Program to perform a conversion on characters in a text file, Assembly Language

Assignment Help:

Write an assembly language program to perform a rot131 conversion on characters in a text file.  The program should read charatcers from the input file, perform the rot13 conversion, and write the characters to the output file.  The program should read the buffer size (large or small), name of an input file, and the name of an output file from the command line.  For example:

ed-vm% ./rot13  -b l  -i file.txt  -o tmp.txt

The provided main program calls the following routines:

? A getOptions() procedure that reads and checks the command line arguments. The command line arguments must be in the format of:  /b= /i= /o= in that order.  The buffer size should be "l" or "s".  To check the file name, attempt to open the file.  If the files open correctly, the routine should return the file descriptor (for each file).  If there is an error, an appropriate error message should be displayed (see examples).

? A readBuff() procedure to read the input file and return the characters in a buffer.  The procedure will use the buffer address, file descriptor, and buffer size flag to read the file.  The procedure should return the characters in the buffer and the count of characters.  The procedure should also return a flag (true/false) for the last read.

? A rotate13() procedure to perform the rot13 function on the characters in the buffer.  The rot13 function is as follows:

NewCharacter = (Character+13) mod 26

The procedure will use the buffer address and character count.

? A writeBuff() procedure to write the modified buffer to the output file.  The procedure will use the buffer address, file descriptor, and count of characters.

To ensure efficient I/O operations, the program must buffer the data read and written.  For this assignment, the buffer size should be either 2 for the 'small' or 60,000 for the 'large' buffer size.

Note, a utility for testing will be made available on the class web page.

Note, your procedures must be placed into a separate file and linked with the provided main.  Refer to the handout for directions how to link multiple files.  Only the procedures file, not the provided main, will be submitted on-line.  As such, you must not change the provided main!

Testing

A batch file to execute the program on a series of pre-defined inputs will be provided.  The test script compares the program output to pre-defined expected output (based on the example I/O).


Related Discussions:- Program to perform a conversion on characters in a text file

8088 timing system diagram-Microprocessor, 8088  Timing System Diagram ...

8088  Timing System Diagram The 8088 address/data  bus is divided  in 3 parts (a) the lower 8 address/data  bits, (b) the middle 8 address bits, and (c) the upper 4 status/

Data copy/transfer instructions-microprocessor, Data copy/transfer Instruct...

Data copy/transfer Instructions MOV: This data transfer instruction transfers data from one register or memory location to another register or memory location. The source can

Solotuon, using 8086 assembly language that interchange upper four bits to ...

using 8086 assembly language that interchange upper four bits to lower four bits. assume that data store in byte memory and it written back to same location. and assume the data as

Summation Program., Write a program to solve problem 9, Summation Program, ...

Write a program to solve problem 9, Summation Program, on page 179 of chapter 5 in the textbook (book:kip Irvine Assembly Language sixth edition)

How to write an assembly program-microprocessor, How to write an assembly p...

How to write an assembly program The initial step in writing an assembly language program is to identify and study the problem. After studying the problem, choose the logical m

Segment registers-microprocessor, Segment Registers The 8086 addresses ...

Segment Registers The 8086 addresses a segmented memory unlike 8085. The complete 1 megabyte memory, which 8086 is capable to address is divided into 16 logical segments.Thusea

C#, * * * * **** ...

* * * * **** * * * * * How can i print this help me pls

8086 minimum mode system and timing-microprocessor, 8086 Minimum mode Syst...

8086 Minimum mode System and Timing In a minimum mode 8086 system, the microprocessor 8086 is operated in minimum mode by strapping its MN/MX pin to logic 1.All the control si

Seg-segment-assemblers directive-microprocessor, SEG : Segment of a Label:...

SEG : Segment of a Label:- The SEG operator is which is used to decide the segment address of the, variable, label or procedure and substitutes the segment base address in plac

Imul-arithmetic instruction-microprocessor, IMUL: Signed Multiplication: T...

IMUL: Signed Multiplication: This instruction multiplies a signed byte by a signed bit in source operand e in the register AL or signed word in source operand by signed word in th

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd