Programs of microprocessor

Programs of microprocessor

Microprocessor cannot understand a program written in Assembly Language .
Instruction are divided into bit pattern of word in byte as op-code(operation code) and operation.

Instruction format has three types
1.One byte-has op-code
2.Two byte- has op-code and address
3.Three byte -has code and address lower byte and address higher byte

opcode

1 byte

 

opcode

Data / address

2 byte

 

opcode

Data / address Lower byte

Data Address Higher byte

3 byte

Instructions set for 8085 are classified into five categories:

1.Data transfer group instructions
2.Arithmetic group of instructions
3.Logical group
4.Branching group

5.Machine Level

 
A program known in Assembler is used to covert a Assembly language program to machine language.

I) Data transfer copy operation:-

1)Load a 8 bit number in a register
2)Copy from Register to Register
3)Copy between Register and Memory
4)Copy between Input/Output Port and Accumulator
5)Load a 16 bit number in a Register pair
6) Copy between Register pair and Stack memory

Example :-
1.Load a 8 bit number 4F in register B                     -MVI   B ,4FH
2.Copy from Register B to Register A                     -MOV  A,B
3.Load a 16 bit number 2050 in Register pair HL     -LXI     H,2050H
4.Copy from Register B to Memory Address 2050  - MOV   M,B
5.Copy between Input/Output Port and Accumulator- OUT   01H      IN   07H

2)Arithmetic operations and instructions

1.Add 8 bit number 32 H to Accumulator                   -ADI   32 H
2.Add contents of Register B to Accumulator            - ADD B
3.Subtract a 8 bit number 32 H from Accumulator      - SUI   32H
4.Subtract contents of Register C from Accumulator   -SUB  C
5.Increment the contents of Register D by 1                - INR   D
6.Decrement the contents of Register E by 1               -  DCR   E

3)Logical and Bit Manipulation operations and instructions

1.Logically AND Register H with Accumulator             -ANA  H
2.Logically OR Register L with Accumulator                 -ORA   L
3.Logically XOR Register B with Accumulator             - XRA   B
4.Compare contents of Register C with Accumulator  -CMP  C
5.Complement Accumulator                                            -CMA
6.Rotate Accumulator Left                                             -RAL

Program to add two 8 bit numbers and store result in register C

1.Get two numbers
a) Load first no. in register D                                        MVI   D , 2H
b)Load  second no. in register E                                    MVI   E , 3H

2.Add them
a) Copy register D to A                                                  MOV  A , D
b) Add register E to A                                                     ADD E

3. Store result
a)Copy A to register  C                                                  MOV  C , A

4.Stop
a)Stop processing                                                            HLT

Program to add 8 bit number result can be more than 8 bits i.e.with carry present

1.Get two numbers
a) Load first no. in register D                                        MVI   D , 2H
b)Load  second no. in register E                                    MVI   E , 3H

2.Add them
a) Copy register D to A                                                  MOV  A , D
b) Add register E to A                                                     ADD E

3. Store result
a)Copy A to register  C                                                  MOV  C , A

4.Carry
a)Use conditional Jump instructions                           JNC  END

5.Change
a)Clear register B                                                          MVI  B ,0H
b)Increment B                                                                INR  B

6.Stop
a)Stop processing                                                         HLT

Solve

SOLVE

 1) Find the odd man out - 49,  81 ,100 ,144
A) 49
B) 81
C) 100
D) 144

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2)Which of the following numbers is different from others?
A) 3156
B) 4164
C) 5255
D) 6366

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3)Fill the blank
5,35,7,42,6 ,_____,9,36,4.
A) 45
B) 48
C) 54
D) 56 

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4)Find odd man out :DFEG,  JKLM,   GHIK,   RTSU
A) DFEG
B)JKLM
C)GHIK
D)RTSU

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5) find odd man out 1,3,10,21,64,129,256,778
1) 10
2) 21
3) 129
4) 256

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6)The missing number in the series is:
0 , 6 , 24 , 60 , 120 , ? , 336 is
A) 240
B) 220
C) 280
D)210

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7)Let a means minus (-) ,b means multiplied by (*), C means divided by (/) and D means plus(+).The value of 90 D 9 a 29 C 10 b 2
A) 8
B) 10
C) 12
D) 14

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8)If MOHAN is represented by the code KMFYL, then COUNT will be represented by
A) AMSLR
B) MSLR
C) MASRL
D)SAMLR

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9)  32     33      37    46     62    ?
Replace ? by
A)85
B) 87
C) 94
D) 99

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10)AZ       DW        GT       JQ     ? .Replace ? from following
A) LD
B) MN
C) MO
D) NM

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11) In certain code language ,PUNE is written as SXQH.How would you write DELHI in this language?
A )  FGNJL
B )  FGNJK
C )  GHOKM
D )  GHOKL

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12)Find odd man out:
A)  6
B ) 18
C )  24
D )  36

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13)Find the wrong term in the following series:
68    66     62    59     56    53
A)  59
B)  62
C)  66
D)  68

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14)If P is the brotherb of Q;Q is the son of R; Sis R's father .What is P of S?
A)  Son
B)  Brother
C)  Grandson
D)  Grandfather

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15) The missing number in the series 40 , 120 , 60 , 180 , 90 , ? , 135
A)  110
B)  270
C)  105
D)  210

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16) The length and breadth of  a rectangle are changed by +20% and by -10% respectively.What is the percentage change in the area of rectangle?
A)  10%
B)  12.5%
C)  8%
D)  15%

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17) Two pipes A and B can fill a tyank in 15 minutes and 20  minutes respectively .Both the pipes are opened together but after 4 minutes,pipe A is turned off. What is the total time required to fill the tank?
A) 10 min.40 sec
B) 11 min.45 sec
C) 12 min.30 sec
D)  14 min.40sec

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18)The odd number from 1 to 45 which are exactly divisible by 3 are arranged in an ascending order .The number at 6^th position is
A)  18
B)  24
C)  33
D)  36

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19) If the radius of a circle is increased by 50%, the perimeter of the circle will increase by

A) 30
B)  85/2
C)  170/3
D)  110

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20) In code language, the word BAD is written as 658 .In the same language, what could be the code for FIG?
A)  9         12       10
B) 10        13        11
C) 8          11       13
D) 10        12       13

More C++ programs

//Program to display Floyd triangle:-

#include<iostream.h>
main()
{
    int line,i,num=1,incr;
    cout<<" How many lines = ";
    cin>>line;
    cout<<"\n \n Floyds Triangle \n \n";
    for(i=1;i<=line;i++)
    {
        for(incr=1;incr<=i;incr++)
        {
            cout<<num;
            cout<<"\t";
            num++;
        }
        cout<<"\n";
    }
}

Output:-

How many lines = 7

Floyds triangle


1
2           3
4           5           6
7           8           9           10
11         12         13         1 4          15
16         17         18         19            20          21
22          23         24         25           26          27          28

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//Program to check whether number is prime or not:-

#include<iostream.h>
main()
{
    int num,temp=0,i;
    cout<<"\n Enter the number =";
    cin>>num;
    if(num>0)
    {
        for(i=2;i<=num-1;i++)
        {
            if(num%i==0)
            {
                cout<<"\n Number is not a Prime number \n \n";
                temp=1;
                break;
            }
        }
        if(temp==0)
        {
            cout<<"\n Number is Prime \n \n";
        }
    }
    else
        cout<<"Enter only Positive numbers";
}

Output:-

Enter the number = 7
Number is Prime

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//Program of destructors in c++:-

#include<iostream.h>
class ABC
{
int *width,*height;
public:
    ABC(int,int);
    ~ABC();
        int area()
    {
            return(*width**height);
    }
};
ABC::ABC(int a,int b)
{
width=new int;
height= new int;
*width=a;
*height=b;
}
ABC::~ABC()
{
    delete width;
    delete height;
   
}
main()
{
    ABC obj1(4,9),obj2(7,5);
    cout<<obj1.area();
    cout<<"\n";
    cout<<obj2.area();
}

Output :-

36
35

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//Program of static member variable and member function:

#include<iostream.h>
class ABC
{
private :static int data;
public: static void getdata()
        {
            ++data;
            cout<<"count is =";
            cout<<data<<"\n";
        }
     void putdata()
        {
            ++data;
            cout<<"now count is ="<<data<<"\n";
        }
 void newfunction()
{
++data;
cout<<"now count is ="<<data;
}
};
int ABC::data=5;
void main()
{
    ABC obj;
    ABC::getdata();
    obj.putdata();
    obj.newfunction();
}

Output:-

count is =6
now count is =7
now count is =8

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Solve the sums of Quiz

Solve the sums of Quiz 

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Database- SQL

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• SQL stands for Structured Query Language. 

• SQL is an ANSI  (i.e. American National Standards Institute  ) standard computer language for accessing and manipulating database systems.
• SQl statements are used to retrieve and update data in a database.
• SQL works with database programs like MS Access ,DB2,Informix ,MS SQL Server, Oracle ,Sybase,etc.
• The first version ,initially called SEQUEL,was designed to manipulate and retrieve data 
• SQL is a Non - procedural query language.
• Sometimes referred to as a Database Gateway Language as an ANSI made it a standard language for all DBMS.

SQL commands are divided into following categories:-
1)Data Definition Language-
                   It uses create,alter,drop,truncate
2)Data Manipulation Language -
                   It uses insert ,update,delete
3)Data Query Language -
                   It uses select
4)Data Control Language
                   It uses grant ,revoke
5)Transaction control Language
                   It uses commit,rollback
Syntax:
Create-  CREATE TABLE tablename(columnname1 data_type (size)[constraints],(columname2 data_type (size)[constraints],....)

Alter- ALTER TABLE tableman ADD(<columnname> data_declaration[constraints],<columnname> data_declaration[constraints],.....);

 To delete  Drop - DROP TABLE <tablename>[CASCADE CONSTRAINTS];

Truncate Table<tablename>;
removes all rows from a table and is high speed data deletion statement.

Click here for solve the quiz on sql

Mathematics

Mathematics

Cryptography and Security

Cryptography and Security

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Cryptography is the art and science of achieving security by encoding messages to make them non-readable.

Cryptanalysis is the technique of decoding messages from a non-readable format back to readable format without knowing how they were initially converted from readable format to non-readable format.

Cryptology is a combination of cryptography and cryptanalysis.

Principles of Security:-

1)Confidentiality
2)Integrity
3)Authentication
4)Access control
5)Availability
6)Non-repudiation
7)Ethical and legal issues

Types of attacks:-

1)General attacks-

Criminal attacks,publicity attacks,legal attacks

2)Technical view-

a)Theoretical concepts-
I)Passive attacks-

i)release of message contents

 ii)traffic analysis

 
II)Active attacks-

i)Interception,

ii)Fabrication,

iii)Modification,

iv)Interruption

3)Program attack-

a)Virus- A virus is a computer program that attaches itself to another legitimate program and causes damage to the computer system or to the network.

During lifetime ,a virus goes through four phases:-
Dormant phase,Propagation phase,Triggering phase ,Execution phase.

b)Worm

A worm does not perform any destructive actions and instead ,only consumes system resources to bring it down.

c)Trojan Horse

It allows an attacker to obtain some confidential information about a computer or a network.

d)Specific attacks-

Sniffing and spoofing-cause packet level attacks.

Phishing-It is new attack which attempts to fool legitimate users to provide their confidential information to fake sites

Pharming  /DNS spoofing -attack involves changing the DNS entries so that users are redirected to an invalid site,while they thinking that they have connected to the right site.

My Biz card

Logic Gates

The logic gates have one or many input and only one output.The gates used to form integrated circuits (IC's) for digital circuit with combination and sequential of gates.this gates are of mainly 6 types .The three are basic gates and others are the prepared from basic gates called as derived gates.They are as follow:-
I ) Basic gates :-
                               i ) OR gate
                               ii) AND gate
                               iii)NOT gate
II) Derived gates :-
                              i ) NOR gate
                              ii) NAND gate
                              iii)EX-OR gate
The Logic gates have the following details:
1)Logic:-It is the statement.
2)Truth table:-It shows the input provided and output get.It is combination of input provided  making rows in the table with inputs and output column.
3)Boolean equation :-It is the mathematical representation with the equation.
4)Symbol:-It is the logical symbol of the logic gate.

1) OR gate

    Logic :- If A is true or B is true  then Y is true

    Boolean Equation :- Y=A+B

   Truth table:

A	B	Y
0	0	0
0	1	1
1	0	1
1	1	1

Symbol:-

2) AND gate

    Logic :- If A is true and B is true  then Y is true

    Boolean Equation :- Y=A.B or Y=AB

   Truth table:

A	B	Y
0	0	0
0	1	0
1	0	0
1	1	1

Symbol:-

 
3) NOT gate

    Logic :- If A is true then Y is false and A is false  then Y is true

    Boolean Equation :- Y=A.B or Y=AB

   Truth table:

A	Y
0	1
1	0

Symbol :-

4) NAND gate
   Truth table:

A	B	Y
0	0	1
0	1	1
1	0	1
1	1	0

Boolean Equation :- Y=A.B 
 
Symbol:-

5) NOR gate
Boolean Equation :- Y=A+B
Truth table:

A	B	Y
0	0	1
0	1	0
1	0	0
1	1	0

Symbol :-

6)EX-OR gate:
Y=A.B+A.B
Logic :When A is true or Bis true but not both then Y is true.
Truth table:

A	B	Y
0	0	0
0	1	1
1	0	1
1	1	0

Symbol :-

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De Morgan's  Law :-
First law :-
Statement:-The compliment of sum is equal to product of compliments.

OR

                   The NOR gate is equivalent to bubbled NAND gate.
 Boolean Equation :-  A+B  = A.B
Second Law :-
Statement:-The compliment of product is equal to sum of compliments.

OR

                   The NAND gate is equivalent to bubbled OR gate.
 Boolean Equation :-  A.B  = A+B 

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 Half Adder:-
For addition of two bits half adder is used .
Truth table:

A	B	Carry	SUM
0	0	0	0
0	1	0	1
1	0	0	1
1	1	1	0

 From the above table we can conclude that gates used in half adder are AND gate for carry and EX-OR gate for Sum.

Number System

Number System
The number system of a number is determined by the Radix of a number.Radix means the base of the number which is obtained by the number of symbols present in that number System.They are as follow:-
 

Sr.No.	Number System	Radix	Symbols
1.	Binary number System	2	0,1
2.	Octal Number System	8	0,1,2,3,4,5,6,7
3.	Decimal Number System	10	0,1,2,3,4,5,6,7,8,9
4.	Hexadecimal Number System	16	0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

1.Binary Number System:- The number System with Radix 2 are said to be binary numbers.It contains only 2 symbols 0 and 1.For Examples (1110)₂,(1001)₂,(1101)₂,(1010)₂,(1010.111)₂,etc.
2.Octal Number System:- The number System with Radix 8 are said to be octal numbers.It contains 8 symbols 0,1,2,3,4,5,6 and 7.For Examples (257)₂,(346)₈,(412)₈,(728)₈,(72.39)₈,etc.
3.Decimal Number System:- The number System with Radix 10 are said to be decimal numbers.It contains 10 symbols 0,1,2,3,4,5,6,7,8 and 9.For Examples (123)₁₀,(786)₁₀,(157)₁₀,(928)₁₀,(28.625)₁₀,etc.
4.Hexadecimal Number System:- The number System with Radix 16 are said to be hexadecimal numbers.It contains only 16 symbols 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E and F.Here A represent for number 10,B represents for number 11,C represents for number 12,D represents for number 13,E represents for number 14,F represents for number 15,For Examples (1ABC)₁₆,(25DE)₁₆,(85C.3D)₁₆,(0AB)₁₆,etc.
Note that for hexadecimal number should start with symbols of numbers only that is 0,1,2,3,4,5,6,7,8,9 and not with A,B,C,D,E,F .If there is a possibility of number to star with alphabets then write 0(zero before that number example (0AB)₁₆,(0ADE)_16,etc.

Conversion of numbers Systems:
# Convert Binary to Decimal:- To convert the binary number into decimal we have to multiply each binary number with 2 and then take the raise to power from Least Significant Bit (LSB) side starting with 0 and so on.
1.(1010)₂   = 1*2³ + 0*2²+ 1*2¹ + 0*2⁰
                  = 1*8 + 0*4 +1*2 + 0*1
                  = 8 + 0 + 2 + 0
                  =(10)₁₀
2.(101)₂    = 1*2² + 0*2¹ + 1*2⁰
                 = 1*4 + 0*2 + 1*1
                 =  4 + 0 +1
                 = (5)₁₀
3.(1001)₂   = 1*2³ + 0*2²+ 0*2¹ + 1*2⁰
                  = 1*8 + 0*4 +0*2 + 1*1
                  = 8 + 0 + 0 +1
                  =(9)_10
4..(1111)₂   = 1*2³ + 1*2²+ 1*2¹ + 1*2⁰
                  = 1*8 + 1*4 +1*2 + 1*1
                  = 8 + 4 + 2 +1
                  =(15)₁₀
But for fractional numbers that is for the numbers with decimal point we have to multiply fractional numbers with 2 raise to power of -1 from decimal point.
1. (1111.101) =1*2³ + 1*2²+ 1*2¹ + 1*2⁰ + 1*2^-1+ 0*2^-2 + 1*2^-3
                      = 1*8 + 1*4 + 1*2 + 1*1+ 1*1/2¹+ 0*1/2² + 1*1/2²
                      = 1*8 + 1*4 + 1*2 + 1*1+ 1*1/2 + 0*1/4 + 1*1/8
                      = 1*8 + 1*4 +1*2 + 1*1+1*0.5 + 0*0.25 + 1*0.125
                      = 8 + 4 + 2 + 1 + 0.5 + 0 + 0.125
                      =(15.625)₁₀
# Convert Decimal to Binary :- To convert the decimal number to binary divide the number by 2 to get integers note down the remainders in front of them simultaneously and then write remainders in reverse order from bottom to top.
  1.(157)₁₀
Solution : -

2	157	Remainders
2	78	1
2	39	0
2	19	1
2	9	1
2	4	1
2	2	0
2	1	1
	0	1

Answer=(11011101)₂

2.(9)₁₀
Solution : -

2	9	Remainders
2	4	1
2	2	0
2	1	0
	0	1

Answer=(1001)₂
Note:Binary counting can be done by taking only those values of numbers from similarly like starting from decimal numbers only those numbers which contains symbols 0 and 1.That is 0,1,2,3,4,5,6,7,8,9,10,11,12,13,............,99,100,101,102,103,....,110,111,...and so on.Cancel the numbers that contains symbols other than 0 and 1.So the counting will be:

Decimal Numbers	Binary numbers	Hexadecimal Numbers
0	0	0
1	1	1
2	10	2
3	11	3
4	100	4
5	101	5
6	110	6
7	111	7
8	1000	8
9	1001	9
10	1010	A
11	1011	B
12	1100	C
13	1101	D
14	1110	E
15	1111	F

# Convert octal to decimal:- To convert the octal number into decimal we have to multiply each octal  number with 8 and then take the raise to power from Least Significant Bit (LSB) side starting with 0 for integers and so on.

1. (157)₈ = 1*8² + 5*8¹ + 7*8⁰
              = 1*64 + 5*40 + 7*1
              = 64 + 200 +7
              = (271)₁₀
2.(459)₈ = 4*8² + 5* 8¹ + 9*8⁰
              = 4*64+5*8+9*1
              =256 + 40 +9
              =(305)₁₀
 But for fractional numbers that is for the numbers with decimal point we have to multiply fractional numbers with 8 raise to power of -1 from decimal point and so on -2,-3,.. .
 1.(351.123)₈   =  3*8² + 5*8¹ + 1*8⁰ + 1*8^-1 + 2*8^-2 + 3*8^-3
                        =  3*64 + 5*8 + 1*1 + 1*1/8¹ + 2*1/8² + 3*1/8³
                        =3*64 + 5*8 + 1*1 + 1*1/8 + 2* 1/64 + 3* 1/512
                        =192 + 40 + 1 + 1*0.125 + 2*0.015625 + 3* 0.001953125
                        = 192 + 40 + 1 + 0.125 + 0.03125 +  0.005859375
                        = (233.162109375)₁₀
# Convert decimal to octal:- To convert the decimal number to octal divide the number by 8 to get integers note down the remainders in front of them simultaneously and then write remainders in reverse order from bottom to top.
1.(157)₁₀
Solution : -

8	157	Remainders
8	19	5
8	2	3
	0	2

Answer=(532)₈

1.(304)₁₀
Solution : -

8	304	Remainders
8	38	0
8	4	6
	0	4

Answer=(064)₈