09.12.35 Uppbygging tölva

Programming project 4

On the exam you will not have access to a computer, so you should first try to solve the problems on paber, before entering them and trying out.

1. Consider the following 8086 program:

                   clc
           upp:    mov     ax, [si]
                   adc     [di], ax
                   inc     si
                   inc     si
                   inc     di
                   inc     di
           loop    upp
     

   Answer the following questions on the code:
   
   

   a) Describe what the program does (with words or in pseudocode).
   
   

   b) Why is the instruction inc used to increment si and di, instead of adding 2 with the command add?

2. [Exam '98]
   1. Below is a program to copy CX bytes from DS:SI to ES:DI. Explain carefully the purpose of each instruction

      	shr       cx, 1
      	rep movsw
      	adc       cx, cx
      	rep movsb
         

   2. Explain what the following program does. (Hint: The total size of the instructions from L1 down to the end of the program is 20 bytes)

      		lea	di, L1
      	L1:	mov	cx, 10
      		mov	bx, di
      	L2:	mov	ax, [cs:di]
      		mov	[cs:di+100], ax
      		add	di, 2
      		loop	L2
      		add	di, 80
      		jmp	bx
         

3. [Exam 2000] Explain what the following 8086 program does. The input is in BX and the output in AX.

   		xor	ax, ax
   		mov	cx, 16
   	lyk:	ror	bx, 1
   		rcl	ax, 1
   		loop	lyk
     

4. [Exam 2000] Write the for loop in the following C++ function by using inline IA-32 assembly language in Visual C++. You do not have the rewrite the head of the function, the declaration, the input statement, or the return statement.

   	double fall( ) {
   	    double x;
   	
   	    cin >> x; 
   	    for(int i=0; i<100 && x<100.0; i++)
   	        x = 1.3 * x;
   	
   	    return x;
   	} 
     

5. Write the following program in IA-32 assembly language with floating point instructions so that the code could be placed inline in a C++ program. Assume that a and b are defined as double variables:

   	if ( a <= 10.0 )
   	    b = 5.0*a;
   	else
   	    b = b - a/2.0;
     

hh (hja) hi.is, october 31st, 2001.