Lectures schedule 2 When ? Topic Lectu re October 20, 2013 Introduction to C Programming in Unix Environment - I 1 October 27, 2013 Introduction to C Programming in Unix Environment - II 2 November 3, 2013 Introduction to Assembly 3 November 17, 2013 November 10, 2013 Functions and System Calls (Assembly) 4 Midterm A December 8, 2013 Unix Processes 5 December 15, 2013 Programs Execution 6 December 22, 2013 Introduction to script languages (Python) 7
Transcript
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Lectures schedule
When ? Topic Lecture
October 20, 2013 Introduction to C Programming in Unix Environment - I
1
October 27, 2013 Introduction to C Programming in Unix Environment - II
2
November 3, 2013 Introduction to Assembly 3
November 17, 2013November 10, 2013
Functions and System Calls (Assembly) 4
Midterm A
December 8, 2013 Unix Processes 5
December 15, 2013 Programs Execution 6
December 22, 2013 Introduction to script languages (Python) 7
January 5, 2014 Web programming 8
Midterm B
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From last lab...
msg is a pointer that points to memory which is in the data segment (read only part)
Abed Asi - ESPL3
char msg2[]= “text”;msg2[0] = ‘n’;
char* msg = “text”;msg[0] = ‘n’;
msg2 is an array of chars that are on the stack
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A User’s View of Computer Systems
Abed Asi - ESPLCourtesy: Guide to Assembly Language Programming in Linux, Sivarama P. Dandamudi, 2005
Define directive takes one of this basic forms DB Define Byte ; allocates 1 byte DW Define Word ;allocates 2 bytes DD Define Doubleword ;allocates 4 bytes DQ Define Quadword ;allocates 8 bytes DT Define Ten Bytes ;allocates 10 bytes
Examples …
Abed Asi - ESPL
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Data Allocation – Examples
Abed Asi - ESPL
sorted DB ‘y’ sorted DB ‘79H’ sorted DB 1111001B
Allocate two bytes of contiguous storage and initialized it to 25159 value DW 25159 automatically converted to its 16-bit hex. equivalent (6247H) address: x x+1
47 62
sorted DB ‘y ’
Pentium uses little-endian
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Uninitialized Data
Reserve space for uninitialized data
Reserve directives RESB Reserve a byte RESW Reserve a Word RESD Reserve a Doubleword RESQ Reserve a Quadword REST Reserve Ten bytes
response RESB 1 buffer RESW 100 total RESD 1
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Multiple Definitions
A string
message DB ‘H’DB ‘E’DB ‘L’DB ‘L’DB ‘O’
more compactly:
message DB ‘HELLO’
Abed Asi - ESPL
Array with 8 elements
message DW 0DW 0DW 0DW 0DW 0
can be abbreviated:
marks DW 0,0,0,0,0marks TIMES 8 DW 0
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Addressing Modes
Assembly language instructions require operands
Possible locations of the operands (addressing mode): in a register internal to the processor in the instruction itself in the main memory (usually in the data segment)
mov instruction copies the content of the source register into the dest. register syntax: mov destination, source ;example: mov EAX, EBX works with 16-bit and 8-bit registers memory-to-memory transfer is not supported!
Abed Asi - ESPL
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Register Addressing Mode
Using processor’s internal registers Very efficient. Why ?
mov EAX, EBXmov BX, CXmov AL, CL
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Immediate Addressing Mode Data is part of the instruction Data is located in the code segment not in the data
segment The immediate data is always a constant
mov AL, 75
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Direct Addressing Mode
response DB ‘Y’ ;allocates a byte, initializes to Ytable1 TIMES 20 DD 0 ;allocates 80 bytes, initializes to 0name1 DB ‘Jim Ray’ ; 7 bytes -----------------------------------------------------------------------------------mov AL, [response] ; copies Y into AL registermov [response], ‘N’ ; N is written into responsemov [name1], ‘K’ ; writes K as the first character of name1move [table1],56 ; 56 is written in the first element – table1[0] = 56
mov EBX, table1 copies the address of table1 into EBX
mov EBX, [table1] what is the difference ?
Abed Asi - ESPL
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Indirect Addressing Mode
How to access second element of table1 ? This mode is required for variable with several elements
Word equivalent of 100 ? maybe a byte equivalent of 100 ? Clarify this by using a type specifier
mov WORD [EBX], 100 mov [EBX], WORD 100mov BYTE [EBX], 100 mov [EBX], BYTE 100
Abed Asi - ESPL
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Type Specifiers
Abed Asi - ESPL
Bytes addressed Type Specifier
1 BYTE
2 WORD
4 DWORD
8 QWORD
10 TBYTE
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Arithmetic Instructions – INC and DEC
increment/decrement the operand by 1
The operand can be either in register or in memory
inc destination ;might be 8-, 16-, 32-bitdec destination ;might be 8-, 16-, 32-bit
Abed Asi - ESPL
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ADD
Used to add 8-, 16-, 32-bit operands
add destination, source ;dest = dest + source
inc EAX is preffered to add EAX,1
Less memory, same speed
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SUB and CMP
SUB used to subtract 8-, 16-, 32-bit operands
sub destination, source ;dest = dest - source
CMP compares two operands (equal, not equal, ..)
CMP behavior is similar to SUB except that the result is not saved So, how can we retrieve the comparison result ?
CMP is typically used with a conditional JUMP inst.
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Unconditional Jump
mov EAX, 1inc again:
inc EAXjmp inc_againmov EBX, EAX…
When to stop ?
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Conditional Jump Jump if the specified condition is satisfied
j<cond> label ;identifies the condition
The condition being tested is the result of the last arithmetic or logic operation
read_char:mov DL,0. . . (code for reading a character into AL). . .cmp AL,0DH ;compares the character to CRje CR_received ; if equal, jump to CR_receivedinc CL ;otherwise, increment CL andjmp read_char ; go back to read another char.
CR_received:mov DL, AL
Abed Asi - ESPL
but, the CMP doesn’t save the result, so what really happens ?!!
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Conditional Jump
Abed Asi - ESPL
Meaning Mnemonic
jump if equaljump if zero
jejz
jump if not equaljump if not zero
jnejnz
jump if greaterjump if not less of equal
jgjnle
jump if greater or equaljump if not less
jgejnl
jump if lessjump if not greater or equal
jljnge
jump if less or equaljump if not greater
jlejng
jump if carry (i.e. if CF = 1) jc
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Multiplication
Two multiplication instr. : mul and imul
mul source
The source operand can be in a register or in memory
