System Level Programming by 'C' Programming Language

In the header file dos.h there are two important structures and union (Remember this structure)

1. struct BYTEREGS {

unsigned char al, ah, bl, bh;

unsigned char cl, ch, dl, dh;

   };

2. struct WORDREGS {

unsigned int ax, bx, cx, dx;

unsigned int si, di, cflag, flags;

   };

3. union REGS {

struct WORDREGS x;

struct BYTEREGS h;

  };

There is function int86() which has been defined in  dos.h header file. It is general 8086 software interrupt interface. It will better to explain it by an example.

Write a c program to display mouse pointer?
Answer:

#include

#include

void main()

{

union REGS i,o;

i.x.ax=1;

int86(0x33,&i,&o);

getch();

}

Explanation: To write such program interrupt table is necessary. 

Complete interrupt table
It is a small part of interrupt table. It has four field input, output, service number and purpose,now see the first line which is inside the rectangle. To display the mouse pointer assign ax equal to 1 i.e. service number while ax is define in the WORDREGS

struct WORDREGS {

unsigned int ax, bx, cx, dx;

unsigned int si, di, cflag, flags;

};

and WORDRGS is define in the union REGS

union REGS {

struct WORDREGS x;

struct BYTEREGS h;

};

So to access the ax first declare a variable of REGS i.e. REGS i,o;
To access the ax write i.x.ax (We are using structure variable i because ax is input, see interrupt table)
So to display mouse pointer assign the value of service number:
i.x.ax=1;
 

To give the this information to microprocessor, We use int86 function. It has three parameters
1. Interrupt number i.e. 0x33

2. union REGS *inputregiste i.e. &i

3. union REGS *outputregiste i.e. &o;

So write: int86 (0x33, &i, &o);