I'm curious to know if it would be possible to create a computer that uses binary that can go from 0000 up to 9999 by having true and false be 1 and 0, but add numbers 2-9 to get more possibilities for numbers. Is binary code only consisting of 0's and 1's for simplicity? Is it because for some reason computers can only understand True and False?
Binary code starts with 0 (0000) and increases to 1 (0001) to 2 (0010) and 10 (1010). Could is be possible for a computer to recognize 0's and 1's but then go to 2's and other numbers? For example, 0000 = 0, 0001 = 1, 0002 = 2, 0009 = 9 then 0010 = 10, and so on.
If this isn't possible somehow, please explain why and give a general explanation of how computers work because I'm interested and want to learn more. If this isn't used because it's inefficient, please epxlain what makes it inefficient and what makes 0's and 1's more efficient.
Thank you.
I expect that it would be possible to create a computer like this but I searched online and couldn't find out why binary code can't have numbers other than 0's and 1's.
Answer to myself for future reference:
Binary is based on Boolean Algebra because it's a base 2 system, and Decimal is a base 10 system that goes from 0-9 instead of 0 or 1 like Binary which is a base 2 system. Computers easily understand binary because its based off on and off states (0 or 1) with 0 being off and 1 being on. Computers use logic gates which are composed of a multitude of transistors that use boolean logic to store data for the computer. Binary makes hardware convenient for computers. Other number systems are used for other purposes different from Binary's purpose. For example, hexadecimal is used to represent numbers that are large in a more simple way that decimal is able to, take the number one million for example,in decimal, it would be 1000000, in binary it would be 11110100001001000000, and in hexadecimal it would be F4240. This is why the Binary number system is based off boolean alegbra and why computers use binary and not other number systems.
It is based on how the data is stored. Each piece of data stored in your memory can have only two values. Think of your memory as a number of glasses which can either be empty or full. Which means the data is stored as bunch of 1s and 0s. This is the result of moving from analog systems to digital, analog values can be between 0 and 1. In analog systems for example, you can have 0.25 or 0.7. But since computers became digital, the logic became binary.
It will be really beneficial to research the history of computers and learn how they evolved over time, if you are interested in this topic.
First of all, this is not a duplicate question of this because it hasn't answered my questions below.
I searched for many resources and ended with no clear perception of how to perform signed number division using 2's complement, specifically for the case where one of the divisors or dividend or both are negative.
I read the Signed Number Division section from chapter 2 of Digital Fundamentals by Floyed and all of its examples and problems. I even read through internet resources that explain division, but they only used positive numbers (as explained below):
, but what about 100/-25 or -100/25 or -100/-25?
What are the rules to follow?
Can anyone give a simple explanation on, for example, how to divide 14 by -7?
The rules states that if the numbers have different sign bits that the quotient and remainder gotten when both are divided(this is after the have both been turned positive) will be negative.So for 14/-7, they will have different sign bits when turned to binary so this means your quotient and remainder will be negative. Now you leave 14 in positive form and turn -7 to it's corresponding positive binary number and carry out the division with the positive forms of both numbers, when you get your quotient and remainder from dividing the positive form of both numbers you turn both the quotient and remainder negative.
How do I write 0xFA in signed mantissa. I converted it to binary = 1111_1010. Not sure where to go from here.
The question is "If the register file has 8 bits width total, write the following in signed mantissa."
Also, an explanation of signed mantissa would be great!
So what you have to work with is a byte of data with an unknown type, apparently.
In order to write a number in signed mantissa (see Significand) one would expect that your dealing with a floating point type such as single or double. However you've only got a single byte.
A single is 8 bytes so surely it can't be that and double is double trouble. Also a half requires 16 bits. The only logical alternative type would be SByte but in that case you will never get any numbers that have any mantissa (significant digits) after the decimal. In fact there is no decimal. So Perhaps this is a trick question?
If you go on the assumption of SByte, you get -6x10^0
Just in case you want proof, or if your curious how this looks during debug:
private void SByte2Dec()
{
sbyte convertsHexToSByte = Convert.ToSByte("0xFA", 16);
Single yourAnswer = Convert.ToSingle(convertsHexToSByte);
label1.Text = Convert.ToString(youranswer);
}
In this example I had a windows form with nothing but label1 on it.
Then I put SByte2Dec(); right under InitializeComponent();
The solution is -122. Not sure how to get there...any ideas?
Working backwards from the answer it's simple to see what your professor has done. He is assuming the MSB is the sign bit and the rest is treated like a 7 bit integer. There is a precedent for this, called "Signed Magnitude Representation" but it's not used in modern computing. These days pretty much everyone is using Two's compliment.
I take it this is a beginners course and rather than go though all the trouble of explaining two's compliment and data types your professor is mainly trying to drive home the point of the MSB being a sign bit. If you got the whole sign bit thing and don't know anything else about the way modern computer hardware performs calculations, then you would probably arrive at the same answer.
My guess is that your professor also took to wording the question in a strange way so as to throw you off the path if you tried to Google the answer. If you want to get him back, ask him what the difference between "1000 0000" and "000 0000" is. Also if you or anyone else in the class answered -6 and he counted it wrong, he should be fired. Those students should be awarded bonus points for teaching themselves about two's compliment.
Why would the signed mantissa be -6? I see that the 2's complement is -6 but signed mantissa is different?
I have you read the wiki article I linked to on "Significand"?
The important thing to realize is that "signed mantissa" is not a data type. However, there are (were?) many different machine-specific data types that implemented their own versions of storing floating point numbers before the IEEE standard became widely adopted. These early data types were often referred to as DFP or decimal floating point numbers as opposed to binary floating point. Read this paper and for more in-depth understanding. Also this paper covers the topic quite well.
As I stated earlier, your professor most likely used the terminology "signed mantissa" to throw you off if you went searching the internet for the answer. Apparently you were expected to read between the lines and know that what he was really asking for was a form of decimal floating point, or Signed Magnitude Representation.
"Signed Mantissa" ≠ Two's Compliment
"Signed Mantissa" is to be interpreted as some form of Decimal Floating Point
Where as, Two's Compliment is a form of Binary Floating Point
I'm watching some great lectures from David Malan (here) that is going over binary. He talked about signed/unsigned, 1's compliment, and 2's complement representations. There was an addition done of 4 + (-3) which lined up like this:
0100
1101 (flip 0011 to 1100, then add "1" to the end)
----
0001
But he waved his magical hands and threw away the last carry. I did some wikipedia research bit didn't quite get it, can someone explain to me why that particular carry (in the 8's ->16's columns) was dropped, but he kept the one just prior to it?
Thanks!
The last carry was dropped because it does not fit in the target space. It would be the fifth bit.
If he had carried out the same addition, but with for example 8 bit storage, it would have looked like this:
00000100
11111101
--------
00000001
In this situation we would also be stuck with an "unused" carry.
We have to treat carries this way to make addition with two's compliment work properly, but that's all good, because this is the easiest way of treating carries when you have limited storage. Anyway, we get the correct result, right :)
x86-processors store such an additional carry in the carry flag (CF), which is possible to test with certain instructions.
A carry is not the same as an overflow
In the example you do have a carry out of the MSB. By definition, this carry ends up on the floor. (If there was someplace for it to go, then it would not have been out of the MSB.)
But adding two numbers with different signs cannot overflow. An overflow can only happen when two numbers with the same sign produce a result with a different sign.
If you extend the left-hand side by adding more digit positions, you'll see that the carry rolls over into an infinite number of bit positions towards the left, so you never really get a final carry of 1. So the answer is positive.
...000100
+...111101
----------
....000001
At some point you have to set the number of bits to represent the numbers. He chose 4 bits. Any carry into the 5th bit is lost. But that's OK because he decided to represent the number in just 4 bits.
If he decided to use 5 bits to represent the numbers he would have gotten the same result.
That's the beauty of it... Your result will be the same size as the terms you are adding. So the fifth bit is thrown out
In 2's complement you use the carry bit to signal if there was an overflow in the last operation.
You must look at the LAST two carry bits to see if there was overflow. In your example, the last two carry bits were 11 meaning that there was no overflow.
If the last two carry bits are 11 or 00 then no overflow occurred. If the last two carry bits are 10 or 01 then there was overflow. That is why he sometimes cared about the carry bit and other times he ignored it.
The first row below is the carry row. The left-most bits in this row are used to determine if there was overflow.
1100
0100
1101
----
0001
Looks like you're only using 4 bits, so there is no 16's column.
If you were using more than 4 bits then the -3 representation would be different, and the carry of the math would still be thrown out the end. For example, with 6 bits you'd have:
000100
111101
------
1000001
and since the carry is outside the bit range of your representation it's gone, and you only have 000001
Consider 25 + 15:
5+5 = 10, we keep the 0 and let the 1 go to the tens-column. Then it's 2 + 1 (+ 1) = 4. Hence the result is 40 :)
It's the same thing with binaries. 0 + 1 = 1, 0 + 0 = 0, 1 + 1 = 10 => send the 1 the 8-column, 0 + 1 ( + 1 ) = 10 => send the 1 to the next column - Here's the overflow and why we just throw the 1 away.
This is why 2's complement is so great. It allows you to add / substract just like you do with base-10, because you (ab)use the fact that the sign-bit is the MSB, which will cascade operations all the way to overflows, when nessecary.
Hope I made myself understood. Quite hard to explan this when english is not you native tongue :)
When performing 2's complement addition, the only time that a carry indicates a problem is when there's an overflow condition - that can't happen if the 2 operands have a different sign.
If they have the same sign, then the overflow condition is when the sign bit changes from the 2 operands, ie., there's a carry into the most significant bit.
If I remember my computer architecture learnin' this is often detected at the hardware level by a flag that's set when the carry into the most significant bit is different than the carry out of the most significant bit. Which is not the case in your example (there's a carry into the msb as well as out of the msb).
One simple way to think of it is as "the sign not changing". If the carry into the msb is different than the carry out, then the sign has improperly changed.
The carry was dropped because there wasn't anything that could be done with it. If it's important to the result, it means that the operation overflowed the range of values that could be stored in the result. In assembler, there's usually an instruction that can test for the carry beyond the end of the result, and you can explicitly deal with it there - for example, carrying it into the next higher part of a multiple precision value.
Because you are talking about 4 bit representations. It's unussual compared to an actual machine, but if we were to take for granted that a computer has 4 bits in each byte for a moment, then we have the following properties: a byte wraps at 15 to -15. Anything outside that range cannot be stored. Besides, what would you do with an extra 5th bit beyond the sign bit anyway?
Now, given that, we can see from everyday math that 4 + (-3) = 1, which is exactly what you got.
I'm in a basic Engineering class and we're going through binary conversions. I can figure out the base 10 to binary or hex conversions really well, however the 8bit floating point conversions are kicking my ass and I can't find anything online that breaks it down in a n00b level and shows the steps? Wondering if any gurus have found anything online that would be helpful for this situation.
I have questions like 00101010(8bfp) = what number in base 10
Whenever I want to remember how floating point works, I refer back to the wikipedia page on 32 bit floats. I think it lays out the concepts pretty well.
http://en.wikipedia.org/wiki/Single_precision_floating-point_format
Note that wikipedia doesn't know what 8 bit floats are, I think your professor may have invented them ;)
Binary floating point formats are usually broken down into 3 fields: Sign bit, exponent and mantissa. The sign bit is simply set to 1 if the entire number should be negative, and 0 if the number is positive. The exponent is usually an unsigned int with an offset, where 2 to the 0'th power (1) is in the middle of the range. It's simpler in hardware and software to compare sizes this way. The mantissa works similarly to the mantissa in regular scientific notation, with the following caveat: The most significant bit is hidden. This is due to the requirement of normalizing scientific notation to have one significant digit above the decimal point. Remember when your math teacher in elementary school would whack your knuckles with a ruler for writing 35.648 x 10^6 or 0.35648 x 10^8 instead of the correct 3.5648 x 10^7? Since binary only has two states, this required digit above the decimal point is always one, and eliminating it allows another bit of accuracy at the low end of the mantissa.