Mastering MOD and DIV (“/”)
MOD
DIV
S TAT E MENT S
S U B R O UT INE S
T R ACE TA B L ING P R E D ICTI ONS
U S E O F MO D A ND D I V
Why VB.Net ? The Language resembles Pseudocode - good for teaching and learning fundamentals of programming
Operators
Operator
Mathematical Function
Addition
+
^
*
/
Subtraction
Exponential
MOD
\ (or DIV)
in VB.Net “\” is used.
Example
1+2=3 
4–2=2
2^4 = 16
2^3 = 8 etc
Multiplication
7 * 2 = 14
Division - ordinary
12 / 4 = 3
Modulus (returns remainder from
integer division)
Integer Division – discards the
decimal place
Mod 4 = 3 or 255 Mod 10 = 5
19 \ 4 = 4
or 19 DIV 4 = 4
MOD
In Computing, the modulo operation finds the remainder of division of one number by another
(sometimes called modulus).
For instance, the expression "5 mod 2" would evaluate to
a quotient of 2? and a remainder of 1?
"9 mod 3" would evaluate to
leaves a remainder of 0?
0?
1?
because 5 divided by 2 leaves
because the division of 9 by 3 has a quotient of 3? and
What about when the first no. is smaller!
?
What about 1 Mod 4? Answer: If you divide 1 by 4, you get 0? with a remainder of 1.
That's all the modulus is, the remainder
after division. In this case, 4 CANNOT go into 1 at all,
?
which is why the answer is 0. The remainder is simply the first number, in this case = 1!
MOD
10 MOD 5 =
15 MOD 5 =
12 MOD 5 =
5 MOD 13 =
1 MOD 4 =
3 MOD 10 =
9 MOD 10 =
10 MOD 9 =
0?
0?
2?
5?
1?
3?
9?
1?
Remember, the
MOD is just the
REMAINDER
After DIVISION!
DIV
DIV
10 DIV 5 =
15 DIV 5 =
14 DIV 3 =
2 DIV 10 =
22 DIV 11 =
1 DIV 1 =
Remember, the
is just the
finding the number of divides in long division
2?
3?
4?
0?
2?
1?
i.e. 10 DIV 7 = 1
(7 goes into 10 only ONCE)
In DIV, we are concerned only with the
quotient (not the remainder)
*It is useful to note that in VB.net there
is no DIV - but “/” is used instead
Analyse the following code
n=2
N
T
2
L
0
R
0
2 Mod 10 =
2
Output
0
2/10 = 1/5
(but this is
an integer)
so 0
T+L
0+2
=2
N=R
N=0
=0
L=2
R=0
T=2
N=0
So what’s the point of MOD and DIV
• Both can be used to calculate
various results.
*If you are still not sure about MOD and DIV
– watch this video:
• They basically produce a single
value – so are often used in
Functions.
finds the number of divides in long division i.e. 10 DIV 7 = 1
• They act as Boolean values (e.g.
in MOD – either there IS a
remainder or there IS NOT a
remainder)
• The next slide(s) look at
examples of the use of MOD and
DIV in coding
http://www.youtube.com/watch?v=b5cb_nfDyyM
Analyse this – any idea what it is doing?
Here we are calling the sub with the parameter
“number” – that means the variable or parameter
“number” is being PASSED by Value to the
subroutine
This is a type of recursion - as it is a subroutine
that “CALLS ITSELF” (that is the definition of
recursion = “a function that calls itself”)
Here I have declared an array that can hold 8
integers. (0 to 7 – which gives us 8!)
Analyse this – any idea what it is doing?
This is a familiar “For Loop”, but this time it is going
from 7 to 0 (and stepping down by 1)
Usually For loops go from 0 to 10 (incrementing by 1).
Note the change here.
Here you can see the DIV function being used.
…and here’s MOD!
This is the output of the program. (The input is a
single integer number and the output is……an array
of digits)
…so - - - -what is the output!?
What is this program doing!?
Completing a complex trace table…
• Can feel like climbing a very difficult mountain with no end in sight (especially when the
algorithm is complex)
• It’s actually quite simple
• The keys to success are:
• BE PATIENT
• BE METHODICAL
• BE UTTERLY AND PAINSTAKINGLY ACCURATE (NO CARELESS MISTAKES)
• DO NOT GIVE UP
Trace table it ….(I have amended the array slightly for this example)
Try it with the number input: 2
Number
Myarray(0)
Myarray(1)
Myarray(2)
Myarray(3)
2
Numbertoconvert
i
output
2
3
*This is DIV
0
0
=numbertoconvert \ (2 ^ i )
=2 \ (2 ^ 3)
numbertoconvert =numbertoconvert Mod 2 ^ i
Numbertoconvert = 2 Mod 2 ^ 3
Numbertoconvert = 2 Mod 8
Numbertoconvert = 2
=numbertoconvert \ (2 ^ i )
=2 \ (2 ^ 2)
=2\8
=2\4
=0
=0
2
2
2
numbertoconvert =numbertoconvert Mod 2 ^ i
Numbertoconvert = 2 Mod 2 ^ 2
Numbertoconvert = 2 Mod 4
Numbertoconvert = 2
Part 2 – nearly there!
Try it with the number input: 2
Number
Myarray(0)
Myarray(1)
Myarray(2)
Myarray(3)
2
Numbertoconvert
i
output
2
3
*This is DIV
0
0
2
2
2
*We are here
1
1
0
0
0
0
=numbertoconvert \ (2 ^ i )
=numbertoconvert \ (2 ^ i )
=2 \ (2 ^ 1)
= 0 \ (2 ^ 0)
=2\2
=1
numbertoconvert =numbertoconvert Mod 2 ^ i
Numbertoconvert = 2 Mod 2 ^ 1
Numbertoconvert = 2 Mod 2
Numbertoconvert = 0
=0\1
=0
numbertoconvert =numbertoconvert Mod 2 ^ i
Numbertoconvert = 0 Mod 2 ^ 0
Numbertoconvert = 0 Mod 1
Numbertoconvert = 0
And finally – what is the output!
Try it with the number input: 2
Number
Myarray(0)
Myarray(1)
Myarray(2)
Myarray(3)
2
Numbertoconvert
i
output
2
3
*This is DIV
0
0
2
2
2
1
1
*We are here
0
0
0
0
Myarray(3) = 0
Myarray(2) = 0
The input was: 2
It is simply asking you to OUTPUT the values of
Myarray(i) – starting with Myarray(3) and going
down to MyArray(0)
And the output is: 0 0 1 0
Myarray(1) = 1
Myarray(0) = 0
WHAT IS IT ? A Decimal to Binary Converter!
The one on the left doesn’t work perfectly
because the array only allows for 4 bits. Try coding the
one on the right yourself to see how it works!
[ESC]* Teachers only: Code to cut and paste -students are advised to type it out line by line
If you’re feeling adventurous, here is another
example of a program that uses MOD
• See if you can code it in yourself (use VB.Net Console)