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Computer Architecture and Assembly Language

Computer Architecture and Assembly Language. Data Representation Basics Bit - the basic unit of information: (true/false) or (1/0). Byte structure : a byte has 8 bits. 7. 6. 5. 4. 3. 2. 1. 0.

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Computer Architecture and Assembly Language

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  1. Computer Architecture and Assembly Language

  2. Data Representation Basics • Bit - the basic unit of information: • (true/false) or (1/0) • Byte structure: • a byte has 8 bits 7 6 5 4 3 2 1 0 MSB (Most Significant Bit) LSB (Least Significant Bit)

  3. Registers: CPU contains a unit called “Register file”. This unit contains the registers of thefollowing types: 1. 8-bit general registers: AL, BL, CL, DL, AH, BH, CH, DH 2. 16- bit general registers:AX, BX, CX, DX, SP, BP, SI, Dl 3. 32-bit general registers: EAX, EBX, ECX, EDX, ESP, EBP,ESI, EDI (Accumulator, Base, Counter, Data, Stack pointer, Base pointer, Source index, Destination Index) 4. Segment registers: ES, CS ,SS, DS, FS, GS 5. instruction pointer: EIP Note: the registers above are a partial list. There are more registers.

  4. EIP - instruction pointer: contains offset (address) of the next instruction that is going to be executed. Exists only during run time. The software change it by performing unconditional jump, conditional jump, procedure call, return. AX,BX,CX,DX - 16-bit general registers: contains two 8-bit registers:Example: AH,AL (for AX) EAX - 32-bit general purpose register: lower 16 bits are AX. segment registers: we use a flat memory model –32bit 4GB address space, without segments. So for this course you can ignore segment registers. ESP - stack pointer: contains the address of last used dword in the stack. high byte low byte XH XL

  5. Assembly language program • written in assembly language consists of a series of processor instructions and meta-statements, comments and data • translated by an assemblerinto machine language instructions (binary code) that can be loaded into memory and executed • Example • assembly code: • MOVAL, 61h ; load AL with 97 decimal (61 hex) • binary code: • 1011000001100001 • 10110 a code of instruction 'MOV' • 000 an identifier for a register 'AL' • 01100001 97 decimal (61 hex) • The Netwide Assembler (NASM) is an assembler and for x86 architecture

  6. Basic assembly instructions: Each NASM standard source line contains a combination of the 4 fields: label: (pseudo) instructionoperands ; comment Either required or forbidden by an instruction optional fields Notes: 1. backslash (\) uses as the line continuation character: if a line ends with backslash, the next line is considered to be a part of the backslash-ended line.2. no restrictions on white space within a line.3. a colon after a label is optional. Examples: 1. mov ax, 2 ; moves constant 2 to the register ax2. buffer: resb 64 ; reserves 64 bytes

  7. A typical instruction has 2 operands. • The left operand is the target operand, while the right operand is the source operand • 3 kinds of operands exists: • Immediate, i.e. a value • Register, such as AX,EBP,DL • Memory location; a variable or a pointer. • One should notice that the x86 processor does not allow • both operands be memory locations. • mov [var1],[var2] Instruction arguments

  8. MOV – move data mov r/m8,reg8(copies content of 8-bit register (source) to 8-bit register or 8-bit memory unit (destination) ) mov reg32,imm32(copies content of 32-bit immediate (constant) to 32-bit register) - In all forms of the MOV instruction, the two operands are the same size Examples:mov EAX, 0x2334AAFFmov [buffer], ax Note: NASM doesn’t remember the types of variables you declare. It will deliberately remember nothing about the symbol var except where it begins, and so you must explicitly code mov word [var], 2. Move instructions:

  9. ADD: add integers add r/m16,imm16 (adds its two operands together, and leaves the result in its destination (first) operand) Examples:add AX, BX Basic arithmetical instructions: ADC: add with carry adc r/m16,imm8(adds its two operands together, plus the value of the carry flag, and leaves the result in its destination (first) operand) Examples:adc AX, BX (AX gets a value of AX+BX+CF)

  10. SUB: subtract integers sub reg16,r/m16 (subtracts its second operand from its first, and leaves the result in its destination (first) operand) Examples:sub AX, BX Basic arithmetical instructions (Cont.): SBB: subtract with borrow sbb r/m16,imm8 (subtracts its second operand, plus the value of the carry flag, from its first, and leaves the result in its destination (first) operand) Examples:sbb AX, BX (AX gets a value of AX-BX-CF)

  11. INC: increment integer inc r/m16 (adds 1 to its operand) * doesnot affect the carry flag; affects all the other flags according to the result Examples:inc AX Basic arithmetical instructions (Cont.): DEC: decrement integer dec reg16 (subtracts 1 from its operand) * doesnot affect the carry flag; affects all the other flags according to the result Examples:dec byte [buffer]

  12. NEG, NOT: two's and one's complement neg r/m16 (replaces the contents of its operand by the two's complement negation - invert all the bits, and then add one) not r/m16 (performs one's complement negation- inverts all the bits) Examples: neg AL ; (if AL = (11111110), it becomes (00000010)) 11111110 + 00000010 = 100000000 = 0 not AL ; (if AL = (11111110), it becomes (00000001)) Basic logical instructions:

  13. OR: bitwise or or r/m32,imm32 (each bit of the result is 1 if and only if at least one of the corresponding bits of the two inputs was 1; stores the result in the destination (first) operand) Example:or AL, BL (if AL = (11111100), BL= (00000010) => AL would be (11111110)) Basic logical instructions (Cont.): AND: bitwise and and r/m32,imm32 (each bit of the result is 1 if and only if the corresponding bits of the two inputs were both 1; stores the result in the destination (first) operand) Example:and AL, BL (if AL = (11111100), BL= (11000010) => AL would be (11000000))

  14. CMP: compare integers cmp r/m32,imm8 (performs a ‘mental’ subtraction of its second operand from its first operand, and affects the flags as if the subtraction had taken place, but does not store the result of the subtraction anywhere) Example:cmp AL, BL (if AL = (11111100), BL= (00000010) => ZF would be 0) (if AL = (11111100), BL= (11111100) => ZF would be 1) Compare instruction:

  15. Each instruction of the code has its offset (address from the beginning of the address space). If we want to refer to the specific instruction in the code, we should mark it with a label: my_instruction: add ax, ax… - label can be with or without colon- an instruction that follows it can be at the same or the next line- a code can’t contain two different non-local (as above) labels with the same name Labels definition (basic):

  16. JMP: jump to instruction Usually it takes the form: jmp label *see section B.4.130 JMP in the nasm manual for full specification Tells the processor that the next instruction to be executed is located at the label that is given as part of the instruction. Example: mov eax,1 inc_again: ; in this case it is infinite loop! inc eax jmp inc_again mov ebx,eax ; never reached from this code … Unconditional Jump:

  17. Conditional Jumps: JE,JG, JL, JGE, JLE, JNE: jump to instruction if condition is satisfied Usually it takes the form: j<cond> label*see section B.4.128 JMP in the nasm manual for full specification Execution is transferred to the target instruction only if the specified condition is satisfied. Usually, the condition being tested is the result of the last arithmetic or logic operation. Example:read_char: mov dl,0 . . . (code for reading a character into AL) . . . cmp al, ‘a’ ; compare the character to ‘a’ je a_received ; if equal, jump to a_received inc cl ; otherwise, increment CL and jmp read_char ;go back to read another a_received: …

  18. DB, DW, DD : declaring initialized data DB, DW, DD, DQ (DT, DDQ, and DO) are used to declareinitialized data in the output file. They can be invoked in a wide range of ways: db 0x55 ; just the byte 0x55 db 0x55,0x56,0x57 ; three bytes in succession db 'a',0x55 ; character constants are OK db 'hello',13,10,'$‘ ; so are string constants dw 0x1234 ; 0x34 0x12 dw 'a' ; 0x41 0x00 (it's just a number) dw 'ab‘ ; 0x41 0x42 (character constant) dw 'abc' ; 0x41 0x42 0x43 0x00 (string) dd 0x12345678 ; 0x78 0x56 0x34 0x12 (dword) Example var: dd 0 ; define variable ‘var’ of size dword, initialized by 0

  19. DT, DDQ, and DO : declaring initialized data dq 0x1122334455667788 ; 8 bytes ddq 0x112233445566778899aabbccddeeff00 ; 16 bytes do 0x112233445566778899aabbccddeeff00 ; 16 bytes dd 1.234567e20 ; floating-point constant dq 1.234567e20 ; double-precision float dt 1.234567e20 ; extended-precision float

  20. Assignment 0 You get a simple program that receives a string from the user. Then, it calls to a function (that you’ll implement in assembly) that receives one string as an argument and should do the following: 1. Convert every letter uppercase letter to lower case later and every lowercase letter to upper case latter. 2. Convert ‘(’ into ‘[’ and ‘[‘ into ‘{‘. 3. Convert ‘)’ into ‘]’ and ‘[‘ into ‘{‘. 4. Convert each digit n to a character which follows by n places in the ascii table. 5. Count the number of characters which aren’t uppercase or lowercase letter. e.g. “53: [heLL() WorLd]!" → “:6:{HEll[]wORlD}!“ Returns 8 The function shall return the number of characters which aren’t uppercase or lowercase letter (the output should be just the number) . The characters conversion should be in-place.

  21. section .data ; data section, read-write an: DD 0 ; this is a temporary var section .text ; our code is always in the .text section global do_str ; makes the function appear in global scope extern printf ; tell linker that printf is defined elsewhere ; (not used in the program) do_str: ; functions are defined as labels push ebp ; save Base Pointer (bp) original value mov ebp, esp ; use base pointer to access stack contents pushad ; push all variables onto stack mov ecx, dword [ebp+8] ; get function argument ;;;;;;;;;;;;;;;; FUNCTION EFFECTIVE CODE STARTS HERE ;;;;;;;;;;;;;;;; mov dword [an], 0 ; initialize answer label_here: ; Your code goes somewhere around here... inc ecx ; increment pointer cmp byte [ecx], 0 ; check if byte pointed to is zero jnz label_here ; keep looping until it is null terminated ;;;;;;;;;;;;;;;; FUNCTION EFFECTIVE CODE ENDS HERE ;;;;;;;;;;;;;;;; popad ; restore all previously used registers mov eax,[an] ; return an (returned values are in eax) mov esp, ebp pop ebp ret

  22. To assemble a file, you issue a command of the form > nasm -f <format> <filename> [-o <output>] [ -l listing] Example: > nasm -f elf mytry.s -o myelf.o It would create myelf.o file that has elf format (executable and linkable format).We use main.c file (that is written in C language) to start our program, and sometimes also for input / output from a user. So to compile main.c with our assembly file we should execute the following command: gcc –m32 main.c myelf.o -o myexe.out The -m32 option is being used to comply with 32- bit environment It would create executable file myexe.out.In order to run it you should write its name on the command line: > myexe.out Running NASM

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