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Understanding Assembler: Hand Assembling and Creating Assembler Code, Slides of Microprocessor and Interfacing

Punjab Engineering College Microprocessor and Interfacing

An overview of hand assembling assembler code and creating assembler code files. It covers the process of translating high-level language (hll) to assembler, editing assembler language programs, and the role of an assembler in producing machine code. Understanding hand assembly is essential for computer engineers to gain a complete understanding of microprocessor or microcontroller architecture.

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Lecture Overview

•Editing and Assembler

–So far have covered the assembler language

instructions and translating a program to

assembler

–Now

•Creating the assembler code file

•What goes on during the execution of the assembler

•REF: Chapter 4.10 and 4.11

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Discover Slides of Microprocessor and Interfacing Punjab Engineering College

Partial preview of the text

Download Understanding Assembler: Hand Assembling and Creating Assembler Code and more Slides Microprocessor and Interfacing in PDF only on Docsity!

Lecture Overview

• Editing and Assembler

– So far have covered the assembler languageinstructions and translating a program to

assembler

– Now

• • Creating the assembler code fileWhat goes on during the execution of the assembler

• REF: Chapter 4.10 and 4.

Assembler language files

• Translated HLL to assembler.

• Edit the assembler language program into aneditor.

– This is sometimes called a source file

• What is done with the source file? – It is input as a data file to a program called and

– assembler.Most times you simply use a simple text editor and set

– up the various fields at tabs.The file is save as name.asm typically.

And a machine code listing

Another example

Previous example

• The code – ORG $

– – unkwnguess FCBRMB 1 47 location for guess varset unknown value

– – incr RMBORG 1 $C000 location for increment var

– – LDAA #50STAA guess

– – LSRASTAA incr first guess is 50divide by 2

– – LSRBLDAB incrset to 25

Previous ex continued

– – tol CMPABEQ unknwndone

– – * guess is too highBLT low

– – incadj SBALSRB subtract incrementincr = incr/

– – BCCINCB ceilgd carry was 1 so make ceil

– – ceilgdlow BRAABA tol add increment

– – done BRASTAA incadjguess done

Hand assembly continued

• The lines hand assemble

– ORG $C

– LDAA

– addr contents

– $C000 86 32

– $32 is 0011 0010

– and hexidecimal

– for 50 (32+16+2)

The next instruction

– STAA guess

– $C002 $97 11

– As address of

– Guess is $

And then

• LDAB incr

• $C007 D6 12

• LSRB

• $C009 54

And then

• tol CMPA unkwn

• $C00A 91 10

• As unkwn is at memory

• Location $

• tol will have value of

• $C00A

And then

– SBA

– $C010 10

– incadj LSRB

– $C011 54

And then

– – $C012 24 rrjumpBCC ceilgd

– – $C014 5CINCB

– – $C015ceilgd 20 back_to_tolBRA tol

– will have value of $C014 so rrjump gets value of $01ceilgd had value $C015 and at time of BCC execution, the PC

– back to location $C00A from location $C017 is a negative 13back to tol will need a negative offset. The PC for calculation

– bytes.Or back_to_tol^ 0001 0111 – 0000 1010 = gets $F3 WHY?^ 0000 1101

The full code

– (^) Loc contentsORG $0010 comment
– (^) $0010 47 ORGunkwn FCB 47 $
– $0011 00$0012 00 guessincr RMB 1RMB 1
– (^) $C000 86 64ORG LDA #100 $C
– $C002 97 11$C004 44 LSRASTAA guess
– $C005 97 12$C007 D6 12 LDAB incrSTAA incr
– $C009$C00A 91 10 54 LSRBtol CMPA unkwn
$C00C 27 0E BEQ done

  $C00F 2D 09$C010 10 BLT lowSBA   $C011$C012 24 01 54 incadj LSRBBCC ceilgd   $C014 5C$C015 20 F3 (^) ceilgd BRAINCB tol   $C017 18$C018 20 F7 low ABABRA incadj  $C01A 97 10 done STAA guess   Symbol tableunkwn $0010 tol $C00A low $C   guessincr $0012$0011 incadj $C011ceilgd $C015 done $C01A

Understanding Assembler: Hand Assembling and Creating Assembler Code, Slides of Microprocessor and Interfacing

Related documents

Partial preview of the text

Download Understanding Assembler: Hand Assembling and Creating Assembler Code and more Slides Microprocessor and Interfacing in PDF only on Docsity!

Lecture Overview

• Editing and Assembler

– So far have covered the assembler languageinstructions and translating a program to

assembler

– Now

• • Creating the assembler code fileWhat goes on during the execution of the assembler

• REF: Chapter 4.10 and 4.

Assembler language files

• Translated HLL to assembler.

• Edit the assembler language program into aneditor.

– This is sometimes called a source file

• What is done with the source file? – It is input as a data file to a program called and

– assembler.Most times you simply use a simple text editor and set

– up the various fields at tabs.The file is save as name.asm typically.

And a machine code listing

Another example

Previous example

• The code – ORG $

– – unkwnguess FCBRMB 1 47 location for guess varset unknown value

– – incr RMBORG 1 $C000 location for increment var

– – LDAA #50STAA guess

– – LSRASTAA incr first guess is 50divide by 2

– – LSRBLDAB incrset to 25

Previous ex continued

– – tol CMPABEQ unknwndone

– – * guess is too highBLT low

– – incadj SBALSRB subtract incrementincr = incr/

– – BCCINCB ceilgd carry was 1 so make ceil

– – ceilgdlow BRAABA tol add increment

– – done BRASTAA incadjguess done

Hand assembly continued

• The lines hand assemble

– ORG $C

– LDAA

– addr contents

– $C000 86 32

– $32 is 0011 0010

– and hexidecimal

– for 50 (32+16+2)

The next instruction

– STAA guess

– $C002 $97 11

– As address of

– Guess is $

And then

• LDAB incr

• $C007 D6 12

• LSRB

• $C009 54

And then

• tol CMPA unkwn

• $C00A 91 10

• As unkwn is at memory

• Location $

• tol will have value of

• $C00A

And then

– SBA

– $C010 10

– incadj LSRB

– $C011 54

And then

– – $C012 24 rrjumpBCC ceilgd

– – $C014 5CINCB

– – $C015ceilgd 20 back_to_tolBRA tol

– will have value of $C014 so rrjump gets value of $01ceilgd had value $C015 and at time of BCC execution, the PC

– back to location $C00A from location $C017 is a negative 13back to tol will need a negative offset. The PC for calculation

– bytes.Or back_to_tol^ 0001 0111 – 0000 1010 = gets $F3 WHY?^ 0000 1101

The full code

Lecture summary

• Have looked at hand assembly

• a very useful skill