Learning assembly

[rjwilldohisbest]

Hello

I have the art of assembly tutorial and another one and I appear to be having trouble with converting binary into decimal.

Here's an example of what I mean:
**********************************************************
BIN to DEC coversions self exercises


BIN: 11001
This one from the tutorial and comes out to be correct:
1*2^0 + 1*2^1 + 0*2^2 + 0*2^3 + 1*2^4

= 1 + 2 + 0 + 0 + 16 //--> 19 DEC

*******************************************************************
These answers should be flip-flopped
41 should be the first example and 37 should be the second example?

BIN: 101001

1*2^0 + 0*2^1 + 1*2^2 + 0*2^3 + 0*2^4 + 1*2^5

= 1 + 0 + 4 + 0 + 0 + 32 //--> 37 DEC


100101

1*2^0 + 0*2^1 + 0*2^2 + 1*2^3 + 0*2^4 + 1*2^5

= 1 + 0 + 0 + 8 + 0 + 32 //--> 41 DEC

********************************************************************
One I tried out on my own.
Another one that comes out wrong. Should be 78, not 57?
BIN: 1001110

1*2^0 + 0*2^1 + 0*2^2 + 1*2^3 + 1*2^4 + 1*2^5 + 0*2^6

= 1 + 0 + 0 + 8 + 16 + 32 + 0 //--> 57 DEC



I do not know what I may be doing wrong. I do examples on paper and they come out different, but the example from the tutorial comes out right. I am a total beginner at this and if anyone knows what I might be doing wrong, please let me know.

Thanks for your time.

RJ

 

[Bertv100]

How do you split up a decimal number? From high significance to low significance!
e.g.: 1324 = 1*1000 + 3*100 + 2*10 + 4*1

How do you split up a binary number? Also from high significance to low significance!
e.g.: 11001 = 1*16 + 1*8 + 0*4 + 0*2 + 1*1

What does high and low significance mean? A digit is high significant if the value of a number changes relatively much when making small changes to this digit.
e.g.: take 43816, add 1 to MSD (most significant digit), then we get 53816, which has a much bigger value than 43816 (we added 10000 to the number, just by adding 1 to the MSD!).
The LSD (least significant digit) is then of course the rightmost digit.

The only difference between binary and decimal is that you use another base number, 2 in stead of 10, and this has two consequences:
1) you can only use 2 different digits (people have chosen this to be zero and one) in stead of 10.
2) you must define values on the base of powers of 2 in stead of powers of 10
e.g.: 1101d = 1*1000 + 1*100 + 0*10 + 1*1
1101b = 1*8 + 1*4 + 0*2 + 1*1 Regards,
Bert Vingerhoets
vingerhoetsbert@hotmail.com

http://student.vub.ac.be/~bvingerh/

Don't worry what people think about you. They're too busy wondering what you think about them.

 

[rjwilldohisbest]

Hi Bert

This explaination simplifies things better as the base 2 and ten systems goes, now -- the 11001 part, how is this converted to a decimal equivalent?

11001 = 1*16 + 1*8 + 0*4 + 0*2 + 1*1

The way you show this looks easier to comprehend than how the tutorial I'm learning explains it.

The tutorial had a way of adding these somehow together to form a decimal equivalent. The first example I displayed was considered correct, the others after that aren't for some reason.

I read that going in the other direction i.e. converting decimal to binary -- you need to keep dividing from the full decimal number by 2's and if there is no remainder then that number becomes a O and if the number doesn't divide equally it somehow becomes a 1? This division keeps going until there is nothing left of the decimal number and what is leftover is the binary string. I think this is right?

maybe the tutorial I'm learning is explaining it differently but is the same thing, or maybe I need a different tutorial?

Thanks for your time. The layout you showed looks better to me and looks much simpler.

RJ

 

[Bertv100]

bin->dec:
each position of a binary digit corresponds to a value that can be interpreted as a decimal value, if the digit is a 1, this means that the value at that position is part of the result (i.e. must be added to the result).
Example:

positions:    8    7    6    5    4    3    2    1
dec value:  128   64   32   16    8    4    2    1
pwr of 2 :  2^7  2^6  2^5  2^4  2^3  2^2  2^1  2^0

              0    0    0    1    1    0    0    1
            = 0  + 0  + 0 + 16  + 8  + 0  + 0  + 1
            = 25

dec->bin:
when converting a decimal value X to its corresponding binary value, you can use the following procedure:

substract the highest possible power of 2 from X
take a pencil and write 1
(*)if the second highest power of 2 is smaller than X, substract and write 1, else write 0
repeat from (*) to end until the power of 2 is 1

Example:
X = 25
highest power of 2, smaller than X = 16 (2^4)
X-16 = 9
bitstring = '1'
next power of 2 is 8 (2^3)
X >= 8?
yes -> X-8 = 1; bitstring = '11'
next power of 2 is 4 (2^2)
X >= 4?
no -> bitstring = '110'
next power of 2 is 2 (2^1)
X >= 2?
no -> bitstring = '1100'
next power of 2 is 1 (2^0)
X >= 1?
yes -> X-1 = 0; bitsring = '11001'
Regards,
Bert Vingerhoets
vingerhoetsbert@hotmail.com

http://student.vub.ac.be/~bvingerh/

Don't worry what people think about you. They're too busy wondering what you think about them.

 

[rjwilldohisbest]

Thanks Bert

Your examples are very easy to follow. I did a quiz exercises on one of those sites on math coversions and I passed it, was very surprised.

Thanks for your time Bert

RJ