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Program to show the Keyboard Status

Posted By: Easy Tutor     Category: C Programming     Views: 8293

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

Hello Friends,

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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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