In the Roots 1 program, it is possible for the answer to be negative.  a = 1, b = 4, , c  = 1.  but you don't get -15!!! 

 

What do you get?

Every time , i tried to compute numbers in which  b^2 < 4ac, I  get a postive back.

Why?

What can you do to fix it?

 

I haven't solve the problem but let see if  somebody can help me with it. I was testing some numbers to see why i couldn't get negative so I try this: Let a = 4, b = 8, and c = 10, we get:

    x = b^2 - 4ac = (8^2) - (4*4*10) = 64 - 160 = -96

 

    but my code gives me:

 

    x = b^2 - 4ac = (8^2) - (4*4*10)  = 160 //Weird  right? Now if  you look carefully  256 - 96 = 160. Also, 2^8 = 256 and  the al register  is 8 bit too.

 

    So this lead me  to look the hex representation of  160 an -96, and guess what? both have the same value:

    160  = A0h = 10100000b2.

   -96    = A0h = 10100000b2.

This is correct.  see the comment - rg

    So my ouput was right! the problem is how PutDec figure out if it is 160 or - 96 .  Also  on signed 8 bit integer , you cannot represent 160 , but  -96 it is possible.

 

    If you read chapter 1, there 3 ways of reprensenting negative numbers, they are:  by Magnitude, by 1 complement, and by 2 complement. but how can we do this?

    Because if you get a positive value which is correct, then you would get its negation by complementing it right?

    I hope we can  solve this cool  problem!

 

    Armando Fonseca.

 

Updated(March 04, 2008)

PROBLEM: Every time , i tried to compute numbers in which  b^2 < 4ac, I  get a postive back.

 

SOLUTION:

Remark1-->Depending  on how you wrote the code there might be  a small difference why my program and yours  were not giving  negative numbers when it was supposed to. I would give a general answer , so you and our  classmates can relate to it as well.

 

Regardless what is on ax, when you do an 8 bit multiplication let say:

 

                    imul   mem/reg  ; Assuming there is an integer on the al register.

 

The product becomes a 16 bit  number ! Let me give you an example:

 

 Let's multiply   -32 * 3 = -96 <==>  E0h*3h = A0h ,According to page 66 of the textbook the product  becomes a 16b number(al * mem/reg =  ax) , and if you se above A0h = 160 too.

I did some testing , and i found that  our imul by default  storage in the ax = 00A0h instead of  FFA0h. So how can we solve it?

 

Two ways:

       Solution1: one mov to the ah 255 --> mov ah, 255

       Solution2: use the sign extension instruction . For this 8 bit number use  cbw.

 

Remark2--> I'm not sure if they are the right ones because i'm a student too, but this two solutions solve my problem.

Remark 3--> Actually , i did think about the solution 1 before today.'s class.  I askedProfessor Glass on class if this has to do with the sign extension, but he told me is simpler than than.... may be , I he tought i was asking something else.

 

Armando Fonseca

 

 

Solution 2 is the correct / best way.  that is why the CBW instruction is there.  Solution 1 is basically what the CBW instruction does for negative numbers.  Realize that Solution 1 only works for negative where CBW would work for both positive and negative numbers filling with 0's or 1's accordingly.

 

As for remark 3, :-)  I said that there was a simpler solution to what you stated the fix should be not that your explaination that it was sign extension problem. 

 

good. problem is closed  :-)

 

PS: Some of you may have noticed that one of these wiki questions was on the exam :-)

 

 

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