I cant seem to find answer about how to convert hexadecimal to decimal with Two's Complement example AD100002 i already know how to do the calculation for unsigned but not for signed 2's complement.
Any help will do or exampe)
(thank's for your time)
(Sorry for bad english).
{unsigned} ->(AD100002 = (10 × 16⁷) + (13 × 16⁶) + (1 × 16⁵) + (0 × 16⁴) + (0 × 16³) + (0 × 16²) + (0 × 16¹) + (2 × 16⁰) = 2903506946).
The basic explanation of two's complement on a binary/hex value is to flip every digit and then add 1. For example, say we had the following value:
0xA5
The first thing to do is convert the value to a binary number:
0xA5 -> 10100101
To perform two's complement, flip all the bits:
10100101
||
\/
01011010
and then add 1:
01011011
Converting this binary number to a decimal number yields 91. Therefore, the two's compliment of the hex value "0xA5" is -91.
(If you're treating the hex value as a signed bit representation of an integer, only perform the two's complement if the most-significant bit in the binary representation is 1. If it's 0, treat the rest of the bits as normal.)
Related
I want to preface this question by stating that it at first appears that this is a duplicate of several other questions here on SO, but none of the answers to those questions answered my question and thus I am asking it.
What is the relationship between a binary integer interpreted as a positive integer versus that exact same binary integer interpreted as a negative integer in terms of decimal? Let's take for example the integer 5:
5 is 101
-5 is 11111011
11111011 is 251 when interpreted as an unsigned number.
The question is, what is the decimal relationship between -5 and 251? Is there a direct relationship aside from the action which occurs in the binary number system? Meaning, is there some rule in decimal that we can directly map any given decimal integer to the decimal integer for which the identical binary integer would be when converted from positive to negative and vice versa?
Note that -5 is not actually 11111011 in binary--that is the binary representation in eight bits. If you use a different number of bits you get a different binary representation. For example, if you use 16 bits, as is often done, you get 1111111111111011, which is 65531.
This then is the key. In eight bits we consider 2^8 which is 256. (That caret stands for exponentiation.) We then see that -5 is represented by 256 - 5.
So the final answer is this: for a given positive integer n that is to be represented in b binary bits, the number -n is then represented by
(2 ^ b) - n
At least, if the number of bits b is large enough. Is that clear? Much more can be said than that, but you are better off reading more about two's-complement notation in a book or large web page.
Assuming that we work with two complement, we have that a vector of n bits:
n = <a(n-1), a(n-2), ..., a(2), a(1), a(0)>
is interpreted as:
n_signed = -[a(n-1) * 2^(n-1)] + sum(i=0, i=n-2) {a(i) * 2^(i)}
whereas, treating it as unsigned leads to:
n_unsigned = sum(i=0, i=n-1) {a(i) * 2^(i)}
Hence the difference is:
n_unsigned - n_signed = 2 * [a(n-1) * 2^(n-1)]
Hope it helps (and sorry for the poor formatting).
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What is 2's Complement Number?
Why do we take 1's Complement and add 1 to it? Why don't we subtract 1 after taking 1's Complement?
Why do computers use 2's Complement?
What is 2's Complement Number?
Complementary number system is used to represent negative numbers. So,
2's Complement number system is used to represent negative numbers.
UPDATE
Q: What “2’s Complement System” says?
A: The negative equivalent of binary number is its 2’s complement. (1’s Complement + 1)
Note: 1 extra bit is required to represent the sign of a number. MSB (Most Significant Bit) is used as sign bit. If MSB is 0, then the number is positive. If MSB is 1, then the number is negative.
1’s Complement Value 2’s Complement
011 +3 011
010 +2 010
001 +1 001
000 +0 000
111 -0 000
110 -1 111
101 -2 110
100 -3 101
-4 100
How '100' (3 bits) is -4?
MSB is used as sign, if 1, its negative, if 0 it is positive.
-1 * 2^2 + 0*2^1 + 0*2^0 = -4 + 0 + 0 = -4
Similarly 101 (3 bits) is -3
-1 * 2^2 + 0*2^1 + 1*2^0 = -4 + 0 + 1 = -3
Observations:
• In 1’s complement, using 3 bits, we represented 2^3 = 8 numbers i.e from -3 to +3.
• In 1’s complement, -0 and +0 are having 2 representation. (+0 is ‘000’ and -0 is ‘111’).
But mathematically +0 and -0 are same.
• In 2’s complement, using 3 bits, we represented only 2^3 = 8 numbers i.e from -4 to +3.
• In 2’s complement, -0 and +0 are having same representation.
• Since +0 and -0 in 2’s complement is having same representation,
we are left out with one more combination which is ‘100’ = -4.
Why do we take 1's Complement and add 1 to it? Why don't we subtract 1 after taking 1's Complement?
Refer "Why Inversion and Adding One Works" topic in the below link. If
I start explaining, this post will grow big.
http://www.cs.cornell.edu/~tomf/notes/cps104/twoscomp.html
Why computer uses 2' Complement?
Cos' of less hardware. If the computer is using 2' Complement means,
it does subtraction using addition circuit. So, less hardware!!!
As seen in the above example, +0 and -0 have same representation. (1's complement and sign magnitude representation have 2 different representation for +0 and -0).
(Not an important) You will be able to represent one extra number using 2's complement. (in the above example its -4 which is '100' in binary using 3 bits).
From Wikipedia Two's Complement :
The two's-complement system has the advantage that the fundamental arithmetic operations of addition, subtraction, and multiplication are identical to those for unsigned binary numbers (as long as the inputs are represented in the same number of bits and any overflow beyond those bits is discarded from the result). This property makes the system both simpler to implement and capable of easily handling higher precision arithmetic. Also, zero has only a single representation, obviating the subtleties associated with negative zero, which exists in one's-complement systems.
The reason why we use two's complement rather than one's complement is to make arithmetic as simple as possible.
Consider that in one's complement, 1111 1111 and 0000 0000 are the same number - by subtracting one, we have ended up with...the same number. This is too big of a pain to have to think about - so instead we use two's complement, where 1111 1111 is -1 - by subtracting one, we successfully subtract one. Hooray!
(A secondary advantage is that we can represent one more unique number in two's complement than in one's complement — in two's complement -128 to +127 instead of one's complement -127 to +127.)
I have to convert floating point number from binary to usable decimal number.
Of course my floating point number has been separated into bytes, so 4 bytes total.
1 2 3 4
[xxxxxxxx][xxxxxxxx][xxxxxxxx][xxxxxxxx]
These 4 bytes are already converted to decimal, so I have e.g.
1 2 3 4
[0][10][104][79]
Now Mantissa is held in three parts, two rightmost bytes (3 & 4) and in byte 2 but without the MSB bit (that one is easy to mask out, so let's assume we have a nice decimal number there as well). So three decimal numbers.
Is there an straightforward mathematical conversion to a floating point mantissa for these three decimal numbers?
This is along the lines: if I needed to get an integer, the formula would be
10 * 65536 + 104 * 256 + 79.
Call these bytes a, b, and c. I assume a has already been masked, so it contains only the bits of the significand and none of the exponent, and that the number is IEEE-754 32-bit binary floating-point, with bytes taken with the appropriate endianness.
If the raw exponent field is 1 to 254 (thus, not 0 or 255), then the significand is:
1 + a*0x1p-7 + b*0x1p-15 + c*0x1p-23
or, equivalently:
(65536*a + 256*b + c) * 0x1p-23 + 1.
If the raw exponent field is 0, then remove the 1 from the sum (the number is subnormal or zero). If the raw exponent field is 255, then the floating-point value is infinity (if a, b, and c are all 0) or a NaN (otherwise).
I cannot be of much help, since it has been a while since I did conversions, but I hope you find this tutorial useful.
The 2's complement of a number which is represented by N bits is 2^N-number.
For example: if number is 7 (0111) and i'm representing it using 4 bits then, 2's complement of it would be (2^N-number) i.e. (2^4 -7)=9(1001)
7==> 0111
1's compliment of 7==> 1000
1000
+ 1
-------------
1001 =====> (9)
While calculating 2's complement of a number, we do following steps:
1. do one's complement of the number
2. Add one to the result of step 1.
I understand that we need to do one's complement of the number because we are doing a negation operation. But why do we add the 1?
This might be a silly question but I'm having a hard time understanding the logic. To explain with above example (for number 7), we do one's complement and get -7 and then add +1, so -7+1=-6, but still we are getting the correct answer i.e. +9
Your error is in "we do one's compliment and get -7". To see why this is wrong, take the one's complement of 7 and add 7 to it. If it's -7, you should get zero because -7 + 7 = 0. You won't.
The one's complement of 7 was 1000. Add 7 to that, and you get 1111. Definitely not zero. You need to add one more to it to get zero!
The negative of a number is the number you need to add to it to get zero.
If you add 1 to ...11111, you get zero. Thus -1 is represented as all 1 bits.
If you add a number, say x, to its 1's complement ~x, you get all 1 bits.
Thus:
~x + x = -1
Add 1 to both sides:
~x + x + 1 = 0
Subtract x from both sides:
~x + 1 = -x
The +1 is added so that the carry over in the technique is taken care of.
Take the 7 and -7 example.
If you represent 7 as 00000111
In order to find -7:
Invert all bits and add one
11111000 -> 11111001
Now you can add following standard math rules:
00000111
+ 11111001
-----------
00000000
For the computer this operation is relatively easy, as it involves basically comparing bit by bit and carrying one.
If instead you represented -7 as 10000111, this won't make sense:
00000111
+ 10000111
-----------
10001110 (-14)
To add them, you will involve more complex rules like analyzing the first bit, and transforming the values.
A more detailed explanation can be found here.
Short answer: If you don't add 1 then you have two different representations of the number 0.
Longer answer: In one's complement
the values from 0000 to 0111 represent the numbers from 0 to 7
the values from 1111 to 1000 represent the numbers from 0 to -7
since their inverses are 0000 and 0111.
There is the problem, now you have 2 different ways of writing the same number, both 0000 and 1111 represent 0.
If you add 1 to these inverses they become 0001 and 1000 and represent the numbers from -1 to -8 therefore you avoid duplicates.
I'm going to directly answer what the title is asking (sorry the details aren't as general to everyone as understanding where flipping bits + adding one comes from).
First let motivate two's complement by recalling the fact that we can carry out standard (elementary school) arithmetic with them (i.e. adding the digits and doing the carrying over etc). Easy of computation is what motivates this representation (I assume it means we only 1 piece of hardware to do addition rather than 2 if we implemented subtraction differently than addition, and we do and subtract differently in elementary school addition btw).
Now recall the meaning of each of the digit's in two's complements and some binary numbers in this form as an example (slides borrowed from MIT's 6.004 course):
Now notice that arithmetic works as normal here and the sign is included inside the binary number in two's complement itself. In particular notice that:
1111....1111 + 0000....1 = 000....000
i.e.
-1 + 1 = 0
Using this fact let's try to derive what the two complement representation for -A should be. So the problem to solve is:
Q: Given the two's complement representation for A what is the two's complement's representation for -A?
To do this let's do some algebra using values we know:
A + (-A) = 0 = 1 + (-1) = 11...1 + 00000...1 = 000...0
now let's make -A the subject expressed in terms of numbers expressed in two's complement:
-A = 1 + (-1 - A) = 000.....1 + (111....1 - A)
where A is in two's complements. So what we need to compute is the subtraction of -1 and A in two's complement format. For that we notice how numbers are represented as a linear combination of it's bases (i.e. 2^i):
1*-2^N-1 + 1 * 2^N-1 + ... 1 = -1
a_N * -2^N-1 + a_N-1 * 2^N-1 + ... + a_0 = A
--------------------------------------------- (subtract them)
a_N-1 * -2^N-1 + a_N-1 -1 * 2^N-1 + ... + a_0 -1 = A
which essentially means we subtract each digit for it's corresponding value. This ends up simply flipping bits which results in the following:
-A = 1 + (-1 - A) = 1 + ~ A
where ~ is bit flip. This is why you need to bit flip and add 1.
Remark:
I think a comment that was helpful to me is that complement is similar to inverse but instead of giving 0 it gives 2^N (by definition) e.g. with 3 bits for the number A we want A+~A=2^N so 010 + 110 = 1000 = 8 which is 2^3. At least that clarifies what the word "complement" is suppose to mean here as it't not just the inverting of the meaning of 0 and 1.
If you've forgotten what two's complement is perhaps this will be helpful: What is “2's Complement”?
Cornell's answer that I hope to read at some point: https://www.cs.cornell.edu/~tomf/notes/cps104/twoscomp.html#whyworks
It's been a while since my assembly class in college (20 years to be exact).
When someone gives you a number, say 19444, and says that X is bits 15 through 8 and Y are bits 7 through 0... how do I calculate values of X and Y?
I promise this is not homework, just a software guy unwisely trying to do some firmware programming.
First of all convert the input number to hexadecimal:
19444 => 0x4BF4
Hex is convenient because every 4 binary bits are one hex digit. Hence, every 2 hex digits are 8 bits, or a byte. Now assuming traditional little-endian notation (look it up!), bits 7 downto 0 are the low byte, bits 15 downto 8 are the high byte:
[7:0] => 0xF4
[15:8] => 0x4B
Using your preferred language, you can get the least significant byte by using a bitwise AND:
Y = 19444 & 0xff
or, the more mathematical:
Y = 19444 % 256
Now, for the most significant byte you can use bit shifts (if the number is larget than two byte, apply the first stage again):
X = 19444 >> 8
(The following assumes C notation). In general, to access the value in bits N through M, where N is the smaller value and the bits are numbered from 0, use:
(value >> N) & (1U << (M - N + 1)) - 1;
So for bits 0..7, use:
(value >> 0) & (1U << 8) - 1
and for bits 8..15, use:
(value >> 8) & (1U << 8) - 1
Note that for the case where "N through M" is the entire width of the type, you can't use the shift as written.
Also, mind the byte order (wheter the most significant byte comes first).
When given bit positions (like "15 through 8"), by convention bit 0 is the least significant bit of the binary number. If you're dealing with a 16-bit number, then bit 15 is the most significant bit.
One hexadecimal digit corresponds to 4 binary digits. So hex FF is 11111111 in binary. Bitwise AND is often used to "mask out" a certain collection of bits.
Nearly all processors provide some form of bitwise shifting. For example, shifting 1010001 right by 4 bits gives you 101.
Combining all this, in C you would typically do something like this:
unsigned short int num;
unsigned char x, y;
num = 19444;
y = num & 0xff; //use bitwise AND to get 8 least-sig bits
x = num >> 8; //right-shift by 8 bits to get 8 most-sig bits