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Program to show the Keyboard Status

Posted By: Easy Tutor     Category: C Programming     Views: 5438

Write a program to show the Keyboard Status.

Code for Program to show the Keyboard Status in C Programming

 # include <stdio.h>
 # include <conio.h>
 # include   <dos.h>

 int main( )
 {
    int iCount=15;
    int iMaxValue=128;
    unsigned char far* ptrKeyboard=(unsigned char far*)0x00400017;

    clrscr( );

    printf("\nKeyboard Status Byte (Hex) = %X",*ptrKeyboard);
    printf("\nKeyboard Status Byte (Binary) = ");

    for(iCount=7;iCount>=0;iCount--)
    {
       if(iCount==3)
      printf(" ");

       if((*ptrKeyboard&iMaxValue)==iMaxValue)
      printf("1");

       else
      printf("0");

       iMaxValue/=2;
    }

    printf("\n\nWhere:");
    printf("\n Bit 7    Insert");
    printf("\n Bit 6    Caps Lock");
    printf("\n Bit 5    Num Lock");
    printf("\n Bit 4    Scroll Lock");
    printf("\n Bit 3    Alt Key");
    printf("\n Bit 2    Ctrl Key");
    printf("\n Bit 1    Left Shift");
    printf("\n Bit 0    Right Shift");

    getch( );
    return 0;
 }
  
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Easy Tutor
Easy Tutor author of Program to show the Keyboard Status is from United States. Easy Tutor says

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Happy Programming :)

 
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Joachim Stephan from United Kingdom Comment on: Oct 31
can you check this for me please



#include <stdio.h>
#include <stdlib.h>
#define row 3
#define column 3

void main()
{
     int A[row][column], B[row][column], C[row][column];
     int i,j,k;
	 int choice;

         for (i=0;i<3;++i)
         {
            for (j=0;j<3;++j)
			 {
				printf("enter values in matrix A");
				scanf("%d",&A[i][j]);
             }
         }

         for (i=0;i<3;++i)
         {
           for (j=0;j<3;++j)
		   {
             printf("enter values in matrix B");
             scanf("%d",&B[i][j]);
           }
		 }


             printf("choose one of the following:\n");
			 printf("1. ADDITION:\n");
			 printf("2. SUBSTRACTION:\n");
			 printf("3. MULTIPLICATION:\n");

			 printf("%d",choice);
			scanf("%d",&choice);

             if(choice==1)
			 {
				 //ADDITION
				 for (i=0;i<3;++i)
					{
						for (j=0;j<3;++j)
						{
							C[i][j]=A[i][j]+B[i][j];
						}
					}
			 }

             else if(choice==2)
			 {
				 //SUBSTRACTION
				 for (i=0;i<3;++i)
				 {
					for (j=0;j<3;++j)
					{
						C[i][j]=A[i][j]+B[i][j];
   					}
				 }
			 }

			 else if(choice==3)
			 {
				 //MULTIPLICATION

				 for (i=0;i<3;++i)
				 {
					for (j=0;j<3;++j)
					{
						C[i][j]=0;

						for (k=0;k<3;++k)
							C[i][j]=C[i][j]+(A[i][j]*B[i][j]);
					}
				 }
			 }

			 else
			 if (choice==4)
				printf("Invalid Operation:\n");
}


			  //RESULT

              /*printf("the resultant matrix is C",C);
              for (i=0;i<3;++i)
              {
               for (j=0;j<3;++j)
               {
                   C[i][j]
                printf("C is %s",C);
                }
                }
                system("pause");
                }*/



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