I was following 'A tour of GO` on http://tour.golang.org.
The table 15 has some code that I cannot understand. It defines two constants with the following syntax:
const (
Big = 1<<100
Small = Big>>99
)
And it's not clear at all to me what it means. I tried to modify the code and run it with different values, to record the change, but I was not able to understand what is going on there.
Then, it uses that operator again on table 24. It defines a variable with the following syntax:
MaxInt uint64 = 1<<64 - 1
And when it prints the variable, it prints:
uint64(18446744073709551615)
Where uint64 is the type. But I can't understand where 18446744073709551615 comes from.
They are Go's bitwise shift operators.
Here's a good explanation of how they work for C (they work in the same way in several languages).
Basically 1<<64 - 1 corresponds to 2^64 -1, = 18446744073709551615.
Think of it this way. In decimal if you start from 001 (which is 10^0) and then shift the 1 to the left, you end up with 010, which is 10^1. If you shift it again you end with 100, which is 10^2. So shifting to the left is equivalent to multiplying by 10 as many times as the times you shift.
In binary it's the same thing, but in base 2, so 1<<64 means multiplying by 2 64 times (i.e. 2 ^ 64).
That's the same as in all languages of the C family : a bit shift.
See http://en.wikipedia.org/wiki/Bitwise_operation#Bit_shifts
This operation is commonly used to multiply or divide an unsigned integer by powers of 2 :
b := a >> 1 // divides by 2
1<<100 is simply 2^100 (that's Big).
1<<64-1 is 2⁶⁴-1, and that's the biggest integer you can represent in 64 bits (by the way you can't represent 1<<64 as a 64 bits int and the point of table 15 is to demonstrate that you can have it in numerical constants anyway in Go).
The >> and << are logical shift operations. You can see more about those here:
http://en.wikipedia.org/wiki/Logical_shift
Also, you can check all the Go operators in their webpage
It's a logical shift:
every bit in the operand is simply moved a given number of bit
positions, and the vacant bit-positions are filled in, usually with
zeros
Go Operators:
<< left shift integer << unsigned integer
>> right shift integer >> unsigned integer
Related
I have recently sat a computing exam in university in which we were never taught beforehand about the modulus function or any other check for odd/even function and we have no access to external documentation except our previous lecture notes. Is it possible to do this without these and how?
Bitwise AND (&)
Extract the last bit of the number using the bitwise AND operator. If the last bit is 1, then it's odd, else it's even. This is the simplest and most efficient way of testing it. Examples in some languages:
C / C++ / C#
bool is_even(int value) {
return (value & 1) == 0;
}
Java
public static boolean is_even(int value) {
return (value & 1) == 0;
}
Python
def is_even(value):
return (value & 1) == 0
I assume this is only for integer numbers as the concept of odd/even eludes me for floating point values.
For these integer numbers, the check of the Least Significant Bit (LSB) as proposed by Rotem is the most straightforward method, but there are many other ways to accomplish that.
For example, you could use the integer division operation as a test. This is one of the most basic operation which is implemented in virtually every platform. The result of an integer division is always another integer. For example:
>> x = int64( 13 ) ;
>> x / 2
ans =
7
Here I cast the value 13 as a int64 to make sure MATLAB treats the number as an integer instead of double data type.
Also here the result is actually rounded towards infinity to the next integral value. This is MATLAB specific implementation, other platform might round down but it does not matter for us as the only behavior we look for is the rounding, whichever way it goes. The rounding allow us to define the following behavior:
If a number is even: Dividing it by 2 will produce an exact result, such that if we multiply this result by 2, we obtain the original number.
If a number is odd: Dividing it by 2 will result in a rounded result, such that multiplying it by 2 will yield a different number than the original input.
Now you have the logic worked out, the code is pretty straightforward:
%% sample input
x = int64(42) ;
y = int64(43) ;
%% define the checking function
% uses only multiplication and division operator, no high level function
is_even = #(x) int64(x) == (int64(x)/2)*2 ;
And obvisouly, this will yield:
>> is_even(x)
ans =
1
>> is_even(y)
ans =
0
I found out from a fellow student how to solve this simplistically with maths instead of functions.
Using (-1)^n :
If n is odd then the outcome is -1
If n is even then the outcome is 1
This is some pretty out-of-the-box thinking, but it would be the only way to solve this without previous knowledge of complex functions including mod.
I can see in Apple's documentation that enumerations are sometimes defined like this
enum {
UICollectionViewScrollPositionTop = 1 << 0,
UICollectionViewScrollPositionBottom = 1 << 1
}
What does the << mean?
It's the bitwise shift left operator. It's used to create values having a single bit set, very common when combination through bitwise OR is intended.
For those values, you might later say:
const int top_and_bottom = UICollectionViewScrollPositionTop | UICollectionViewScrollPositionBottom;
which would result in top_and_bottom being set to 3 (binary 112).
Here it is simply left bit shift. So this means 1<<0 = 1 for instance. And 1<<1 is two. Maybe the author chose this way to initialize the enumeration to emphasize on the fact that UICollectionViewScrollPositionTop has only the least significant bit on and UICollectionViewScrollPositionBottom has only the second to least significant bit on. I guess the usage for this enumeration is to somehow later form bitmasks.
<< stands for left shift.
It shifts the binary to specified bits, as 4<<1 will be 8 and 4<<2 will be 16.
Each left shift makes the value multiplied by 2.
1<<0 will be 1 while 1<<1 will be 2.
Check here
I am currently trying to figure out how to multiply two numbers in fixed point representation.
Say my number representation is as follows:
[SIGN][2^0].[2^-1][2^-2]..[2^-14]
In my case, the number 10.01000000000000 = -0.25.
How would I for example do 0.25x0.25 or -0.25x0.25 etc?
Hope you can help!
You should use 2's complement representation instead of a seperate sign bit. It's much easier to do maths on that, no special handling is required. The range is also improved because there's no wasted bit pattern for negative 0. To multiply, just do as normal fixed-point multiplication. The normal Q2.14 format will store value x/214 for the bit pattern of x, therefore if we have A and B then
So you just need to multiply A and B directly then divide the product by 214 to get the result back into the form x/214 like this
AxB = ((int32_t)A*B) >> 14;
A rounding step is needed to get the nearest value. You can find the way to do it in Q number format#Math operations. The simplest way to round to nearest is just add back the bit that was last shifted out (i.e. the first fractional bit) like this
AxB = (int32_t)A*B;
AxB = (AxB >> 14) + ((AxB >> 13) & 1);
You might also want to read these
Fixed-point arithmetic.
Emulated Fixed Point Division/Multiplication
Fixed point math in c#?
With 2 bits you can represent the integer range of [-2, 1]. So using Q2.14 format, -0.25 would be stored as 11.11000000000000. Using 1 sign bit you can only represent -1, 0, 1, and it makes calculations more complex because you need to split the sign bit then combine it back at the end.
Multiply into a larger sized variable, and then right shift by the number of bits of fixed point precision.
Here's a simple example in C:
int a = 0.25 * (1 << 16);
int b = -0.25 * (1 << 16);
int c = (a * b) >> 16;
printf("%.2f * %.2f = %.2f\n", a / 65536.0, b / 65536.0 , c / 65536.0);
You basically multiply everything by a constant to bring the fractional parts up into the integer range, then multiply the two factors, then (optionally) divide by one of the constants to return the product to the standard range for use in future calculations. It's like multiplying prices expressed in fractional dollars by 100 and then working in cents (i.e. $1.95 * 100 cents/dollar = 195 cents).
Be careful not to overflow the range of the variable you are multiplying into. Your constant might need to be smaller to avoid overflow, like using 1 << 8 instead of 1 << 16 in the example above.
For example I have a float 55.2f and want to round it down such that the result can be divided by two without rest.
So 55.2 would become 54 as that is the nearest smaller "step" that can be divided by two. Is there a function for this or must I write my own algorithm?
If your result must remain a float, you can do:
float f=55.2f;
f=floorf(f/2.f)*2.f;
First convert to an integral type, such as int or long, and then clear the lowest bit.
float f = 55.2f;
int i = (int)f & ~1;
Explanation
~ means the bitwise inverse, i.e. all the 0 bits become 1, and vice versa.
So, if 1 has the bit pattern
0...0001
then ~1 is
1...1110
(Here I'm using ... to represent all the in-between bits depending on how big an integer is on your platform.)
When you & (bitwise AND) your integer with 1...1110, you are preserving the value of each bit apart from the lowest bit, which is forced to 0. See this description of the bitwise AND operator if you still don't get it.
By forcing the lowest bit to be 0, you are rounding the number down to the nearest even number.
You can write your own algorithm, for example with bitwise operators.
The following code works with clearing the last bit of your number. An even number has indeed the last bit not set.
int
f(float x)
{
return (int)x & ~1;
}
How about long int f = lrintf(x / 2);, where x is your float?
You could also just say int f = x / 2;, but some people have argued that that's more expensive, because the C standard mandates a specific rounding mode which may or may not be native to the CPU. The lrintf function on the other hand uses the CPU's native rounding mode. You need to #include <math.h>.
int x = n / 3; // <-- make this faster
// for instance
int a = n * 3; // <-- normal integer multiplication
int b = (n << 1) + n; // <-- potentially faster multiplication
The guy who said "leave it to the compiler" was right, but I don't have the "reputation" to mod him up or comment. I asked gcc to compile int test(int a) { return a / 3; } for an ix86 and then disassembled the output. Just for academic interest, what it's doing is roughly multiplying by 0x55555556 and then taking the top 32 bits of the 64 bit result of that. You can demonstrate this to yourself with eg:
$ ruby -e 'puts(60000 * 0x55555556 >> 32)'
20000
$ ruby -e 'puts(72 * 0x55555556 >> 32)'
24
$
The wikipedia page on Montgomery division is hard to read but fortunately the compiler guys have done it so you don't have to.
This is the fastest as the compiler will optimize it if it can depending on the output processor.
int a;
int b;
a = some value;
b = a / 3;
There is a faster way to do it if you know the ranges of the values, for example, if you are dividing a signed integer by 3 and you know the range of the value to be divided is 0 to 768, then you can multiply it by a factor and shift it to the left by a power of 2 to that factor divided by 3.
eg.
Range 0 -> 768
you could use shifting of 10 bits, which multiplying by 1024, you want to divide by 3 so your multiplier should be 1024 / 3 = 341,
so you can now use (x * 341) >> 10
(Make sure the shift is a signed shift if using signed integers), also make sure the shift is an actually shift and not a bit ROLL
This will effectively divide the value 3, and will run at about 1.6 times the speed as a natural divide by 3 on a standard x86 / x64 CPU.
Of course the only reason you can make this optimization when the compiler cant is because the compiler does not know the maximum range of X and therefore cannot make this determination, but you as the programmer can.
Sometime it may even be more beneficial to move the value into a larger value and then do the same thing, ie. if you have an int of full range you could make it an 64-bit value and then do the multiply and shift instead of dividing by 3.
I had to do this recently to speed up image processing, i needed to find the average of 3 color channels, each color channel with a byte range (0 - 255). red green and blue.
At first i just simply used:
avg = (r + g + b) / 3;
(So r + g + b has a maximum of 768 and a minimum of 0, because each channel is a byte 0 - 255)
After millions of iterations the entire operation took 36 milliseconds.
I changed the line to:
avg = (r + g + b) * 341 >> 10;
And that took it down to 22 milliseconds, its amazing what can be done with a little ingenuity.
This speed up occurred in C# even though I had optimisations turned on and was running the program natively without debugging info and not through the IDE.
See How To Divide By 3 for an extended discussion of more efficiently dividing by 3, focused on doing FPGA arithmetic operations.
Also relevant:
Optimizing integer divisions with Multiply Shift in C#
Depending on your platform and depending on your C compiler, a native solution like just using
y = x / 3
Can be fast or it can be awfully slow (even if division is done entirely in hardware, if it is done using a DIV instruction, this instruction is about 3 to 4 times slower than a multiplication on modern CPUs). Very good C compilers with optimization flags turned on may optimize this operation, but if you want to be sure, you are better off optimizing it yourself.
For optimization it is important to have integer numbers of a known size. In C int has no known size (it can vary by platform and compiler!), so you are better using C99 fixed-size integers. The code below assumes that you want to divide an unsigned 32-bit integer by three and that you C compiler knows about 64 bit integer numbers (NOTE: Even on a 32 bit CPU architecture most C compilers can handle 64 bit integers just fine):
static inline uint32_t divby3 (
uint32_t divideMe
) {
return (uint32_t)(((uint64_t)0xAAAAAAABULL * divideMe) >> 33);
}
As crazy as this might sound, but the method above indeed does divide by 3. All it needs for doing so is a single 64 bit multiplication and a shift (like I said, multiplications might be 3 to 4 times faster than divisions on your CPU). In a 64 bit application this code will be a lot faster than in a 32 bit application (in a 32 bit application multiplying two 64 bit numbers take 3 multiplications and 3 additions on 32 bit values) - however, it might be still faster than a division on a 32 bit machine.
On the other hand, if your compiler is a very good one and knows the trick how to optimize integer division by a constant (latest GCC does, I just checked), it will generate the code above anyway (GCC will create exactly this code for "/3" if you enable at least optimization level 1). For other compilers... you cannot rely or expect that it will use tricks like that, even though this method is very well documented and mentioned everywhere on the Internet.
Problem is that it only works for constant numbers, not for variable ones. You always need to know the magic number (here 0xAAAAAAAB) and the correct operations after the multiplication (shifts and/or additions in most cases) and both is different depending on the number you want to divide by and both take too much CPU time to calculate them on the fly (that would be slower than hardware division). However, it's easy for a compiler to calculate these during compile time (where one second more or less compile time plays hardly a role).
For 64 bit numbers:
uint64_t divBy3(uint64_t x)
{
return x*12297829382473034411ULL;
}
However this isn't the truncating integer division you might expect.
It works correctly if the number is already divisible by 3, but it returns a huge number if it isn't.
For example if you run it on for example 11, it returns 6148914691236517209. This looks like a garbage but it's in fact the correct answer: multiply it by 3 and you get back the 11!
If you are looking for the truncating division, then just use the / operator. I highly doubt you can get much faster than that.
Theory:
64 bit unsigned arithmetic is a modulo 2^64 arithmetic.
This means for each integer which is coprime with the 2^64 modulus (essentially all odd numbers) there exists a multiplicative inverse which you can use to multiply with instead of division. This magic number can be obtained by solving the 3*x + 2^64*y = 1 equation using the Extended Euclidean Algorithm.
What if you really don't want to multiply or divide? Here is is an approximation I just invented. It works because (x/3) = (x/4) + (x/12). But since (x/12) = (x/4) / 3 we just have to repeat the process until its good enough.
#include <stdio.h>
void main()
{
int n = 1000;
int a,b;
a = n >> 2;
b = (a >> 2);
a += b;
b = (b >> 2);
a += b;
b = (b >> 2);
a += b;
b = (b >> 2);
a += b;
printf("a=%d\n", a);
}
The result is 330. It could be made more accurate using b = ((b+2)>>2); to account for rounding.
If you are allowed to multiply, just pick a suitable approximation for (1/3), with a power-of-2 divisor. For example, n * (1/3) ~= n * 43 / 128 = (n * 43) >> 7.
This technique is most useful in Indiana.
I don't know if it's faster but if you want to use a bitwise operator to perform binary division you can use the shift and subtract method described at this page:
Set quotient to 0
Align leftmost digits in dividend and divisor
Repeat:
If that portion of the dividend above the divisor is greater than or equal to the divisor:
Then subtract divisor from that portion of the dividend and
Concatentate 1 to the right hand end of the quotient
Else concatentate 0 to the right hand end of the quotient
Shift the divisor one place right
Until dividend is less than the divisor:
quotient is correct, dividend is remainder
STOP
For really large integer division (e.g. numbers bigger than 64bit) you can represent your number as an int[] and perform division quite fast by taking two digits at a time and divide them by 3. The remainder will be part of the next two digits and so forth.
eg. 11004 / 3 you say
11/3 = 3, remaineder = 2 (from 11-3*3)
20/3 = 6, remainder = 2 (from 20-6*3)
20/3 = 6, remainder = 2 (from 20-6*3)
24/3 = 8, remainder = 0
hence the result 3668
internal static List<int> Div3(int[] a)
{
int remainder = 0;
var res = new List<int>();
for (int i = 0; i < a.Length; i++)
{
var val = remainder + a[i];
var div = val/3;
remainder = 10*(val%3);
if (div > 9)
{
res.Add(div/10);
res.Add(div%10);
}
else
res.Add(div);
}
if (res[0] == 0) res.RemoveAt(0);
return res;
}
If you really want to see this article on integer division, but it only has academic merit ... it would be an interesting application that actually needed to perform that benefited from that kind of trick.
Easy computation ... at most n iterations where n is your number of bits:
uint8_t divideby3(uint8_t x)
{
uint8_t answer =0;
do
{
x>>=1;
answer+=x;
x=-x;
}while(x);
return answer;
}
A lookup table approach would also be faster in some architectures.
uint8_t DivBy3LU(uint8_t u8Operand)
{
uint8_t ai8Div3 = [0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, ....];
return ai8Div3[u8Operand];
}