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Thursday, November 27, 2014

Binary Search in C using Recursion

 Sumit Kar     November 27, 2014     Algorithm, C - Programming, Data Structure     No comments   


#include<stdio.h>

void bins(int lb,int ub,int item,int a[]);
void main()
{
int i,n,item,a[20];
printf("Enter the size of an array: ");
scanf("%d",&n);

printf("Enter the elements of the array: " );
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}

printf("Enter the number to be search: ");
scanf("%d",&item);


bins(0,n-1,item,a);

}

void bins(int lb,int ub,int item,int a[])
{
int mid,flag=0;

if(lb<=ub)
{
mid=(lb+ub)/2;
if(item==a[mid])
{
flag=flag+1;
}
else if(item<a[mid])
{
return bins(lb,mid-1,item,a);
}
else
return bins(mid+1,ub,item,a);
}


if(flag==0)
printf("Number is not found.");
else
printf("Number is found at %d", mid+1);
}


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Using the concept of Inheritance write a C++ Program to calculate the area and perimeter of rectangle

 Sumit Kar     November 27, 2014     C++ Programming, CPP     No comments   



Sumit Kar, Timus Rak,CPP



/* C++ Program to calculate the area and perimeter of rectangles using concept of inheritance. */

#include 
using namespace std;
class Rectangle
{
    protected:
       float length, breadth;
    public:
        Rectangle(): length(0.0), breadth(0.0)
        {
            cout<<"Enter length: ";
            cin>>length;
            cout<<"Enter breadth: ";
            cin>>breadth;
        }

};


/* Area class is derived from base class Rectangle. */

class Area : public Rectangle   
{
    public:
       float calc()
         {
             return length*breadth;
         }

};


/* Perimeter class is derived from base class Rectangle. */

class Perimeter : public Rectangle
{
    public:
       float calc()
         {
             return 2*(length+breadth);
         }
};

int main()
{
     cout<<"Enter data for first rectangle to find area.\n";
     Area a;
     cout<<"Area = "<
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Sunday, November 23, 2014

Java Program to implement Thread

 Sumit Kar     November 23, 2014     Java Programming     No comments   



class   A     extends   
Thread                                                                        


      {


                public  void 
run( )                                                                           


                    {


                                for(  int i = 1 ;  
i <= 5  ;    i++  
)


                                                 System.out.println( "   From thread A :  i 
=   "  +  
i  ) ;


                                 System.out.println( " End of thread
A  " 
)  ;


                     }


       }





class    B    extends    
Thread                                                                       


      {


                public  void 
run( )                                                                           


                    {


                                for(  int j = 1 ;  
j <= 5  ;   j++ )


                                                 System.out.println(   "  
From thread B :    j =   "    
+   j   );


                                 System.out.println(  "  
End of thread B   "  )  ;


                      }


      }





class   C   extends   
Thread                                                                          


    {


                public  void 
run( )                                                                           


                    {


                                for(   int  
k = 1   ;    k 
<= 5  ;    k++ 
)


                                                 System.out.println(  " 
From thread C :     k  =  
"    +    k  );


                                 System.out.println(  "  
End of thread C    "   ) ;


                     }


     }








class   ThreadTest


                    {


                                public  static 
void  main (  String  
args[ ] )


                                    {


                                                A    obA  
=    new     A( ) ;               //
new born state


                                                obA.start(
) ;                                      //
runnable state


                               


                                                new   B( ).start( );                            





                                                new   C( ).start( );





                                System.out.println(  " 
End of main thread   "   ) ;


                                      }




                      }


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Friday, November 21, 2014

8085 Programming : store the nos in a memory loc. In the reverse order in another memory location

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming




               MVI C, 0AH  ; Initialize counter  




               LXI H, 2200H  ; Initialize source memory pointer  




               LXI D, 2309H  ; Initialize destination memory pointer  




BACK:    MOV A, M  ; Get byte from source memory block  




               STAX D  ; Store byte in the destination memory block  




               INX H  ; Increment source memory pointer  




               DCX D  ; Decrement destination memory pointer   




               DCR C  ; Decrement counter   




               JNZ BACK  ; If counter  0 repeat  




               HLT ; Terminate program execution




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8085 Programming : In a set of nos add only the odd nos

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming



                 LDA 2200H  


           MOV C, A  ; Initialize counter  


           LXI H, 2201H  ; Initialize pointer  


           MVI E, 00  ; Sumlow = 0  


           MOV D, E ; Sumhigh = 0  


BACK:      MOV A, M  ; Get the number  


           ANI 01H   ; Mask Bit1 to Bit7  


           JZ SKIP                  ; Do not add if number is even  


           MOV A, E  ; Get the lower byte of sum  


                 ADD M     ; Sum = sum + data  


           MOV E, A   JNC SKIP  ; Store result in E register  


           INR D   ; Add carry to MSB of SUM  


SKIP:        INX H   ; Increment pointer  


           DCR C     ; Decrement counter  


          JNZ BACK  ; Check if counter 0 repeat  


          MOV A, E  


          STA 2300H  ; Store lower byte  


          MOV A, D  


          STA 2301H ; Store higher byte  


         HLT ; Terminate program execution  


  



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8085 Programming : In a set of nos add only the even nos

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming



            LDA 2200H  


      MOV C, A  ; Initialize counter  


            MVI B, 00H ; sum  =  0  


      LXI H, 2201H ; Initialize pointer  


BACK: MOV A, M  ; Get the number  


      ANI, 01H   ; Mask Bit1 to Bit7  


      JNZ SKIP  ; Don’t add if number is ODD  


      MOV A, B  ; Get the sum  


      ADD M  ; SUM = SUM + data  


      MOV B, A  ; Store result in B register  


SKIP:   INX  H        ; increment pointer  


      DCR C ; Decrement counter  


      JNZ BACK  ; if counter 0 repeat  


      STA 2210H ; store sum  


      HLT    ; Terminate program execution  



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8085 Programming : To find the no of 1’s in the byte

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming




            MVI B,00H   




            MVI C,08H  




            MOV A,D  




BACK: RAR  




            JNC SKIP  




            INR B  




SKIP:   DCR C  




            JNZ BACK  




            HLT  




  




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8085 Programming : Block data transfer

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming




              MVI C, 0AH  ; Initialize counter  




              LXI H, 2200H  ; Initialize source memory pointer  




              LXI D, 2300H  ; Initialize destination memory pointer  




BACK:   MOV A, M  ; Get byte from source memory block  




              STAX D  ; Store byte in the destination memory block  




              INX H  ; Increment source memory pointer  




              INX D  ; Increment destination memory pointer   




              DCR C  ; Decrement counter   




              JNZ BACK  ; If counter  0 repeat  




              HLT ; Terminate program execution  




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8085 Programming : Adding 2 Hexa-Decimal nos with carry

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming




             LDA 2200H  




             MOV C,A ; Initialize counter  




             LXI H, 2201H ; Initialize pointer  




             SUB A  ; Sum low = 0  




             MOV B,A ; Sumhigh = 0  




BACK:  ADD M ; Sum = sum + data  




             JNC SKIP  





             INR B   ; Add carry to MSB of SUM  


SKIP:    INX H   ; Increment pointer  


            DCR C ; Decrement counter  


            JNZ BACK  ; Check if counter 0 repeat  


            STA 2300H 


            MOV A,B  ; Store lower byte  


            STA 2301H  ; Store higher byte  


            HLT   ; Terminate program execution








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8085 Programming : Adding 2 Hexa-decimal nos without carry

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming





  LDA 2200


             MOV C, A ;  Initialize counter


             SUB A  ;  sum = 0


             LXI H, 2201H ;  Initialize pointer


BACK:  ADD M   ; SUM = SUM + data


             INX H  ;  increment pointer


             DCR C ;  Decrement counter


             JNZ BACK ;  if counter  0 repeat


             STA 2300H ;  store sum


             HLT







         








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8085 Programming: Adding 2 BCD nos without carry

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming







MOV A, L ; Get lower 2 digits of no. 1  




ADD E ; Add two lower digits  




DAA  ; Adjust result to valid BCD  




STA 2300H ; Store partial result  




MOV A, H ; Get most significant 2 digits of no. 2  




ADC D ; Add two most significant digits   




DAA  ; Adjust result to valid BCD   




STA 2301H ; Store partial result  




HLT ; Terminate program execution  




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8085 Programming : ADDITION OF 16BIT NUMBERS

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   




8085 Programming






ADDITION OF TWO 16BIT NUMBERS SUM 16 BITS OR MORE


Manually strore 1st 16 bit no in the memory location C050 & C051 in reverse order

Manually store 2nd 16 bit no in the memory location C052 & C053 in reverse order

Result is stored in C053, C054 & C055 in reverse order



          LHLD C050

         XCHG

         LHLD C052

         MVI C,00

         DAD D

         JNC AHEAD

         INR C



AHEAD:   SHLD C054

        MOV A,C

        STA C056

        HLT



EXAMPLE-> A645+9B23=014168

STORE-> C050=45,C051=A6,C052=23,C053=9B

Answer-> C054=68,C055=41,C056=01

// 45H,A6H,23H,9BH





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8085 Programming : 8 BIT DECIMAL SUBSTRACTION

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



 8085 Programming




If 2nd no is greater than 1st no then the answer will in 2's complement

Manually strore 1st 8 bit no in the memory location C050

Manually store 2nd 8 bit no in the memory location C051

Result is stored in C052



  LXI H,C051

  MVI A,99

  SUB M

  INR A

  DCX H

  ADD M

  DAA

  STA C052

  HLT

 EXAMPLE-> 99-48=51

 STORE-> C050=99,C051=48



OUTPUT:

C052=51





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8085 Programming: Exchange the contents of memory locations

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming





Exchange the contents of memory locations 2000H and 4000H.






Program 1:



LDA 2000H : Get the contents of memory location 2000H into accumulator



MOV B, A : Save the contents into B register



LDA 4000H : Get the contents of memory location 4000H into accumulator



STA 2000H  : Store the contents of accumulator at address 2000H



MOV A, B : Get the saved contents back into A register



STA 4000H : Store the contents of accumulator at address 4000H









Program 2:



LXI H 2000H : Initialize HL register pair as a pointer to memory location 2000H.



LXI D 4000H : Initialize DE register pair as a pointer to memory location 4000H.



MOV B, M  : Get the contents of memory location 2000H into B register.



LDAX D : Get the contents of memory location 4000H into A register.



MOV M, A : Store the contents of A register into memory location 2000H.



MOV A, B : Copy the contents of B register into accumulator.



STAX D : Store the contents of A register into memory location 4000H.



HLT : Terminate program execution.



Note: In Program 1, direct addressing instructions are used, whereas in Program 2, indirect addressing instructions are used.

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8085 Programming: Store 8 bit data in Memory

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming





Program 1:




MVI A, 52H : Store 32H in the accumulator



STA 4000H : Copy accumulator contents at address 4000H



HLT : Terminate program execution




Program 2:




LXI H : Load HL with 4000H



MVI M : Store 32H in memory location pointed by HL register pair (4000H)



HLT : Terminate program execution



Note: The result of both programs will be the same. In program 1 direct addressing instruction is used, whereas in program 2 indirect addressing instruction is used.

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8085 Programming: 8 BIT MULTIPLICATION: PRODUCT 16-BIT

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming








       LHLD 2501

       XCHG

       LDA 2503

       LXI H,0000

       MVI C,08



LOOP:     DAD H

       RAL

       JNC AHEAD

       DAD D



AHEAD:  DCR C

       JNZ LOOP

       SHLD 2504

       HLT





LSB OF MULTIPLICAND - 84H, MSB OF MULTIPLICAND - 00H ,MULTIPLIER - 56H





ANSWER

AT ADDRESS 2504 - 58H, LSBs OF PRODUCT

AT ADDRESS 2505 - 2CH, MSB sOF PRODUCT





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8085 Programming: 8 BIT DIVISION

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming








      LHLD 2501

      LDA 2503

      MOV B,A

      MVI C,08



LOOP:   DAD H

     MOV A,H

     SUB B

     JC AHEAD

     MOV H,A

     INR L



AHEAD: DCR C

      JNZ LOOP

      SHLD 2504

      HLT







INPUT:



LSB OF DIVIDEND - 9BH, MSB OF DIVIDEND - 48H , DIVISOR - 1AH







ANSWER

AT ADDRESS 2504 - F2H, QUOTIENT

AT ADDRESS 2505 - 07H, REMAINDER





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8085 Programming: 2's COMPLEMENT OF AN 8-BIT NUMBER

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming






The number to be complemented is stored in C050. Answer is stored in C051



  LDA C050

  CMA

  INR A

  STA C051

  HLT



EXAMPLE-> C050=96

Answer-> C051=6A





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8085 Programming: 2's COMPLEMENT OF A 16-BIT NUMBER

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming




The 16bit number is stored in C050,C051. The answer is stored in C052,C053



  LXI H,C050

  MVI B,00

  MOV A,M

  CMA

  ADI 01

  STA C052

  JNC GO

  INR B



GO:   INX H

  MOV A,M

  CMA

  STA C053

  HLT

EXAMPLE-> C050=8C,C051=5B

Answer-> C052=74,C053=A4





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8085 Programming: 1's COMPLEMENT OF AN 8-BIT NUMBER

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming




The number to be complemented is stored in C050. Answer is stored in C051



  LDA C050

  CMA

  STA C051

  HLT





EXAMPLE-> C050=96

Answer-> C051=69





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8085 Programming: 1's COMPLEMENT OF A 16-BIT NUMBER

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming




The 16bit number is stored in C050,C051

The answer is stored in C052,C053



  LXI H,C050

  MOV A,M

  CMA

  STA C052

  INX H

  MOV A,M

  CMA

  STA C053

  HLT

 EXAMPLE-> C050=85,C051=54

 Answer-> C052=7A,C053=AB



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8085 Programming :

 Sumit Kar     November 21, 2014     8085, Assembly Programming     No comments   



8085 Programming



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Thursday, November 20, 2014

Nokia's first device to the post Microsoft era

 Sumit Kar     November 20, 2014     Android, Microsoft, Nokia, Technology     No comments   








Nokia isn't exactly known for its tablets, but having sold its phone-making division to Microsoft it's hoping to catch your eye with its latest compellingly priced Android offering. The Nokia N1 starts at $250 and is set to go on sale in China before the Chinese New Year -- that's February 19, 2015. Information on other international pricing and release dates have yet to be announced, but the US dollar price converts to around £160 or AU$290.





The N1 looks trendy and slim. It's 6.9mm thick and an aluminum frame that comes in dark gray or silver. Its 7.9-inch IPS LCD screen boasts a 4:3 aspect ratio with a 2,048x1,536-pixel resolution. Inside you'll find a 64-bit 2.3GHz Intel Atom Z3580 CPU and PowerVR G6430 GPU with 2GB RAM.





In a somewhat unusual move, the Nokia N1 houses a whopping 32GB of internal storage. Initially this sounds impressive, until you realise it can't be expanded. Nonetheless, some manufactures have the gall to leave you with a lot less and no microSD card slot, so it's a satisfactory trade-off. That amount of space can accommodate a large number of apps, movies and games.





The Nokia N1 is one of the first tablets to sport a USB Type-C socket, which is a reversible USB port that's set to replace all types of USB connections. It's not the sexiest cutting-edge feature, but it's useful future-proofing.





The N1 is also the first to sport Nokia's Android launcher, Z Launcher. Running on top of the latest version of Google's operating system, Lollipop 5.0, the Z Launcher simplifies the home screen with a gridlike display of apps. It customizes the look of the home screen according to where you are, what time it is and what you're doing, as well as making adjustments geared toward your typical activities.





If you want to try out this user interface before you buy the N1, Nokia invites you to download it from Google Play and install it on your current Android tablet or phone.








Source: CNET


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Durga Puja Maps

 Sumit Kar     November 20, 2014     Durga Puja     No comments   











Get the Official Map From Kolkata Police


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WhatsApp add new features to Group Chat : Multiple Admins

 Sumit Kar     November 20, 2014     Android, Technology, WhatsApp, Windows     No comments   



Multiple Admins, WhatsApp Group, WP, Timus Rak





The new Group Chat feature in WhatsApp lets you chat with up to 100 people at once. Previously it was only 50. It is a great way stay in touch with your family, friends and colleagues. Specially for people with larger number of peers, like students in a class or in a big office, it is now easy to add all the people as members of the group. In addition to the change in number of people you can add to a group, there are a few more changes :




  • Every group has one or more admins. Only admins can add or remove participants. Previously you had only one admin for a group. 

  • Now you can see the Received and Read receipt for all the members of the group, even if they have set privacy option.






Other things you must know about Group Chats are:



  • There is no limit to the number of groups you can create.

  • Only admins can make other participants admins.

  • If the last admin leaves the group, a new admin will be randomly assigned.

  • You can always control your own participation in a Group Chat, by staying in or leaving the group when you wish. Remember that only group admins have the ability to add participants. If you have any questions about being added to a group, or prefer not to be added, please speak to a group admin.

  • If you have blocked a contact, you will still receive messages this contact sends into any group you have in common. Likewise, they will still receive messages you send to those groups.



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Firefox dumps Google, signs with Yahoo as default search engine

 Sumit Kar     November 20, 2014     Firefox, Google, Technology, Yahoo     No comments   



Firefox, Google, Timus Rak, Yahoo





Yahoo will replace Google as the default search engine on the Firefox web browser in the US, signalling Yahoo’s resolve to regain some of the ground that it has lost in the most lucrative part of the internet’s ad market.





The five-year alliance announced on Wednesday will end a decade-old partnership in the US between Google and the Mozilla Foundation, which oversees the Firefox browser. The tensions between Google and Mozilla had been rising since Google’s introduction of the Chrome browser in 2008 began to undercut Firefox. Google’s contract with Mozilla expires at the end of November.





Even though Chrome is now more widely used, Firefox still has a loyal audience that makes more than 100bn search requests annually worldwide.





Besides dropping Google in the US, Mozilla is also shifting Firefox to the Baidu search engine in China and to Yandex in Russia. Firefox users still have the option to pull down a tab to pick Google and other search engines as their preferred way for looking up information online.





Yahoo is hoping to impress Firefox users as the company sets out to prove it is still adept at internet search after leaning on Microsoft’s technology for most of the results on Yahoo’s own website for four years.





Financial details of Yahoo’s Firefox contract were not disclosed. In a blogpost, Chris Beard, the Mozilla chief executive, said the new deal offered “strong, improved economic terms” while allowing Mozilla “to innovate and advance our mission in ways that best serve our users and the web”.





Google accounted for 90% or about $274m (£175m) of Mozilla’s royalty revenue in 2012. Mozilla has not released its annual report for last year.





Yahoo plans to unveil a “clean and modern” search engine on Firefox next month and roll out the new model on its own website early next year, according to its chief executive, Marissa Mayer.





The redesign will primarily affect how Yahoo’s search engine’s results are displayed and not the way that requests are processed. The search technology will continue to be provided by Microsoft as part of a 10-year deal Yahoo signed in 2009.








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Source: Mozilla, the gurdian, CNET


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