Hi I have a question about the bit representation in python
when I use bit operation 1<<31, then we can see the bits are
1000 0000 0000 0000 0000 0000 0000 0000
python will print this value as 2147483648
but when I give a variable value like a = -2**31
the bits are also
1000 0000 0000 0000 0000 0000 0000 0000
but python will print -2147483648
so if the bits are the same , how python decide to use 2147483648 or -2147483648 ?
In python integers do not have a limited precision. Which means among other things, that the numbers are not stored in twos compliment binary. The sign is NOT stored in the bit representation of the number.
So all of -2**31, 2**31 and 1<<31 will have the same bit representation for the number. The sign part of the -2**31 is not part of the bitwise representation of the number. The sign is separate.
You can see this if you try this:
>>> bin(5)
'0b101'
>>> bin(-5)
'-0b101'
The representation isn't really the same. You can use int.to_bytes to check it:
(1 << 31).to_bytes(32, 'big', signed=True)
b'\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x80\x00\x00\x00'
(-2 ** 31).to_bytes(32, 'big', signed=True)
b'\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\x80\x00\x00\x00'
Also, be careful about the - operator, which have the lower priority here:
-2 ** 31 == -(2 ** 31)
Related
I am currently testing some things with an accelerometer and its iio buffer and there is something that confuses me.
The sensor does have four different scan elements: x, y, z and a timestamp.
The indices of those values are:
x = 0, y = 1, z = 2 and time = 3. So far so good.
If I enable all available scan elements the order of the entries is set according to the description.
everything enabled:
0000010 f758 011c 3f64 c0b0 be90 0bfe 499f 0004
0000020 f724 0134 3f58 c0b0 3f2f 10ab 499f 0004
But once I have gaps, for example if I disable the scan element for y, the z value jumps onto index 1 and my buffer looks like this:
x, z and time:
0000010 f720 3f70 0000 0000 722a 5c13 4946 0004
0000020 f728 3f74 0000 0000 0958 60c0 4946 0004
z and time:
0000010 3f6c 0000 0000 0000 ca0b 6ef1 48be 0004
0000020 3f44 0000 0000 0000 edf7 739e 48be 0004
only x and z:
0000010 f720 3f48 f748 3f54 f744 3f5c f75c 3f68
0000020 f750 3f78 f738 3f80 f718 3f64 f700 3f50
I could not find further information on this but I am a bit confused and surprised that the scan elements do not respect their given index once the timestamp is activated and there is an index gap. Is this the normal behavior or is this some stuff that the current sensor driver mixes up?
I am allocating memory in the Jetson TX2. It has 8GB of RAM.
I need to specify the maximun GPU memory size available for TensorRT.
max_workspace_size_bytes = (has to be an integer)
I have seen some examples using these "values":
1<<20 = 1048576 (decimal)
= 0001 0000 0000 0000 0000
1<<30 = 1073741824
= 0001 0000 0000 0000 0000 0000 0000
But if I have 8GB of RAM, how can "1048576" or "1073741824" represent a part of RAM?
I have used this to allocate 3GB:
3*(10**9)
But I would like to understand the other way of representing a number.
Perhaps you're having a "Gb vs Gib" problem. Usually, 3 Gigas of RAM refers to 3,221,225,472 bytes instead of 3,000,000,000.
The first value is 3 * (2^10)*(2^10)*(2^10), a nice 3 (11) followed by 30 zeros in binary representation, while the second is 3 * (10^3)*(10^3)*(10^3), which is a mess in binary.
This convention of using powers of 2 instead of powers of 10 is the reason why you'll see people writing a 3Gb as 3 << 30:
3 << 30 == 3 * (1 << 10) * (1 << 10) * (1 << 10)
== 3 * (2**10 * 2**10 * 2**10)
== 3 * (2**30)
There's a related question and a good Wikipedia article about this issue if you want to learn more.
You can sum them up.
((1<<30)+(1<<31))
Or bitwise OR them.
((1<<30) | (1<<31))
Or shift a larger value than 1, e.g. 3.
(3<<30)
3GB = 3,221,225,472
1100 0000 0000 0000 0000 0000 0000 0000
3<<30 = 3GB
I have a problem that asks
"find the K-th bit of the binary representation of an integer N"
where 0 <= K <= 31.
The answer states that when N=1 K=0 , the k-th bit is 1
and also
when N=2 K=1 , the k-th bit is 1 as well. How is this so?
The little endian binary representation of 1 is 0000 0000 0000 0001 and the little endian binary representation of 2 is 0000 0000 0000 0010.
I'm having a topic here which is from "Number Systems" in the subject of "Introduction to Computer Organisation & Architecture"
Then i came across this topic,"Self complementing Codes"
There are 3 parts of it which are as follows:
i)Excess-3 (I understand this part as it requires us to add 3 to BCD)
ii)84-2-1 (I don't understand)
iii)2*421 (I don't understand)
I hope someone could explain how the part ii & iii works.
Thanks alot.
I think this part of "Digital Design" book from Morris Mano will answer your question:
BCD and the 2421 code are examples of weighted codes. In a weighted code, each bit
position is assigned a weighting factor in such a way that each digit can be evaluated by
adding the weights of all the 1’s in the coded combination.
Four Different Binary Codes for the Decimal Digits
_____________________________________________________________
Decimal BCD 2421 Excess‐3 8, 4, -2, -1
Digit 8421
_____________________________________________________________
0 0000 0000 0011 0000
1 0001 0001 0100 0111
2 0010 0010 0101 0110
3 0011 0011 0110 0101
4 0100 0100 0111 0100
5 0101 1011 1000 1011
6 0110 1100 1001 1010
7 0111 1101 1010 1001
8 1000 1110 1011 1000
9 1001 1111 1100 1111
_____________________________________________________________
1010 0101 0000 0001
Unused 1011 0110 0001 0010
bit 1100 0111 0010 0011
combi- 1101 1000 1101 1100
nations 1110 1001 1110 1101
1111 1010 1111 1110
The 2421, the excess‐3 and the 84-2-1 codes are examples of self‐complementing codes. Such
codes have the property that the 9’s complement of a decimal number is obtained
directly by changing 1’s to 0’s and 0’s to 1’s (i.e., by complementing each bit in the pattern). For example, decimal 395 is represented in the excess‐3 code as 0110 1100 1000.
The 9’s complement of 604 is represented as 1001 0011 0111, which is obtained simply
by complementing each bit of the code (as with the 1’s complement of binary numbers).
Digital Design-Fifth edition-By Morris Mano
First of all 84-2-1 & 2421 code are "weighted code" and as well as "self-complementing code" both (because the necessary condition for a code to be self-complementing is that the sum of all of its weight must be equal to 9) i.e. 84-2-1(8+4-2-1=9) and, 2421(2+4+2+1=9).
So, constructing binary equivalent of decimal of number you to need to make sure one thing:
The number's code and its 9's complement's code should have complementary relationship (i.e. the number & its 9's complement code are complements of each other)
For example let's take a 84-2-1 System.
Decimal 0=0000(in 84-2-1 system) & (9's complement of 0=9) than 9(in 84-2-1) should have 1111. Hence 0 & its 9's complement i.e. 9 maintain their self-complement relationship.
Let's take another example:
Decimal 1=0111(in 84-2-1 system) & (9's complement of 1=8 ) that implies (84-2-1) equivalent of 8 should be 1000,hence again the number & its complement preserve their self-complementary relationship.
Similarly, as in 2421 system, all you need to construct the code by making sure that the number and its 9's complement should have maintained their self-complementing relationship.
//((Hello everybody!
i am C# beginner can any one tell me the function of left and right shift operator and their working way w.r.t the following program. I read it somewhere but confuse.
thanks ))
using System;
class clc
{
public static void Main() // the Main method
{
int x = 7, y = 2, z, r;
z = x << y ; //left shift operator
r = x >> y; // right shift operator
Console.WriteLine("\n z={3}\tr={4} ",z,r);
}
}
To understand the shift operations you must understand binary numbers.
Let's take your example for left shift:
z = 7 << 2;
32 bit integer 7 is 0000 0000 0000 0000 0000 0000 0000 0111 in binary. You must move the bits to the left beginning from the right. The bits that are shifted out of either end are discarded.
Shifting it by 1 will result 0000 0000 0000 0000 0000 0000 0000 1110
Shifting it by 1 one more time will result 0000 0000 0000 0000 0000 0000 0001 1100 which is 28 in integer representation.
Read this good wikipedia article Binary number