Practical Session 3

Computer Architecture and Assembly Language

DIV r/m - unsigned integer division

IDIV r/m - signed integer division

Dividend Divisor Quotient Remainder

AX r/m8 AL AH

DX:AX r/m16 AX DX

EDX:EAX r/m32 EAX EDX

Advanced Instructions – Advanced Instructions – divisiondivision

DIV r/m8mov ax,0083h ; dividend

mov bl, 2h ; divisor

DIV bl ; al = 41h, ah = 01h

; quotient is 41h, remainder is 1

DIV r/m16 mov dx,0 ; clear dividend, high mov ax, 8003h ; dividend, low

mov cx, 100h ; divisor

DIV cx ; ax = 0080h, dx = 0003h

; quotient = 80h, remainder = 3

SAL, SAR – Bitwise Arithmetic ShiftBitwise Arithmetic Shift on first operand– vacated bits are filled with zero for SAL– vacated bits are filled with copies of the original high bit original high bit of the source operand for SARSAR

Example:

mov CL , 3

mov AL ,10110111b

sar AL, 1 ; shift right 1 AL = 11011011b

sar AL, CL ; shift right 3 AL = 11111011b

Advanced Instructions – Advanced Instructions – shiftshift

SHL, SHR – Bitwise Logical Shifts Bitwise Logical Shifts on first operand– number of bits to shift is given by second operand– vacated bits are filled with zero – (last) shifted bit enters the Carry Flag

Example: mov CL , 3

mov AL ,10110111b ; AL = 10110111b

shr AL, 1 ; shift right 1 bit AL = 01011011b, CF = 1

shr AL, CL ; shift right 3 bits AL = 00001011b, CF = 0

<instruction> <instruction> r/m8r/m8(16,32) (16,32) 1/CL/imm81/CL/imm8

Note: shift indeed performsdivision / multiplication by 2

ROL, ROR – bitwise rotate (i.e. moves round) on the first operand

Example:

mov CL, 3

mov BH ,10110111b ; BH = 10110111b

rol BH, 01 ; rotate left 1 bit BH = 01101111b

rol BH, CL ; rotate left 3 bits BH = 01111011b

Advanced Instructions - Advanced Instructions - rotaterotate

RCL, RCR –bitwise rotate on first operand and Carry Flag

Example:

mov BH ,10110111b ; BH = 10110111b , CF = 0

rcl BH, 01 ; rotate left 1 bit with CF BH = 01101110b , CF = 1

<instruction> r/m8<instruction> r/m8(16,32)(16,32) 1/CL/imm8 1/CL/imm8

LOOP, LOOPE, LOOPZ, LOOPNE, LOOPNZ – loop with counter (CX or ECX)

Example: mov ax, 1 mov cx, 3 my_ loop: add ax, ax loop my_ loop, cx

1. decrements its counter register (in this case it is CX register)

2. if the counter does not become zero as a result of this operation, it jumps to the given label

Note: if a counter is not specified explicitly, the BITS setting dictates which is used.

The BITS directive specifies whether NASM should generate code designed to run on a processor operating in 16-bit mode, or code designed to run on a processor operating in 32-bit mode. The syntax is BITS 16 or BITS 32.

LOOPE ≡ LOOPZ: jumps if the counter ≠ 0 and Zero Flag = 1

LOOPNE ≡ LOOPNZ: jumps if the counter ≠ 0 and Zero Flag = 0

Advanced Instructions - Advanced Instructions - looploop

Note: LOOP instruction does not set any flags

Stack

• STACK is temporary storage memory areatemporary storage memory area

• every cell every cell in stack is of size 2 / 4 bytes2 / 4 bytes

• ESPESP register points to the top of stacktop of stack

• stack addresses stack addresses go from high to lowhigh to low

Read-only Data Segment

.bss

.data

.text .rodata

ESPESP

Stack Operations

• PUSH PUSH - - push data on stackpush data on stack– decrements ESP by 2 / 4 bytes (according to the operand size)

– stores the operand value at ESP address on stack (in Little Endian manner)

ESP

0x00

0x03

stack

ESP

0x00

0x03

0x12

0xAF

stackESP

stack

push bx pop axESP

stack

pop bx

• POPPOP - load a value from the stackload a value from the stack– reads the operand value at ESP address on stack (in Little Endian manner)

– increment ESP by 2 / 4 bytes (according to operand size)

Example:mov ax, 3

mov bx, 0x12AF

push ax

push bxpop ax ; ax = 0x12AF

pop bx ; bx = 3

ESP

0x00

0x03

stack

push ax

• PUSHADPUSHAD (push all double) - pushes pushes EAXEAX, , ECXECX, , EDXEDX, , EBXEBX, , ESPESP, , EBPEBP, , ESI,ESI, and and EDIEDI onto the stack onto the stack

• PUSHFD PUSHFD (push flags double) - push flags register onto the stackpush flags register onto the stack

ESPstack

ESP

stack

PUSHADEAX

ECX

EDX

EBX

ESP

EBP

ESI

EDI

ESP

stack

PUSHFDEAX

ECX

EDX

EBX

ESPESP

EBP

ESI

EDI

EFLAGS

original ESP value before PUSHAD

• POPADPOPAD (pop all double) - pop a dword from the stack into each one of pop a dword from the stack into each one of (successively) EDIEDI, , ESIESI, , EBPEBP, nothing , nothing (placeholder for ESP), , EBXEBX, , EDXEDX, , ECX,ECX, and and EAXEAX

• POPFDPOPFD (pop flags double) - pop a dword and stores it in EFLAGSpop a dword and stores it in EFLAGS

POPFD

ESP

stack

EAX

ECX

EDX

EBX

ESP

EBP

ESI

EDI

ESPstack

POPHAD

Stack Operations

Function Calls - Stack Frame

main: ; caller codepushad ; backup registerspushfd ; backup flags registerpush dword 2 ; push argument #2push dword 1 ; push argument #1CALL myFunc ; call the function the address of the next instruction of callercaller

; (i.e. return address) is pushed automatically onto the stackmov [answer], eax ; retrieve return value from EAXadd esp, <argumentsSize> ; "delete" function arguments (in our case argumentsSize = 2 * dword = 8 bytes)popfd ; restore flags registerpopad ; restore registers

myFunc: ; callee codepush ebp ; backup EBPmov ebp, esp ; reset EBP to the current ESPsub esp, <localsSize> ; allocate space for localsmov ebx, [ebp+12] ; get second argumentmov ecx, [ebp+8] ; get first argumentmov [ebp-4], ebx ; initialize a local... ; function codemov eax, <return value> ; put return value into EAXmov esp, ebp ; move EBP to ESPpop ebp ; restore old EBPRET ; return from the function

stack

local variables

space

PUSHAD

PUSHFD

2

1

return address (of “mov [answer], eax”)

EBP

…

EBP + 12

EBPEBP

ESPESP

EBP + 8

EBP - 4

Gdb-GNU Debugger

Run Gdb from the console by typing gdbgdb executableFileName

Adding breaking points points by typing:

break label Start debugging by typing:

run parameters (argv)

• si – one step forward

• c – continue to run the code until the next break point.

• q – quit gdb

• p $eax – prints the value in eax

• x $esp+4 – prints the address in esp + 4 hexadecimal and the value (dword) that stores in this address. It is possible to use label instead of esp. Type x again will print the next dword in memory.

Task 1

Write a program that gets a string containing binary digits and convert it to an unsigned hexadecimal number.

You can get up to 32 binary digits.

Example:

Input: 10111011 Output: BB

(Ignore the input character ‘\n’)

Assignment 1

Task 2

Practice parameters passing from assembly to C and vice versa.

Write a C program that:

• prompts for and reads 32-bit decimal integer x from the user

• calls a function ‘calc_func(int x)’ written in ASSEMBLY

Write an assembly function ‘calc_func' that:

• calls a C function 'check' (as defined below) to check if the number is legal

• if x is legal– computes y= quotient((2^x)/x) (e.g. 5/2 = 2)

– calls ‘printf’ to print the result y in decimal

• if x is not legal, calls ‘printf’ to print "x is out of range“

Write a C function ‘check(int x)’ that:

• returns False (0) if x is non-positive or greater than 31

• returns True (1) otherwise

main1.c file for task 1#include <stdio.h>#define BUFFER_SIZE (128)

extern int my_func(char* buf);

int main(int argc, char** argv)

{

char buf[BUFFER_SIZE];

printf("Enter string: ");

fgets(buf, BUFFER_SIZE, stdin);

printf("Got string: %s\n", buf);

my_func(buf);

return 0;

}

section .rodataLC0:

DB "The result is: %s", 10, 0 ; Format stringSection .bssLC1:

resb 32

section .textalign 16global my_funcextern printf

my_func:push ebpmov ebp, esp ; Entry code - set up ebp and esppusha ; Save registers

mov ecx, dword [ebp+8] ; Get argument (pointer to string)

; Your code should be here...; Your code should be here...

push LC1 ; Call printf with 2 arguments: a numberpush LC0 ; and pointer to format string.call printfadd esp, 8 ; Clean up stack after call

popa ; Restore registersmov esp, ebp ; Function exit codepop ebpret

The skeleton of assembly programs in both tasks