When we want to show 8-bit binary representations of a negative number -123 in two’s complement we do:
Number in decimal = -123
Convert -123 to its absolute value: 123
Covert to binary: 01111011
One's complement it: 10000100
Add 1: 10000101
Check:
2^7
2^6
2^5
2^4
2^3
2^2
2^1
2^0
128
64
32
16
8
4
2
1
---
---
---
---
---
---
---
---
1
0
0
0
0
1
0
1
↓
↓
↓
↓
↓
↓
↓
↓
128
0
0
0
0
4
0
1
But since the first bit is to represent a negative sign we take it as -128 instead of 128 and add them all up.
Therefore: (-128)+4+1 = -123
But why does every online 2's complement calculator give the answer as 0111 1011?
Related
I am currently trying to gain a more intuitive understanding of two's complement and its uses; however, I cannot seem to perform subtraction using two's complement correctly. I understand that when a negative number is stored in a signed int variable the procedure is to perform two's complement on the number having the MSB be 1 to represent the negative sign.
So in a 4 bit system 1010 represents -6.
Now I am following this guide on how to subtract two binary #s
For example I have the number 0101 (5 in decimal) and 1010 (-6 in decimal). If I wanted to do the equation 5 - (-6) it would look like 0101 - 1010 in binary. Next I would take the 1010 and perform twos' complement on it to get 0110. Now I take 0101 + 0110 and get 1011. I don't have a carry so I perform two's complement on the result giving me 0101, but this says the answer is -5 when it should be 11.
4 bits can represent 16 different values. With two's complement, they are -8 to 7.
1000 -8
1001 -7
1010 -6
1011 -5
1100 -4
1101 -3
1110 -2
1111 -1
0000 0
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
You can't possibly get an answer of 11. It's too large to fit in 4 bits as a two's complement number. As such, the calculation you are attempting results in an overflow. (See below for how to detect overflows.)
This means 4 bits is not enough to compute 5 - -6. You need at least 5 bits.
0(...0)0101 5
- 1(...1)1010 -6
------------------
0(...0)0101 5
+ 0(...0)0110 6
------------------
0(...0)1011 11
Detecting Overflow in Two's Complement
With two's complement numbers, an addition overflows when the carry into the sign bit is different than the carry out of it.
0101 5
+ 0110 6
-----------
1011 -5 Carry in: 1 Carry out: 0 OVERFLOW!
1011 -5
+ 1010 -6
-----------
0101 5 Carry in: 0 Carry out: 1 OVERFLOW!
0001 1
+ 0010 2
-----------
0011 3 Carry in: 0 Carry out: 0 No overflow
1111 -1
+ 1110 -2
-----------
1101 -3 Carry in: 1 Carry out: 1 No overflow
For example I have the number 0101 (5 in decimal) and 1010 (-6 in decimal). If I wanted to do the equation 5 - (-6) it would look like 0101 - 1010 in binary.
Ok.
Next I would take the 1010 and perform twos' complement on it to get 0110.
Ok. And 0110 binary is 6 decimal is -(-6) decimal.
Now I take 0101 + 0110 and get 1011.
Ok.
I don't have a carry so I perform two's complement on the result giving me 0101,
I take you to mean that as part of the process of interpreting the result, not computing it. The result itself is 1011 (4-bit, two's complement binary).
but this says the answer is -5 when it should be 11.
The answer in the operational system you have chosen, with 4-bit two's complement, is -5, exactly as you have computed. With three data bits and one sign bit, the maximum result your data type can represent is 7 (0111 binary). This underscores the importance of choosing data types appropriate for the computations you want to perform.
Note that 0b1011 is 11 in binary, if interpreted as unsigned or with more than 3 data bits, so both 11 and -5 are the result (which you use depends on how its interpreted).
In terms of modular arithmetic, 4 bits is a base of 24, and:
11 ≡ -5 (mod 16)
Which answer you use depends on whether the nibble is interpreted as signed or unsigned.
Given a binary integer, how can I invert (flip) last n bits using only bitwise operations in c/c++?
For example:
// flip last 2 bits
0110 -> 0101
0011 -> 0000
1000 -> 1011
You can flip last n bits of your number with
#define flipBits(n,b) ((n)^((1u<<(b))-1))
for example flipBits(0x32, 4) will flip the last 4 bits and result will be 0x3d
this works because if you think how XOR works
0 ^ 0 => 0
1 ^ 0 => 1
bits aren't flipped
0 ^ 1 => 1
1 ^ 1 => 0
bits are flipped
(1<<b)-1
this part gets you the last n bits
for example, if b is 4 then 1<<4 is 0b10000 and if we remove 1 we get our mask which is 0b1111 then we can use this to xor with our number to get the desired output.
works for C and C++
I don't understand why an n-bit 2C system number can be extended to an (n+1)-bit 2C system number by making bit bn = bn−1, that is, extending to (n+1) bits by replicating the sign bit.
This works because of the way we calculate the value of a binary integer.
Working right to left, the sum of each bit_i * 2 ^ i,
where
i is the range 0 to n
n is the number of bits
Because each subsequent 0 bit will not increase the magnitude of the sum, it is the appropriate value to pad a smaller value into a wider bit field.
For example, using the number 5:
4 bit: 0101
5 bit: 00101
6 bit: 000101
7 bit 0000101
8 bit: 00000101
The opposite is true for negative numbers in a two's compliment system.
Remember you calculate two's compliment by first calculating the one's compliment and then adding 1.
Invert the value from the previous example to get -5:
4 bit: 0101 (invert)-> 1010 + 1 -> 1011
5 bit: 00101 (invert)-> 11010 + 1 -> 11011
6 bit: 000101 (invert)-> 111010 + 1 -> 111011
7 bit: 0000101 (invert)-> 1111010 + 1 -> 1111011
8 bit: 00000101 (invert)-> 11111010 + 1 -> 11111011
I am using 2' complement to represent a negative number in binary form
Case 1:number -5
According to the 2' complement technique:
Convert 5 to the binary form:
00000101, then flip the bits
11111010, then add 1
00000001
=> result: 11111011
To make sure this is correct, I re-calculate to decimal:
-128 + 64 + 32 + 16 + 8 + 2 + 1 = -5
Case 2: number -240
The same steps are taken:
11110000
00001111
00000001
00010000 => recalculate this I got 16, not -240
I am misunderstanding something?
The problem is that you are trying to represent 240 with only 8 bits. The range of an 8 bit signed number is -128 to 127.
If you instead represent it with 9 bits, you'll see you get the correct answer:
011110000 (240)
100001111 (flip the signs)
+
000000001 (1)
=
100010000
=
-256 + 16 = -240
Did you forget that -240 cannot be represented with 8 bits when it is signed ?
The lowest negative number you can express with 8 bits is -128, which is 10000000.
Using 2's complement:
128 = 10000000
(flip) = 01111111
(add 1) = 10000000
The lowest negative number you can express with N bits (with signed integers of course) is always - 2 ^ (N - 1).
The question seems pretty well formulated
I have a virtual machine which implements only AND, XOR, SHL and SHR, yet I have to do a "OR 0x01" operation.
First of all having a correct bitwise computation for the following two variables is sufficient, because they cover all combinations:
A=0101
B=0011
We want
0101
0011
A or B
0111
for xor we get
0101
0011
A xor B
0110
for and we get
0101
0011
A and B
0001
so if we connect them with an xor we are done.
(A xor B) xor (A and B)
I would just start with
a xor b = ((not a) and b) or (a and (not b))
and unleash some boolean algebra on that until it looks like
a or b = <expression using only and, xor>
Admittedly, this is probably more work to actually do than going the "try every conceivable bit combination" route, but then you did ask for homework solution ideas. :)
The truth table as summarized on Wikipedia here and gasp, basic CS 101 stuff, De Morgan's Law....
AND
0 & 0 0
0 & 1 0
1 & 0 0
1 & 1 1
OR
0 | 0 0
0 | 1 1
1 | 0 1
0 | 0 1
XOR
0 ^ 0 0
0 ^ 1 1
1 ^ 0 1
1 ^ 1 0
A Shift Left involves shifting the bits across from right to left, suppose:
+-+-+-+-+-+-+-+-+
|7|6|5|4|3|2|1|0|
+-+-+-+-+-+-+-+-+
|0|0|0|0|0|1|0|0| = 0x4 hexadecimal or 4 decimal or 100 in binary
+-+-+-+-+-+-+-+-+
Shift Left by 2 places becomes
+-+-+-+-+-+-+-+-+
|7|6|5|4|3|2|1|0|
+-+-+-+-+-+-+-+-+
|0|0|0|1|0|0|0|0| = 0x10 hexadecimal or 16 decimal or 10000 in binary
+-+-+-+-+-+-+-+-+
Shift Right by 1 places becomes
+-+-+-+-+-+-+-+-+
|7|6|5|4|3|2|1|0|
+-+-+-+-+-+-+-+-+
|0|0|0|0|1|0|0|0| = 0x8 hexadecimal or 8 decimal or 1000 in binary
+-+-+-+-+-+-+-+-+
Then it is a matter of combining the bit-wise operations according to the truth table above...
I would just expand DeMorgan's law: A or B = not(not A and not B). You can compute not by XORing with all 1 bits.