I have problem: you know the 2s Complement so you can get the negative number of a positive one with the reverse and adding a one. e.g.
8 Bit
121 = 0111 1001
1st= 1000 0110
+ 0000 0001
---------
1000 0111 --> -121
So now if we have a -0
a zero looks as 8 bit
0000 0000
so a minus 0 should look
1111 1111 + 0000 0001
= 10000 0000
but that is 512
so I think that I've misunderstood something
To expand my previous comment to the question
1111 1111 + 0000 0001 in 8 bit is 0000 0000, the ninth bit is lost because there is no place from it.
And, yes the complement of a negative is a positive
-121 = 1000 0111
1st = 0111 1000
+ 0000 0001
---------
0111 1001 --> 121
Think of them as a circle, at one point there is 0, adding 1 at a time you go up to the opposite point (128 in 8 bit) at that point the sign is switched and the absolute value begin to decrease, e.g.: 128 + 1 = -127, as you continue to add 1 the value go back to 0 and the circle is completed.
So given a number of bit, you only have that much bit, no more, and if you want the value to be signed you really have only x-1 bit for the value, as the most significant bit is used for the sign (0 -> +; 1 -> -)
1 0000 0000b is 256, not 512. Truncated to 8 bits, it's 0.
This is because with two's complement, zero is zero. There is no positive or negative zero.
Compare this to one's complement or sign bit, where positive zero and negative zero are different values.
Related
I'm making a program that would simulate something like a March Madness tournament where 64 teams were to play against each other. I've used fstream to read in the teams from the file. The problem I've having is that I must make sure that there are at least a power-of-2 teams in the file at all times or it wouldn't run properly. How would I implement a function that checks if there is the power of 2 lines?
Check if there is exactly one bit set in the number of teams.
bool powerOfTwo = std::popcount(numberOfTeams) == 1;
A trick with binary numbers is that if n is a power of 2, or 0, then n & (n - 1) is 0. Otherwise it's not.
i.e.
bool isPowerOfTwo(int n) {
return (n & (n - 1)) == 0;
}
This works because:
Powers of 2 have a single 1 bit in binary and the rest of the bits are 0s.
To subtract 1 from a number in binary, you change the last 1 bit and all the bits after it.
If that was the only 1 bit in the entire number, then all the bits that didn't change were 0s. Otherwise they weren't all 0s.
Example:
0000 0000 0010 0000 (32)
& 0000 0000 0001 1111 (31)
= 0000 0000 0000 0000 (0) (so 32 is a power of 2)
0000 0000 0110 0100 (100)
& 0000 0000 0110 0011 (99)
= 0000 0000 0110 0000 (96) (so 100 is not a power of 2)
Consider the i-th column of a binary matrix is denoted with matrix[i]. Let D be the number of columns of the matrix.
My question: What is the result of the following code. In fact, I can't understand the role of the 1llu expression.
matrix[i]^((1llu << D)-1)
This has to be looked at from the binary representation.
1llu means 1 represented as a unsigned long long.
...0000 0000 0000 0001
<< D shift that 1 left D amount of times (bits)
If D==5 then :
...0000 0000 0010 0000
- 1 subtract 1 from the shifted result ( which gives 1's on the positions 0 ~ D-1)
...0000 0000 0001 1111
The bitwise exclusive OR operator (^) compares each bit of its first operand to the corresponding bit of its second operand. If one bit is 0 and the other bit is 1, the corresponding result bit is set to 1. Otherwise, the corresponding result bit is set to 0.
https://learn.microsoft.com/en-us/cpp/cpp/bitwise-exclusive-or-operator-hat?view=vs-2019
It is easy to explain with an example below:
Example 1: 1 << 34
Example 2: 1llu << 34
If the integer size if 32 bits, then Example 1 would produce 0x0000 0000 as 1 will fall off. Whereas, Example 2 would produce 0x0000 0004 0000 0000
So, it should be seen with the context of the what is the type / size of matrix element.
There are two quotes in my textbook (Patterson, 5th edition, Computer Organization) I don't get:
First:
Two’s complement gets its name from the rule that the unsigned sum of an n-bit number and its n-bit negative is 2^n; hence, the negation or complement of a number x is 2^n - x, or its “two’s complement.”
Say I have a 4 bit number 2 = 0010
and it's negative = -2 = 1110
If I add them, I get 10000 (is this overflow? is this bad?) which is 16 right? And 1110 unsigned is 14? Is that what that quote is saying?
and
A third alternative representation to two’s complement and sign and magnitude is called one’s complement. The negative of a one’s complement is found by inverting each bit, from 0 to 1 and from 1 to 0, or x. This relation helps explain its name since the complement of x is 2^n - x - 1.
What does that mean? Can you give an example to help me see why the name is this way?
Say we have the binary representation of 2:
2 = 0000 0000 0000 0000 0000 0000 0000 0010
In one's complement, the negation is:
-2 = 1111 1111 1111 1111 1111 1111 1111 1101
Is that the negative? If so, how is it 2^n - x - 1?
--------------------
I understand two's complement:
Simply invert every 0 to 1 and every 1 to 0, then add one to the result. This shortcut is based on the observation that the sum of a number and its inverted representation must be 111... (32 of them) 111, which represents -1.
So 2 = 0000 0000 0000 0000 0000 0000 0000 0010
if we negate 2:
1111 1111 1111 1111 1111 1111 1111 1101
and add 1:
1111 1111 1111 1111 1111 1111 1111 1110
which is = -2
But I'm a bit confused about one's complement. Is it just two's complement without the addition of 1?
I have a buffer which consists of data in unsigned char and two bytes form a 12 Bit value.
I found out that my system is little endian. The first byte in the buffer gives me on the console numbers from 0 to 255. The second byte gives always low numbers between 1 and 8 (measured data, so higher values up to 4 bit would be possible too).
I tried to shift them together so that I get an ushort with a correct 12 bit number.
Sadly at the moment I am totally confused about the endianess and what I have to shift how far in which direction.
I tried e.g. this:
ushort value =0;
value= (ushort) firstByte << 8 | (ushort) secondByte << 4;
Sadly the value of value is quite often bigger than 12 bit.
Where is the mistake?
It depends on how the bits are packed within the two bytes exactly, but the solution for the most likely packing would be:
value = firstByte | (secondByte << 8);
This assumes that the second byte contains the 4 most significant bits (bits 8..11), while the first byte contains the 8 least significant bits (bits 0..7).
Note: the above solution assumes that firstByte and secondByte are sensible unsigned types (e.g. uint8_t). If they are not (e.g. if you have used char or some other possibly signed type), then you'll need to add some masking:
value = (firstByte & 0xff) | ((secondByte & 0xf) << 8);
I think the main issue may not be with the values you're shifting alone. If these values are greater than their representative bits, they'll create a large value unless "and'd" out.
picture the following
0000 0000 1001 0010 << 8 | 0000 0000 0000 1101 << 4
1001 0010 0000 0000 | 0000 0000 1101 0000
You should notice the first problem here. The first 4 'lowest' values are not being used, and it's using up 16 bits. you only wanted twelve. This should be modified like so:
(these are new numbers to demonstrate something else)
0000 1101 1001 0010 << 8 | 0000 0000 0000 1101
1101 1001 0010 0000 | (0000 0000 0000 1101 & 0000 0000 0000 1111)
This will create the following value:
1101 1001 0010 1101
here, you should note that the value is still greater than the 12 bits. If your numbers don't extend passed the original 8bit, 4 bit size ignore this. Otherwise, you have to use the 'and' operation on the bits to eliminate the left most 4 bits.
0000 1111 1111 1111 & 0000 1001 0010 1101
These values can be created using either 0bXX macros, the 2^bits - 1 pattern, as well as various other forms.
I'm fairly new to bit manipulation and I'm trying to figure out how (1 << 31) - 1 works.
First I know that 1 << 31 is
1000000000000000000000000000
and I know it's actually complement of minimum int value, but when I tried to figure out (1 << 31) - 1, I found an explanation states that, it's just
10000000000000000000000000000000 - 1 = 01111111111111111111111111111111
I was almost tempted to believe it since it's really straightforward. But is this what really happening? If it's not, why it happens to be right?
My original thought was that, the real process should be: the two's complement of -1 is
11111111111111111111111111111111
then (1 << 31) - 1 =
(1)01111111111111111111111111111111
the leftmost 1 is abandoned, then we have maximum value of int.
I'm really confused about which one is right.
It's both! 1 << 31 is:
1000 0000 0000 0000 0000 0000 0000 0000
Subtracting 1 gives:
0111 1111 1111 1111 1111 1111 1111 1111
One of the nice features about the two's complement layout of signed numbers is that addition and subtraction are exactly the same operations as they are for unsigned numbers. So 10000...000 represents a negative number in two's complement, the largest negative number, which is -2,147,483,648 in this case, and subtracting 1 from it causes wrap-around to the largest positive number, 2,147,483,647, but two's complement numbers are arranged so that we can pretend it's an unsigned number instead, so the subtraction is uncomplicated. Subtracting 1 from 10000...000 simply drops the leading 1 to a 0, and borrows a bunch of 1s, same as in decimal you get a bunch of 9s: 10000 - 1 = 9999.
It's also true that mathematically, (a - b) is the same as (a + (-b)), so we can do (1 << 31) + (-1) instead:
1000 0000 0000 0000 0000 0000 0000 0000 (1 << 31)
1111 1111 1111 1111 1111 1111 1111 1111 (-1)
-----------------------------------------
1 0111 1111 1111 1111 1111 1111 1111 1111 +
0111 1111 1111 1111 1111 1111 1111 1111 (truncate)
A 1 is carried out of the high end, and lost once the result is truncated back into a 32-bit integer.
Either way, that pattern, with a single 0 at the high end, then filled by 1s, is the representation of the maximum positive value for a two's complement integer of any width.
There are other ways to generate that pattern if you prefer, such as ~(1 << 31), and (-1 >>> 1) (where >>> means logical shift right) which is agnostic of the width of the integer.