1. CS1010E Programming Methodology Tutorial 1 Basic Data Type and Input/output, Characters and Problem Solving
C14,A15,D11,C08,C11,A02

2. Introduction: Myself My name: Fan Qi Year 2 PhD student
Haven’t code C for years 
Some details are forgotten
Experience in teach C for years as well 
CS1101 (c), CS1010 (c), CS1010E (c), CS3223 …
I usually reply fast... unless I’m AFK

3. Tutorials No marks for attendance or participation 
If you come, do prepare !! Or you will be lost !!
Feel free to ask!
You already paid tuition fee, I won’t charge you any more 
Time is limited: 45mins only I’ll focus on some questions and we can discuss further after class
Google is your best friend!
Most of your doubts are Googlable 

4. Question 1:

5. “\n \r \b \t” … are called Escape Sequence
Printf(“I love\\nprogramming”);
“\n \r \b \t” … are called Escape Sequence
Google them for more information 
Question: what if I want output
“I love\n programming”
into console?

6. Try input: 123, 123.3, 1150869504, can you explain ?
Google IEEE floating point standard!
It will output the integer part of the input
123->123
123.3->123
Try input: 123, , , can you explain ?
What if you change last statement to be:
printf(“Double Value = %d\n”, doubleValue);
Can you explain ?

7. Try : printf(“%d, %d\n”, x, y); What will you see?
When y = x; there is a coercion
When printf(“%d”, x); there is no coercion. Therefore you will see the output is , -2
Try :
printf(“%d, %d\n”, x, y);
What will you see?

8. Answer:
The output is In the above case, because x and y are both int, x/y gives and int value 3.
1. float z = 1.0 * x / y;
2. float z = (float) x / y;
3. float z = 1.0 * (x / y);
4. float z = x / y * 1.0;
5. float z = x / (float)y ;
Which ones will output ?

10. Notice the data type
What if you use:
int digit1, digit2, digit3

12. Question: how do you get ASCII table search online?
Int i = whaterver int < 128;
Printf(“%d=%c”, i, i);

13. Question 3 Least Common Multiplier Analysis (Undestanding the problem)
Design (Devising a plan)
Implementation (Carrying out a plan)
Testing (Looking back)

14. Analysis
The given data are two numbers. Let them be x and y. The unknown we need to find is LCM(x, y)
We can assume that LCM of 2 numbers is always positive. In that case, we can also assume integer values x > y ≥ 0
Other cases of x and y can be easily treated by using their absolute values.

15. Design
Another related problem we have seen is finding GCD of two numbers.
As LCM(x, y) = x*y/GCD(x, y), we can break the current problem into two smaller steps:
Finding GCD(x, y)
Computing x*y/GCD(x, y)

16. Implementation
As the solution can be split into 2 smaller steps, we can now implement it bycarrying out each step:
Step 1: Finding GCD(x, y)
1.1. If y = 0, then the GCD is x and algorithm ends.
1.2. Otherwise, GCD(x, y) = GCD(y, x%y)
Step 2: Finding LCM(x, y)
Assign LCM(x, y) = x*y/GCD(x, y)

17. Testing
We can check the solution by assigning different values to x and y, such as (x = 0, y = 0); (x = 0, y ≠ 0); (x ≠ 0, y ≠ 0); (x = y); (x < y); (x > y); etc...
Corner cases!
For any problem, it is always important to check various situations. All test cases need to give correct results.

18. Is this algorithm... LCM(x, y) = x*y/GCD(x, y) Exact
Is every step deterministic? Same input->same output
Terminating
Will it run forever?
Efficient & Effective
Is the result correct? Is the running time acceptable?
General
Does it work for all valid inputs?
LCM(x, y) = x*y/GCD(x, y)
Exact!
Terminate!
Correct, acceptable
Not all! There is an overflow on x * y
Overflow always exists, x*y encounter overflow very early!
int a =x, b = y;
While(a != b) {
while( a< b) a+=x;
while( a>b) b+= y; }
LCM(x,y) = x / GCD(x,y) * y; ?
How to deal with overflow problem ?

19. Thank you ! See you next week! 