### Cherri, time to put up or shut up

greenspun.com : LUSENET : TimeBomb 2000 (Y2000) : One Thread

Cherri tries to pass herself off as an "embedded systems expert". However, she seems to be unable to answer a simple question about BCD (binary coded decimal) arithmetic in microprocessors used in these systems. Here's her last chance to salvage something of her reputation:

Cherri:

Is it or is it not true that you will get a BCD overflow if you use BCD arithmetic when adding the BCD value "01" to the BCD value "99", using BCD instructions on two-digit BCD data?

Any real expert on microprocessors, embedded or otherwise, would have no trouble answering this question. If you can't answer this question correctly, just admit it and we'll know you're not an expert, but at least not a liar. If you can answer it correctly, please explain your repeated claim that the rules of binary arithmetic have some relevance to BCD overflow.

-- Steve Heller (stheller@koyote.com), January 27, 2000

Steve, can I be your control subject? I'm a shrink, and only have a ~vague impression~ of embeddeds from reading some of the stuff here that wasn't too uh narcoleptic, heh.

So my guess answer would be "no" because y'all have referred to a buffer. My moronic nontech guess is the buffer would catch something as common as 1+99.

So compare Cherri's answer to mine, and if hers is comparably superficial, then Steve wins.

-- Hokie (Hokie_@hotmail.com), January 27, 2000.

Steve,

Help me remember, been so damn long since I had to add in binary and do this low level stuff, how big are the registers in bits. Is it standard

8 bits = 1 byte = nibble(slang)

1 byte = 8 bit 2 byte(s) = word = 16 bit 4 byte(s) = long word = 32 bit 8 byte(s) = quad word = 64 bit

So the question is what is the size of the registers used for arithmetic. I am assuming, I hate it when I do that, todays processors are at least 16 bit.

Thanks,

Iglow Inthedark

-- Iglow Inthedark (u235_36@yahoo.com), January 27, 2000.

Sorry,

Darn formatting,

1 byte = 8 bit

2 byte(s) = word = 16 bit

4 byte(s) = long word = 32 bit

8 byte(s) = quad word = 64 bit

Iglow Inthedark

-- Iglow Inthedark (u235_36@yahoo.com), January 27, 2000.

Steve, why do you have to be SO antagonistic? Just ask the question in a nice way. You are staring to sound a little shrill on the forum. LIke I said before, I like you. But you are bringing a lot of conflict on yourself. Take a lesson from Ed Yourdon and others and chill.

-- JoseMiami (caris@prodigy.net), January 27, 2000.

Uhhh Steve . . . given that Cherri was alot more right about Y2K and embedded chips than you were, I'd say she's already demonstrated that she's at least more of an expert than you. Stop picking at nits and admit that you were wrong. You're coming off sounding like an a-hole with this line of reasoning sir.

-- third party (observer@home.com), January 27, 2000.

Okay, it is several hours later...By now I could have called a freind for the yes/no answer...

-- Hokie (Hokie_@hotmail.com), January 27, 2000.

To explain why I've written my question in such a seemingly antagonistic way: it's because I've been unable to get a straight answer to a simple question. Cherri, a self-proclaimed "embedded systems expert", has been very happy to tell others that they don't know what they're talking about, often in a very overbearing and arrogant way. If she can dish it out, she should be able to take it.

As for the trolls who keep trying to distract attention from this point by bringing up my prior errors in Y2K prediction: I've already said I was wrong about the impact of the rollover. Now if Cherri will just admit the same about her supposed expertise, we can lay this to rest.

-- Steve Heller (stheller@koyote.com), January 27, 2000.

Cherri, so infuriated me, from her first post (that I saw), because she was so full of her/its self, that she/he had just fixed one small area, to redeem all man-kind. I posted a response to her, First, on your knees, to Thank Whom ever, for the knowledge, for which, she was blessed, and second, On Your Knees, to give Thanks, for the fortitude blessings, to carry it through.

-- I Had One (l@stnerve.com), January 27, 2000.

Steve do we have a little extra time on our hands?

-- Billy Vyper (billy_vyper@postmark.net), January 27, 2000.

Jerry Springer is always looking for people angry enough to throw chairs, maybe this "embedded" debate belongs there. I'd say every point now made is moot... The results are +95% in, and the remaining percentage is, as they say in the 4th quarter of a blowout, "garbage time". Mr. Heller, with all due respect, this does not become you.

-- Bemused (and_amazed@you.people), January 27, 2000.

Josemiami,Thired Party and Mr. Bemuse

Gentlemen you have collectively arrived at some very interesting obsrvations and asumptions. And as with that other oriface on the oppisite end of the one that the sounds are coming from..We each all have one of them.

Now where to start; at the first I suppose! Mr.Josemiami, sir you have not been here long enough to have been on the receiving end of Mistress Cherri's "superior (insufferable) attitudes, and you most definatley have not seen her in action, when she pounces upon some one for not being quite up to par on a given subject. Untill you have, plese sit in the bleachers and watch silently sir. You might be rooting for the wrong party; as you most definately are in this case.

Mr. Third Party, evidently you are another one of those "after the fact (in your pale opinion)(roll over) new comer pollys. State your qualifications before you state your unsolisited opinions. With the pollys who had courage of their convictions to come in the forum last year (Mr. Decker, Flint and others). We all at least knew that they where indeed experienced (and just not in having an acess to the internet).

Mr. Bemusee, sir you "asume" too much. The problems have always been the Year two thousand problems; (not the thirty days problems). As for your shallow and unenlightened observations about the embedded systems...They are doing what I thought that they would do, and in the time frame I stated that they would do it in. And sir...Because of human nature (and it's history of destructive reaction). I am still at an infomagic level of alertness.

"As for me...I shall finish the Game"!

~~~~~~~~~~~~~~~~~~~~~~~~~Shakey~~~~~~~~~~~~~~~~~~~~~

-- Shakey (in_a_bunker@forty.feet), January 28, 2000.

What Shakey said :o)

Come on Cherri, stop playing with the Boeings and answer the question pumpkin :o)

-- Andy (2000EOD@prodigy.net), January 28, 2000.

The usual trolls are attacking me to try to distract attention from my point. To them, I have one comment:

I was wrong about Y2K's initial impact, but you are worthless scum. Tomorrow, I may very well be right about something else, but you will still be scum.

-- Steve Heller (stheller@koyote.com), January 28, 2000.

Cherri is a royal pain in the neck and most usually does not contribute any valuable input to Y2K analysis, embeddeds included.

Fully agree with Shakey, although the list of polly a**holes should also include Mr.BS Sane and "I am" BS.

Folks, it's not D2K, it's Y2K, the YEAR 2000 problem

Take care

-- George (jvilches@sminter.com.ar), January 28, 2000.

Hokie,

tooo funnny....Consumer is passing Hokie another lifeline. LOL

-- consumer (shh@aol.com), January 28, 2000.

If it is a yes or no, it would have to be yes, although I would not assume that was it in real life, I would have to read a schematic to see what it was supposed to do in under "overflow" conditions. Such as where "web pages" had their date changed to 19100 and some might be set up to increment correctly. It is never good to assume what it will do, it is always better to check or you can have it come back to bite you.

Back to my initial point, which has nothing to do with BCD, decimal overflow does not cause digital overflow.

As the expression "embedded system" is a new one, I did not take a course called "embedded system engineering" or anything like that, I learned and have worked on "non computer analog/digital devices". "Embedded" is a catchword that has become popular during the past few years.

It really is sad that it is so important to you to "disprove" my ability.

Your arrogance in stating that if you ask me a question and I answer incorrectly (in your opinion) that somehow my 30 years of experience will cease to be disprove or exist is rather pathetic exhibition of your insecure ego.

Does your ego demand that I be disproved so you can feel better about yourself? Why should it bother you that someone else may know things and be capable of doing things that you are not? Why is all of this so important to you? I did not start out here by telling everyone my background, as I felt it was not important, and did nothing to "prove" that I knew what I was talking about. If I say Paula Gordon has no experience and does not know what she is talking about when it comes to "embedded", anyone here can see that for themselves if they want to, simply by going to her home page and reading her biography. The field of digital electronics is a highly technical one that takes a lot of work to learn and many years of experience to get proficient in. It is not a field a person can learn on the web in weeks/months or bluff their way through. It is based in physics. Programming can be done in any manner of ways, but still much follow the rules of digital logic.

The reason that so many people believed there would be more problems at the rollover that actually occurred is that too many people who did not have knowledge of embedded speculated about it without knowing how it worked.

-- Cherri (sams@brigadoon.com), January 28, 2000.

Just for fun,

Lets take a 8 bit binary number like, oh say 99(dec), and then add 1(dec). We get this-

01100011 = 99

00000001 = 1

01100100 = 100

I saw no "binary overflow", the largest decimal number an unsigned 8 bit(1 byte) binary number can hold is 255. Actually it is 256 but who counts zeros but us computer scientists.

What fun!

-- Iglow Inthedark (u235_36@yahoo.com), January 28, 2000.

I had so much fun with the last one, lets do it again.

Lets take a 8 bit binary number like, oh say 255(dec), and then add 1(dec). We get this-

11111111 = 255

00000001 = 1

00000000 = 256

I saw "binary overflow", the largest decimal number an unsigned 8 bit(1 byte) binary number can hold is 255. Remember we can only use a 8 bit(1 byte) register.

What fun! This will conclude the elementary examples of binary addition.

-- Iglow Inthedark (u235_36@yahoo.com), January 28, 2000.

Iglow,

Want to move on to subtraction, division and multiplication?

:o)

-- Cherri (sams@brigadoon.com), January 28, 2000.

Go Cherri, go!

-- I am (LMAO2@pathetic.idiots), January 28, 2000.

For everyone interested.

You can find a nifty little calculator in win95 or win98 that will help in understanding binary conversions and arithmetic. You must set it to scientific mode to get the binary and decimal modes. You can also get hex and octal.

Try punching in a number in decimal mode and then you can convert it to binary by switching modes. Great fun.

Iglow

-- Iglow Inthedark (u235_36@yahoo.com), January 28, 2000.

Hi,

Original question was looking for an answer in Binary Coded Decimal (BCD) not binary. Most early microprocessors included separate instructions to process in BCD and therefore, the overflow is different than in binary.

Early 6502, 6800 processors provided these instructions in their hardware instruction sets. Handy for certain functions too.

wally

-- wally wallman (wally_yllaw@hotmail.com), January 28, 2000.

Wally,

Thank you for pointing that out. I guess my little examples were really just for input to the subject.

If you read my earlier question to the start of this thread you will find that I was looking for additional information about the size of the registers. I am aware of BCD and it's weirdness. I was taught to convert BCD to decimal, decimal to binary and then do the math. Reverse the process to get BCD.

I am sure you are very familiar with BCD and I am just a Database junkie so I do not get to the guts of low level stuff like BCD. If my memory serves me correctly, every decimal digit requires 4 BCD bits and you must add leading 0's to make up the 4 bits. So,

21(decimal) = (00100001)BCD

I think, man it has been waaaaayyyyy too long since I did any of this stuff. Time to dig out the old comp. sci. books.

Iglow

-- Iglow Inthedark (u235_36@yahoo.com), January 28, 2000.

A couple of notes.

My little conversion process for BCD to binary was used for teaching, so I am sure you could have stopped at BCD to decimal and done the math.

The register size question was necessary for me to conclude if the orginal problem would result in "overflow".

I am not a "chip" programmer, just looking for enough information to try to provide myself and the group with more information.

Thank god for "4GL's". I will never have to write another sort routine again(not really true).

Gday,

Iglow

-- Iglow Inthedark (u235_36@yahoo.com), January 28, 2000.

Okay, now that you've explained it further, I think I've finally figured out what your original comment meant, and accept your answer. Sorry about the misunderstanding.

-- Steve Heller (stheller@koyote.com), January 28, 2000.

Steve,

I am still looking for the size of the register in bits to determine an answer to your original question.

Once again I am forced to assume if you are dealing with 8 bit registers:

99(decimal) = 10011001

1(decimal) = 00000001

100(decimal) = unable to fit into an 8 bit register(overflow)

If the register on the chip is 16 bit then no overflow. So the original question still stands. What is the size of the register used internally within the chip? 8 bit? 16 bit? 32 bit? 64 bit?

Gday,

Iglow

-- Iglow Inthedark (u235_36@yahoo.com), January 28, 2000.

Iglow:

I was assuming a one-byte (two-digit) register, as I thought was implied in my original question.

-- Steve Heller (stheller@koyote.com), January 28, 2000.

Steve,

Thanks, gotcha.

-- Iglow Inthedark (u235_36@yahoo.com), January 28, 2000.