Let's say I have the two's complement form of the hexadecimal A41B and I want to convert it to decimal. Normally, I would convert it to binary, flip all the bits, add one, and then convert it to decimal and put the correct sign (positive or negative). Is there a faster way to do this by hand?
You can:
convert it to decimal
subtract 2N from the result, where N is your word size (eg. for 16-bit words, subtract 65536)
In your example,
0xA41B = 42011
42011 - 65536 = -23525
Before doing the subtraction, you would check that the sign bit is in fact 1. Otherwise you would simply have a positive number.
From dave's corner:
=MOD(HEX2DEC(A7)+2^15,2^16)-2^15
for 16 bit numbers.
If you for some reason don't want to or can't perform the subtraction that the other answers suggest, you can treat negative numbers like this:
Flip the bits
Add one
Convert to decimal as if the number were positive
Add a negative sign
For example: -5 is 11111011 in 8-bit 2's complement. Flipping the bits and adding one is 00000101, which is 5 in decimal. Add negative sign and you have -5.
Related
I am writing a generic routine for converting fixed-point numbers between decimal and binary representations.
For positive numbers the processing is simple, however when things come to negative ones I found divergent sources. Someone says there is a single bit used to hold the sign while others say the whole number should be represented in a pseudo integer using 2's complement even it is negative.
Please anyone tell me which source is correct or is there a standard representation for signed fixed point numbers?
Additionally, if the 2's complement representation was correct then how to represent negative numbers with zero integer part. For example -0.125?
Fixed-point numbers are just binary values where the place values have been changed. Assigning place values to the bits is an arbitrary human activity, and we can do it in any way that makes sense. Normally we talk about binary integers so it is convenient to assign the place value 2^0 = 1 to the LSB, 2^1=2 to the bit to the left of the LSB, and so on. For an N bit integer the place value of the MSB becomes 2^(N-1). If we want a two's-complement representation, we change the place value of the MSB to -2^(N-1) and all of the other bit place values are unchanged.
For fixed-point values, if we want F bits to represent a fractional part of the number, then the place value of the LSB becomes 2^(0-F)
and the place value of the MSB becomes 2^(N-1-F) for unsigned numbers and -2^(N-1-F) for signed numbers.
So, how would we represent -0.125 in a two's-complement fixed-point value? That is equal to 0.875 - 1, so we can use a representation where the place value of the MSB is -1 and the value of all of the other bits adds up to 0.875. If you choose a
4-bit fixed-point number with 3 fraction bits you would say that
1111 binary equals -0.125 decimal. Adding up the place values of the bits we have (-1) + 0.5 + 0.25 + 0.125 = -0.125. My personal preference is to write the binary number as 1.111 to note which bits are fraction and which are integer.
The reason we use this approach is that the normal integer arithmetic operators still work.
It's easiest to think of fixed-point numbers as scaled integers — rather than shifted integers. For a given fixed-point type, there is a fixed scale which is a power of two (or ten). To convert from the real value to the integer representation, multiply by that scale. To convert back again, simply divide. Then the issue of how negative values are represented becomes a detail of the integer type with which you are representing your number.
Please anyone tell me which source is correct...
Both are problematic.
Your first source is incorrect. The given example is not...
the same as 2's complement numbers.
In two’s complement, the MSB's (most significant bit's) weight is negated but the other bits still contribute positive values. Thus a two’s complement number with all bits set to 1 does not produce the minimum value.
Your second source could be a little misleading where it says...
shifting the bit pattern of a number to the right by 1 bit always divide the number by 2.
This statement brushes over the matter of underflow that occurs when the LSB (least significant bit) is set to 1, and the resultant rounding. Right-shifting commonly results in rounding towards negative infinity while division results in rounding towards zero (truncation). Both produce the same behavior for positive numbers: 3/2 == 1 and 3>>1 == 1. For negative numbers, they are contrary: -3/2 == -1 but -3>>1 == -2.
...is there a standard representation for signed fixed point numbers?
I don't think so. There are language-specific standards, e.g. ISO/IEC TR 18037 (draft). But the convention of scaling integers to approximate real numbers of predetermined range and resolution is well established. How the underlying integers are represented is another matter.
Additionally, if the 2's complement representation was correct then how to represent negative numbers with zero integer part. For example -0.125?
That depends on the format of your integer and your choice of radix. Assuming a 16-bit two’s complement number representing binary fixed-point values, the scaling factor is 2^15 which is 32,768. Multiply the value to store as an integer: -0.125*32768. == -4096 and divide to retrieve it: -4096/32768. == -0.125.
I'm wondering how you could convert a two's complement in fix-point arithmetic to a decimal number.
So let's say we got this fix-point arithmetic in two's complement: 11001011 with bit numbering, with 2 positions behind the decimal point, and want form it to a decimal number.
We already know that the decimal will be negative because the first bit is a 1.
2 positions behind decimal point, so we have 110010 11.
Convert that from two's complement to normal form (sub by 1, invert):
110010 10 (i sub by 1 here)
001101 01 (i inverted here)
001101 in decimal is 13
01 in decimal is 1
So in the end we get to -13.1. Is that correct or there isn't even a way to convert this?
The simplest method is just to convert the whole value to an integer (ignoring the fixed point, initially), then scale the result.
So for your example where you have a 6.2 fixed point number: 110010 10:
Convert as integer:
11001010 = -54
Divide by scale factor = 2^2:
-54 / 4 = -13.5
Note that the fractional part is always unsigned. (You can probably see now that 10 would give you + 0.5 for the fractional part, i.e. 00 = 0.0, 01 = +0.25, 10 = +0.5, 11 = +0.75.)
A small note (but very important in understanding your question) - when you say "decimal point", do you really mean decimal? or do you mean "binary" point.
Meaning, if it's a decimal point, you can position it after you convert to decimal, to see how many decimal digits should remain to the right of the point, but if you mean binary point, it means in the binary representation how many bits represent the fraction part.
In your example, it seems that you meant binary point, and then the integer part is 001101(bin) = 13(dec) and the fraction part is 0.01(bin)=0.25(dec), because the first bit to the right of the point represents 1/2, the second represents 1/4 and so on, and the whole thing is then negated.
Total result then will be -13.25
I want to design a hardware which will give 2's complement of input 4-bit binary number. But i stuck at very first input: 0000.Because the method i generally use to find 2's complement is to first find 1's complement of binary number and then adding 1 into it.But if I do same with 0000, then it will give me 5-bit number 10000. that is the problem.
You have to cut off the upper bits if you're working with an n-bit value. The two's complement of the four-bit 0000 is the last four bits of 10000, or 0000. Otherwise the ones' complement of 100 would be:
1111111111...ad infinitum...1111111011
rather than the correct 011.
That (the cutting off of extraneous bits) makes sense since the negation of zero is zero (though that could be arguable for ones' complement or sign-magnitude, where -0 is a distinct possibility).
If you design hardware, see hardware circuits for produce 4-bit complement number.
(Not sure if I am allow to ask a question like this but will delete if asked)
Convert -25 into binary representation using 8 bits and 2's complement format to represent a negative number. So far I got 11001.
I tried to google methods but I am not sure how to do it using the 8 bits way.
This question is not using code, just conversions but from what I did was
convert it to binary which was 11001 then I added 3 zero's to make it 8bits (assuming that is correct) 00011001 then I did 1's complement 11100110 and 2's complement by adding 1 which equals 11100111.
I am not sure if that is correct.
Two's complement is probably one of the more straightforward operations on binary numbers. In short, you'll want to take the following actions to convert a decimal number into two's complement form:
Write down the binary representation of the positive version of your number. In this case, 25 should be represented as: 00011001
Next, flip all the digits: 11100110
Add one: 11100111
Sit back, grab a drink, and bask in the glory of the newly-created two's complement representation of a decimal number.
Source
1.assume the negative sign and get the binary of the positive integer.adopt 8bit notation
25--00011001
2.get the two's complement.
11100111
+ 1
=11111000
3.relax and enjoy the two's complement
What is the advantage of 2's complement over 1's complement in negative number representation in binary number system? How does it affect the range of values stored in a certain bit representation of number in binary system?
The primary advantage of two's complement over one's complement is that two's complement only has one value for zero. One's complement has a "positive" zero and a "negative" zero.
Next, to add numbers using one's complement you have to first do binary addition, then add in an end-around carry value.
Two's complement has only one value for zero, and doesn't require carry values.
You also asked how the range of values stored are affected. Consider an eight-bit integer value, the following are your minimum and maximum values:
Notation Min Max
========== ==== ====
Unsigned: 0 255
One's Comp: -127 +127
Two's Comp: -128 +127
References:
http://en.wikipedia.org/wiki/Signed_number_representations
http://en.wikipedia.org/wiki/Ones%27_complement
http://en.wikipedia.org/wiki/Two%27s_complement
The major advantages are:
In 1's there is a -0 (11111111) and a +0 (00000000), i.e two value for the same 0. On the other hand, in 2's complement, there is only one value for 0 (00000000). This is because
+0 --> 00000000
and
-0 --> 00000000 --> 11111111 + 1 --> 00000000
While doing arithmetic operations like addition or subtraction using 1's, we have to add an extra carry bit, i.e 1 to the result to get the correct answer, e.g.:
+1(00000001)
+
-1(11111110)
-----------------
= (11111111)
but the correct answer is 0. In order to get 0 we have to add a carry
bit 1 to the result (11111111 + 1 = 00000000).
In 2's complement, the result doesn't have to be modified:
+1(00000001)
+
-1(11111111)
-----------------
= 1 00000000
Negative integers :
2's complement makes sense to be used for negative integers. 1's complement is just a computation technique which might be helpful to evaluate 2's complement. The real (defeated) rival of 2's complement was the sign-magnitude representation for negative integers.
No overflow : 1's complement has no special usage for negative integers. 2's complement makes sense because it can be used in natural addition and subtraction arithmetic without any need to change the bits. Providing that no overflow occurs, the sign bit of the result is just the right value. The bit number promotion in this notation is straight forward, for example, to promote an 8-bit signed integer to 16, we could simply repeat the sign bit of integer value in the high byte of it.
Sign magnitude : On the contrary, the sign-magnitude notation is just the way that human uses to represent negative integers. The bit number promotion and addition subtraction arithmetic is a bit mess with this notation.
Advantages of Two’s Complement #1
In Two’s Complement representation, the value zero is
uniquely represented by having all bits set to zero:
**
Advantages of Two’s Complement #2
**
When you perform an arithmetic operation (for example,
addition, subtraction, multiplication, division) on two
signed integers in Two’s Complement representation, you
can use
exactly the same method
as if you had two
unsigned integers (that is, nonnegative integers with no sign
bit) ...
EXCEPT
, you throw away the high carry (or the high
borrow
for subtraction)
Advantages of Two’s Complement #3
This property of Two’s Complement representation is so
incredibly handy that virtually every general
purpose
computer available today uses Two’s Complement.
Why? Because, with Two’s Complement, we don’t need
special algorithms
(and therefore extra circuitry) for
arithmetic operations that involve negative values.
Another major advantage of Two's complement over signed bit representation is 2's complement representation is easy to manipulate in hardware
2s complement isn't for representing a negative number it's an inverse.
Means you can do A + B' (where B' is the 2s complement of B) to give A - B, means you can do everything with an adder and not need a substracter