
the list - then as each new number is added to the list
compare it to the largest-so-far and the smallest-so-far
adjusting those values if necessary these int objects
are actually called highestTest and lowestTest in the
program on page 252 This program will find the 
largest and smallest value in a collection of numbers
and then compute the range to be (largest - smallest)

src/p252.cpp
#include <iostream.h>
int main()
{
  int test, highestTest, lowestTest;
  int testsToCheck;
  cout << "Enter number of test scores: ";
  cin >> testsToCheck;
  cout<<endl << "Enter test scores 1 per line" << endl;
  // Input first test to record it as highest and lowest
  cin >> test;
  highestTest = test;
  lowestTest = test;
  // We have processed one check, so start counter at 2.
  int counter = 2;
  while(counter <= testsToCheck)
  {
    cin >> test;
    if(test > highestTest)
      highestTest = test;
    else if(test < lowestTest)
      lowestTest = test;
    counter = counter + 1;
  }
  int range = highestTest - lowestTest;
  cout << "Range: " << range;
  return 0;
}

It may be more convenient to let the computer figure
out how many data entries there will be rather than
asking the user for that number - The machine is good
at counting if it just had some way of knowing when
we were done entering numbers - sometimes it is 
possible to use a special data value called a "sentinel"
or signal - if there is a value which is guaranteed not
to be normal data then we could use that as is done
in the program on page 256 which computes the 
average of a collection of positive integers using a
sentinel value of minus one

src/p256.cpp watches for a special data value to stop
// Use a sentinel of -1 to terminate a loop.
// A sentinel is a value that cannot be within the range
// of valid data. It is the final input indicating there
// is no more data. The extraction of this
// sentinel value is the event that terminates the loop.
#include <iostream.h>
int main()
{
  // PRE:  The user enters at least one valid int != -1
  const int sentinel = -1;
  int testScore;
  float accumulator = 0;
  int n = 0;  // Incremented for every input != -1
  // Display a prompt
  cout << "Enter test scores or " << sentinel
       << " to quit" << endl;
  // This next statement is used to obtain 
  // a value for the loop-test
  cin >> testScore;
  while(testScore != sentinel)
  {
    accumulator = accumulator
                  + testScore;  // Update the accumulator
    n = n + 1;                  // Update total inputs
    cin >> testScore;     // Input next test or the sentinel
  }
  cout << endl
       << "Average of " << n << " tests = "
       << (accumulator / n)
       << endl;
  return 0;
}
make p256

then execute with:
 p256
Enter test scores or -1 to quit
50
55
60
65
70
-1

Average of 5 tests = 60
Note the order of the statements in and around the
while loop - the existence of the two cin statements
and where the counter and accumulator are incremented
What would happen if we entered -1 at the first cin?

cin >> testScore ; this returns true if successful and
false (0) if it fails so we could use:
while ( (cin >> testScore)&&(testScore!=sentinel))
Note what happens if you type in a letter like "q"
which is not a valid int
Any input asked for once cin gets into a bad state
will fail unless the input stream is reset with the
flush(cin); function from ourstuff.h shown next:

src/p260.cpp
// Demonstrate what happens when an input stream 
// is in an error state
#include <iostream.h>
#include "ourstuff.h"  // for flush(cin)

int main()
{
  double testScore;
  double sum = 0;
  double n = 0;

  cout << "Enter test scores or -1 to quit" << endl;

  while( ((cin >> testScore)!= 0) && (testScore != -1) )
  {
    sum = sum + testScore;
    n = n + 1;
  }
  cout << "Ave: " << (sum / n) << endl;

  // Demonstrate flush(cin) to allow further input
  // ...
  flush(cin);
  cout << "Enter another testScore: ";
  cin >> testScore;
  cout << "testScore: " << testScore << endl;
  cout << "and one more: ";
  cin >> testScore;
  cout << "testScore: " << testScore << endl;
  return 0;
}
p260
Enter test scores or -1 to quit
50
55
60
-1
Ave: 55
Enter another testScore: 40
testScore: 40
and one more: 55
testScore: 55
When using input from a data file we can use the
eof or end-of-file function to terminate our entry loop
We input several one word strings from a text file

cat src/p262.cpp
#include <fstream.h>
#include <iostream.h>
#include "ourstr.h"

int main()
{
  ifstream inFile("input.dat");
  cout << "Before loop, eof = " << inFile.eof() << endl;
  string s;
  while(! inFile.eof())
  {
    inFile >> s;
    s.toUpper();
    cout << "s = " << s << endl;
  }
  cout << " After loop, eof = " << inFile.eof() << endl;
  return 0;
}

cat src/input.dat
Several strings
on two lines.pitbull 39>

p262
Before loop, eof = 0
s = SEVERAL
s = STRINGS
s = ON
s = TWO
s = LINES.
 After loop, eof = 1

LESSON 11

But look out if the input file is not in the current
directory - this will cause an error also and an infinite
loop - use control C to interrupt the program in this
case - from home control S and control Q will start
and stop output but not the program

To use eof with keyboard input such as cin you will
need to enter control D on a unix system but control Z
on DOS - you will also need to use flush(cin) after
the eof is encountered as it will mess up subsequent
keyboard input