Lecture 10: a New Interrupt
invoking function
 Software Engineering:

pointers and structures
 Embedded Systems:
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A new interrupt invoking function
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Pointers
 Syntax
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data_type * pointer_name;
 An example:
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int * my_pointer;
//my_pointer is a pointer to an integer
char * another_pointer
//another_pointer is a pointer to a char
//it can also used as an array of char;
another_pointer=“Deng”;
//
cout << another_pointer << endl;
//it is equivalent to cout<<“Deng”<<endl;
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Pointers
 int * my_pointer;
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my_pointer name of the pointer
&my_pointer
address of the pointer
*my_pointer
value of object pointed to by my_pointer (an int here).
 In comparison
 int my_int;

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my_int
&my_int
name and value of the int
address of the int
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Define a struct in Turbo C++
 Syntax
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struct
{
•
•
•
•
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};
struct_type
variable-type variable_name;
variable-type variable_name;
………
variable-type variable_name;
//don’t forget this “;”
 Declare a variable of user defined struct-type

struct_type struct_name;
//after struct_type is defined
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An Example:
 struct

student
{
• char name[20];
• long id;
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};
//name for student
//id for student
//don’t forget this “;”
student stud1;
//define variable stud1 to be type student
stud1.name[0]=‘J’;
stud1.name[1]=‘a’;
stud1.name[2]=‘c’;
stud1.name[3]=‘k’;
stud1.id=88888888;
cout << “The ID of” << stud1.name <<“ is “ << stud1.id<<endl;
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Pointer to struct
 Syntax

structure_type
* pointer_name;
 An example:
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student * stud1P;
//define variable stud1P to be pointer to student
stud1P->name[0]=‘J’;
//
stud1P->name[1]=‘a’;
stud1P->name[2]=‘c’;
stud1P->name[3]=‘k’;
stud1P->id=88888888;
cout << “The ID of” << stud1P->name <<“ is “ << stud1P->id<<endl;
int * my_pointer;
//my_pointer is a pointer to an integer
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Pointer to struct

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
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student * stud1P;
//define variable stud1P to be pointer to student
stud1P
the pointer
*stud1P
the struct pointed to by the pointer
(*stud1P).id the variable id for the struct pointed to by the pointer
stud1P->id simplified version of (*stud1P).id
the followings are similarly interpreted.
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stud1P->name[0]=‘J’;
stud1P->name[1]=‘a’;
stud1P->name[2]=‘c’;
stud1P->name[3]=‘k’;
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A Function to get a key and its key number
 struct CharInt {char x; int y;}; //define a struct type
 CharInt get_key_number () {
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char a; int b;
CharInt tmp;
_AH=0x0;
geninterrupt(0x16);
a=_AL;
b=_AH;
tmp.x=a; tmp.y=b;
return tmp;
}
//return type of the function is CharInt
//service number 0x00
//interrupt 0x16
//_AL is the key and _AH is the keynumber
//assign them to a variable of type CharInt
//return the value
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Problems caused by Pseudo-variables
Pseudo-variables refer to CPU registers which are used by
other programs which may run at the same time. One must
assign values right before using them and read values right
after obtaining them, especially when we program in C.
Function “geninterrupt” has no direct provision for register
manipulation so we have to use pseudo-variables. Therefore,
there is no way to guarantee that pseudo-variables will retain
their values during a succession of assignment operations
before/after a “geninterrupt” invocation.
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More Reliable Access to Registers
Segments registers can be obtained more reliably by
the following approach:
 struct SREGS {unsigned int es,cs,ss,ds;}
 void segread (struct SREGS *SegReg)
The function segread simply copies the values of the
four segment registers into the corresponding structure
SegReg items.
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Three More Structures
 struct WORDREGS { unsigned int
ax,bx,cx,dx,si,di,cflags,flags;}
 struct BYTEREGS {unsigned char
al,ah,bl,bh,cl,ch,dl,dh;}
 union REGS {struct WORDREGS x; struct
BYTEREGS h;}
The item “cflags” of the WORDREGS structure reflects the value
of the carry flag in item “flags”: “cflag” is zero when this flag is
not set. DOS frequently sets this flag to indicate an error condition.
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An Alternative Function for Invoking Interrupts
 int int86(int N,union REGS *in, union REGS *out)
 //precondition: N is assigned to be the interrupt number
to invoke. The relevant service number and the data are
assigned in the REGS *in
 //postcondition: the interrupt with the specified service
is invoked and the output data are in the REGS*out.
NOTE: Function int86 is used to invoke interrupt when the request
for the desired interrupt service does not use special settings
of the segment registers. Its source code consists of a call to
“segread” to load the structure “SegReg”, and then make a call
to invoke interrupt.
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Input/Output of int86
From structure
To registers
 In
 SegReg
• AX,BX,CX,DX,SI,DI
• DS,ES
From registers
To structure
• AX,BX,CX,DX,SI,DI,Flag
• DS,ES
• Out
• SegReg
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Revised Function for Output a Char
 void output_a_char(int x); //prototype
 //precondition: x is the ascii code of a character
 //postcondition: the character is output to screen
 void output_a_char(int x) //definition
 {REGS*in; REGS*out;
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in->h.ah=0x0E;
in->h.al=x;
int86(0x10,in,out);
 }
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Revised Function to get a key and its key number
 struct CharInt {char x; int y;}; //define a struct type
 CharInt get_key_number () {
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REGS*in; REGS*out;
CharInt tmp;
REGS.h->ah=0x0;
int86(0x16,in,out);
tmp.x=out->h.al;
tmp.y=out->h.ah;
return tmp;
}
//return type of the function is CharInt
//service number 0x00
//interrupt 0x16
//_AL is the key and _AH is the keynumber
//assign them to a variable of type CharInt
//return the value
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Check any key is hit (for control)
 int key_ready() {//return 1 if a key is ready, 0 otherwise
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long int x;
REGS in, out;
//page 358 of chapter12 of the handout
REGS.h->ah=0x1;
//service number 0x00
int86(0x16, &in, &out);
//interrupt 0x16
x=out.x.flags;
//get flag register
if (x&(0x40)==0) {return 1;} //if ZF==0 a key is ready
else return 0;
//else no key
 }
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Write a function to read a key
 char getch() {
 {
//return char for the key hit
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REGS in; REGS out;
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//page 358 of chapter12 of the handout
in.h.ah=0x0;
//service number 0x00
int86(0x16,in,out);
//interrupt 0x16
return out.h.al;
//AL is the key
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 }
 //alternatively, one may simply do:
 //{char t; cin >> t; return t;}
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Control by typing keys
 char k;
 If (key_ready())
 {
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k=getch();
switch k
{
• case ‘l’: do_something; break;
• case ‘r’: do_something_else; break;
• case ‘u’: do_another_thing; break;
• case ‘d’: do_different_thing; break;
}
}
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More about assignment 2
 Some suggestions are given below:
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add a constructor in the class Elevator. This will
reduce the length of the main function and makes the
program clear.
Write a function for “pressing buttons”. Recall that
we have 24 buttons to be pressed.
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More about assignment 2
 How to open a door?
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Draw a door by drawing 50 vertical lines.
Use a member variable door to control the door.
• Door is an integer, the value is 25 --3.
– 25-- the door is closed
– 3-- the door is open.

Open the door
• just draw black lines from 25 to door.
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Close the door
• draw more and more yellow lines.
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More about assignment 2
 When to open the door?
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After your door REALLY stops.
You have to study the function stop().
 When to close the door
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before the lift moves, i.e., before x.k is changed.
 However to show buttons inside the elevator
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modify the member function selevator() to show
bottons according to array[].
if (array[i]==1) print a dot/circle
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More about assignment 2
 How to show buttons outside the elevator
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do it according to the two arrays corresponding to the
buttons. Let up[] and down are the arrays.
A ordinary function can be defined.
Call the function in main().
if (up[i]==1) print red square else print green
square
if (down[i]==1) print read circles else print green
circles.
Repeat the above for different i’s.
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More about assignment 2
 How to test if the lift should be stopped?
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Very hard
First, just consider the buttons inside the elevator
two subcases should be considered
• lift is going up (look at x.sign)
• lift is going down
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When buttons outside the elevators are considered,
we have to decide that which elevator stops. (This is
the last step to do.)
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