Implement an assembly language program, Assembly Language

Assignment Help:

A good starting point for your program is the toupper.asm program shown in class. It already queries the user for input and sets up a loop that looks at each character of the input. The source code for toupper.asm 

The bytes entered by the user are ASCII characters. You should convert this to a numerical digit by subtracting the character '0'.

Recall that the last character might be an X and represents the value 10. You will need to special case this. For simplicity, let's not worry about an X appearing in the middle of the input.

You should set up your loop to iterate 10 times and NOT for the length of the string (as is done in toupper.asm) because that will include the newline character at the end of the user input.

Since all the numbers are small, you can use the 8-bit portions of the general purpose registers: AH, AL, BH, BL, CH, CL, DH and DL. This is convenient because you have more registers to play with and because you won't have to mix 8-bit and 32-bit arithmetic.

Develop your program incrementally. After each step, use the debugger to check that you have accomplished the desired goal.

Set up the loop to iterate 10 times, each time storing the next character in an 8-bit register (say AL).

Convert AL into a "number".

Add the special case where AL might be X.

Compute t in another 8-bit register.

Compute t % 11.

Compute sum % 11 in another 8-bit register.

Print out the correct message to the user.

Extra Credit

For 10 points extra credit, submit a separate assembly language program that prints out the value of the check digit given the first 9 digits of an ISBN number. A sample run of your program might look like:

linux2% ./a.out

Enter first 9 digits of ISBN #: 320154197

Check digit: 4

linux2%

linux2% ./a.out

Enter first 9 digits of ISBN #: 045777370

Check digit: 7

linux2%

linux2% ./a.out

Enter first 9 digits of ISBN #: 044101125

Check digit: X

linux2%


Related Discussions:- Implement an assembly language program

8086 microprocessors, program to find negative and positive integers from g...

program to find negative and positive integers from given signed numbers with output and explanation of every instructions.

Memory interface-microprocessor, Memory Interface              ...

Memory Interface                                                                  Figure: Memory Modulation design The memory of a computer contain of number of memo

Dma-how dma works-microprocessor, DMA DMA stands for Direct Memory ...

DMA DMA stands for Direct Memory Access It is uses same Address/Data lines on ISA bus It controls the ISA bus instead of the processor ("bus master") Floppy

Can you write this Program for me please? , $NOMOD51 $NOSYMBOLS ;**********...

$NOMOD51 $NOSYMBOLS ;***************************************************************************** ; Spring 2013 Project ; ; FILE NAME : Project.ASM ; DATE : 3/30/20

Machine level programs-microprocessor, Machine Level Programs In this s...

Machine Level Programs In this section, a few machine levels programming instance, rather then, instruction sequences are presented for comparing the 8086 programming with that

Execution unit and bus interface unit-microprocessor, Execution Unit (EU) a...

Execution Unit (EU) and Bus Interface Unit (BIU) : 8086 consist of two processors called EU and BIU. Two Processors can work parallel. This improves speed of execution. BIU fi

Rics/cisc architecture-microprocessor, RICS/CISC Architecture An essent...

RICS/CISC Architecture An essential aspect of computer architecture is the design of the instruction set for the processor.  The instruction set selected for a specific compute

The pin diagram of 8088-microprocessor, Pin diagram of 8088 : The pin ...

Pin diagram of 8088 : The pin diagram of 8088 is shown in given figure. Most of the 8088 pins and their functions are exactly similar to the corresponding pins of 8086.  Hence

#title:Shifitng of memory, Ask 2. Exchange higher byte of AX and higher byt...

Ask 2. Exchange higher byte of AX and higher byte of BX registers by using memory location 0160 in between the transfer. Then stores AX and BX registers onto memory location 0174 o

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

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