I am running a lua script in redis something like below:
eval "return tostring(tonumber(ARGV[1]))" 0 1538409827183989630
which should return 1538409827183989630 but is returning this 1.538409827184e+18 which is dropping last few nano second digits(Its a timestamp in nano seconds)
What is the correct way to do this assuming i need nano seconds precision cause i am comparing timestamps..
Lua numbers have two subtypes, integer and float which Lua chooses automatically.
One of the few if not the only situation where you have to care about that difference is when you want to convert a number to a string.
print(1538409827183989630) will print 1538409827183989630
print(1538409827183989630.0) will print as 1.538409827184e+18
If you want to make sure you get the same output you'll have to explicitly format the string.
local int = 1538409827183989630
local float = 1538409827183989630.0
print(string.format("%d", int))
print(string.format("%d", float))
output:
1538409827183989630
1538409827183989504
You'll notice that there will be a difference between both numbers due to the float -> integer conversion.
Related
I have written a function that outputs a double, upto 25 decimal
places. I am trying to print it as a formatted output from Raku.
However, the output is incorrect and truncated.
See MWE:
my $var = 0.8144262510988963255087469;
say sprintf("The variable value is: %.25f", $var)
The above code gives The variable value is: 0.8144262510988963000000000 which is not what is expected.
Also, this seems weird:
my $var = 0.8144262510988963255087469;
say $var.Str.chars; # 29 wrong, expected 27
I tested the same in C:
#include <stdio.h>
int main() {
double var = 0.8144262510988963255087469;
printf("The variable value is: %.25lf \n", var);
return 0;
}
However, it works fine. Given the identical nature of sprintf and printf, I expected this C example to work in Raku too. Seems like %lf is not supported.
So is there a workaround to fix this?
I think this is actually a bug in how Rat literals are created. Or at least as WAT :-).
I actually sort of expect 0.8144262510988963255087469 to either give a compile time warning, or create a Num, as it exceeds the standard precision of a Rat:
raku -e 'say 0.8144262510988963255087469'
0.814426251098896400086204416
Note that these are not the same.
There is fortunately an easy workaround, by creating a FatRat
$ raku -e 'say 0.8144262510988963255087469.FatRat'
0.8144262510988963255087469
FWIW, I think this is worthy of creating an issue
From your question:
I have written a function that outputs a double, upto 25 decimal places.
From google:
Double precision numbers are accurate up to sixteen decimal places
From the raku docs :
When constructing a Rat (i.e. when it is not a result of some mathematical expression), however, a larger denominator can be used
so if you go
my $v = 0.8144262510988963255087469;
say $v.raku;
#<8144262510988963255087469/10000000000000000000000000>
it works.
However, do a mathematical expression such as
my $b = $a/10000000000000000000000000;
and you get the Rat => Num degradation applied unless you explicitly declare FatRats. I visualise this as the math operation placing the result in a Num register in the CPU.
The docs also mention that .say and .put may be less faithful than .raku, presumably because they use math operations (or coercion) internally.
Sorry to be the bearer of bad news, but 10**25 > 2 **64, but what you report as an issue is correct & (fairly) well documented behaviour given the constraints of double precision IEEE P754.
from GNU gawk's page
https://www.gnu.org/software/gawk/manual/html_node/Checking-for-MPFR.html
they have a formula to check arbitrary precision
function adequate_math_precision(n) { return (1 != (1+(1/(2^(n-1))))) }
My question is : wouldn't it be more efficient by staying within integer math domain with a formula such as
( 2^abs(n) - 1 ) % 2 # note 2^(n-1) vs. 2^|n| - 1
Since any power of 2 must also be even, then subtracting 1 must always be odd, then its modulo (%) over 2 becomes indicator function for is_odd() for n >= 0, while the abs(n) handles the cases where it's negative.
Or does the modulo necessitate a casting to float point, thus nullifying any gains ?
Good question. Let's tackle it.
The proposed snippet aims at checking wether gawk was invoked with the -M option.
I'll attach some digression on that option at the bottom.
The argument n of the function is the floating point precision needed for whatever operation you'll have to perform. So, say your script is in a library somewhere and will get called but you have no control over it. You'll run that function at the beginning of the script to promptly throw exception and bail out, suggesting that the end result will be wrong due to lack of bits to store numbers.
Your code stays in the integer realm: a power of two of an integer is an integer. There is no need to use abs(n) here, because there is no point in specifying how many bits you'll need as a negative number in the first place.
Then you subtract one from an even, integer number. Now, unless n=0, in which case 2^0=1 and then your code reads (1 - 1) % 2 = 0, your snippet shall always return 1, because the quotient (%) of an odd number divided by two is 1.
Problem is: you are trying to calculate a potentially stupidly large number in a function that should check if you are able to do so in the first place.
Since any power of 2 must also be even, then subtracting 1 must always
be odd, then its modulo (%) over 2 becomes indicator function for
is_odd() for n >= 0, while the abs(n) handles the cases where it's
negative.
Except when n=0 as we discussed above, you are right. The snippet will tell that any power of 2 is even, and any power of 2, minus 1, is odd. We were discussing another subject entirely thought.
Let's analyze the other function instead:
return (1 != (1+(1/(2^(n-1)))))
Remember that booleans in awk runs like this: 0=false and non zero equal true. So, if 1+x where x is a very small number, typically a large power of two (2^122 in the example page) is mathematically guaranteed to be !=1, in the digital world that's not the case. At one point, floating computation will reach a precision rock bottom, will be rounded down, and x=0 will be suddenly declared. At that point, the arbitrary precision function will return 0: false: 1 is equal 1.
A larger discussion on types and data representation
The page you link explains precision for gawk invoked with the -M option. This sounds like technoblahblah, let's decipher it.
At one point, your OS architecture has to decide how to store data, how to represent it in memory so that it can be accessed again and displayed. Terms like Integer, Float, Double, Unsigned Integer are examples of data representation. We here are addressing Integer representation: how is an integer stored in memory?
A 32-bit system will use 4 bytes to represent and integer, which in turn determines how larger the integer will be. The 32 bits are read from most significative (MSB) to less significative (LSB) and if signed, one bit will represent the sign (the MSB typically, drastically reducing the max size of the integer).
If asked to compute a large number, a machine will try to fit in in the max number available. If the end result is larger than that, you have overflow and end up with a wrong result or an error. Many online challenges typically ask you to write code for arbitrary long loops or large sums, then test it with inputs that will break the 64bit barrier, to see if you master proper types for indexes.
AWK is not a strongly typed language. Meaning, any variable can store data, regardless of the type. The data type can change and it is determined at runtime by the interpreter, so that the developer doesn't need to care. For instance:
$awk '{a="this is text"; print a; a=2; print a; print a+3.0*2}'
-| this is text
-| 2
-| 8
In the example, a is text, then is an integer and can be summed to a floating point number and printed as integer without any special type handling.
The Arbitrary Precision Page presents the following snippet:
$ gawk -M 'BEGIN {
> s = 2.0
> for (i = 1; i <= 7; i++)
> s = s * (s - 1) + 1
> print s
> }'
-| 113423713055421845118910464
There is some math voodoo behind, we will skip that. Since s is interpreted as a floating point number, the end result is computed as floating point.
Try to input that number on Windows calculator as decimal, and it will fail. Although you can compute it as a binary. You'll need the programmer setting and to add up to 53 bits to be able to fit it as unsigned integer.
53 is a magic number here: with the -M option, gawk uses arbitrary precision for numbers. In other words, it commandeers how many bits are necessary, track them and breaks free of the native OS architecture. The default option says that gawk will allocate 53 bits for any given arbitrary number. Fun fact, the actual result of that snippet is wrong, and it would take up to 100 bits to compute correctly.
To implement arbitrary large numbers handling, gawk relies on an external library called MPFR. Provided with an arbitrary large number, MPFR will handle the memory allocation and bit requisition to store it. However, the interface between gawk and MPFR is not perfect, and gawk can't always control the type that MPFR will use. In case of integers, that's not an issue. For floating point numbers, that will result in rounding errors.
This brings us back to the snippet at the beginning: if gawk was called with the -M option, numbers up to 2^53 can be stored as integers. Floating points will be smaller than that (you'll need to make the comma disappear somehow, or rather represent it spending some of the bits allocated for that number, just like the sign). Following the example of the page, and asking an arbitrary precision larger than 32, the snippet will return TRUE only if the -M option was passed, otherwise 1/2^(n-1) will be rounded down to be 0.
This might sound crazy but I'm working with floor numbers of a building which has -0 as floor and -0A, B, and C so and so.
My user in entering floor data randomly. In the end I'm supposed to sort the array of these floor numbers. What i found is that even if i enter -0 as floor number and try to sort it.. it sorts it as 0 because for computer -0 is still 0.
How do i define -0 and -1<-0<0?
By definition, you are no longer working with integers as -0 != 0 makes no sense in the realm of integers.
So, yes, you're going to have to define your own type and implement your own sorting rules. Simply storing them as strings and then implementing a sorting block to sort an array of them is straightforward, though.
You could go down the path of using floats so you could have floor 0 and floor -0.1, then round for display. But that sort of shenanigans will lead the maintainer after the maintainer after you to call you unpleasant names (which is sometimes OK). :)
All answers so far are suggesting your own type/functions and probably strings.
Strings will work but you can take an idea from floating-point and store them as sign (boolean or even bit flag) and magnitude (unsigned integer type, 8 or 16 bits should be sufficient).
Comparison is simply compare signs and then compare magnitudes if required.
You could use a struct for such a type which would give you the same value semantics as integer and real types and avoid object allocation.
If there is also a letter ("-0 as floor and -0A, B, and C so and so") that can be a third field in the struct, probably a char, and you could still have value semantics.
HTH
I've got a program, which compute a several variables and then these variables are writing in to the output file.
Is it possilbe, that when my program can't get a correct results for my formula, it does'nt terminate?
To clarify what I do, here is part of my code, where the variable of my interest are compute:
dx=x(1,i)-x(nk,i)
dy=y(1,i)-y(nk,i)
dz=z(1,i)-z(nk,i)
call PBC(dx,dy,dz)
r2i=dx*dx+dy*dy+dz*dz
r2=r2+r2i
r2g0=0.0d0
r2gx=0.0d0
dx=x(1,i)-x(2,i)
call PBC(dx,dy,dz)
rspani=dsqrt(dx*dx)
do ii=1,nk-1
rx=x(ii,i)
ry=y(ii,i)
rz=z(ii,i)
do jj=ii+1,nk
dx=x(jj,i)-rx
dy=y(jj,i)-ry
dz=z(jj,i)-rz
call PBC(dx,dy,dz)
r21=dx*dx+dy*dy+dz*dz
r21x=dx*dx
r2g=r2g+r21
r2gx=r2gx+r21x
r2g0=r2g0+r21
rh=rh+1.0d0/dsqrt(r21)
rh1=rh1+1.0d0
ir21=dnint(dsqrt(r21)/dr)
p(ir21)=p(ir21)+2.0D0
dxs=dsqrt(r21x)
if(dxs.gt.rspani) rspani=dxs
end do
and then in to the output I just write these variables:
write(12,870)r2i,sqrt(r2i),r2g0,r2gx/(nk*nk)
870 FORMAT(3(f15.7,3x),f15.7)
The x, y, z are actully generate via a random number generator.
The problem is that my output contains, correct values for lets say 457 lines, and then a one line is just "*********" when I use mc viewer and then the output continues with correct values, but let's say 12 steps form do cycle which compute these variables is missing.
So my questions are basic:
Is it possible, that my program can't get a correct numbers, and that's why the result is not writing in to the program?
or
Could it this been caused due to wrong output formating or something related with formating?
Thank you for any suggestion
********* is almost certainly the result of trying to write a number too large for the field specified in a format string.
For example, a field specified as f15.7 will take 1 spot for the decimal point, 1 spot for a leading sign (- will always be printed if required, + may be printed if options are set), 7 for the fractional digits, leaving 6 digits for the whole part of the number. There may therefore be cases where the program won't fit the number into the field and will print 15 *s instead.
Programs compiled with an up to date Fortran compiler will write a string such as NaN or -Inf if they encounter a floating-point number which represents one of the IEEE special values
I need to Convert a string say '12/12/2013 14:30:56.583' to be converted in Date Format like 2013-12-12 14:30:56.583 in Informix database.
I Used following function
to_date('12/12/2013 14:30:56.583',"%d/%m/%Y %H:%M:%S.")
But its not accepting Milliseconds , Milliseconds are important to the resulting value.
The database version is important. The behaviour of %F was recently (11.70.xC8 and 12.10.xC2) changed. In previous versions the "." dot must probably be omitted as well as the "n" qualifier.
Regards
If you check the manual you will see is missing the milliseconds at the string format.
source: http://www-01.ibm.com/support/knowledgecenter/api/content/SSGU8G_12.1.0/com.ibm.sqls.doc/ids_sqs_1542.htm
%S Second as a 2-digit integer (00 through 61). The second value can
be up to 61 (instead of 59) to allow for the occasional leap second
and double leap second.
%Fn The value of the fraction of a second, with precision specified by
the unsigned integer n. The default value of n is 2; the range of n is
0 ≤ n ≤ 5. This value overrides any width or precision that is
specified between the % and F characters.
So, this probably will work:
to_date('12/12/2013 14:30:56.583',"%d/%m/%Y %H:%M:%S.%F3")
I’m not that familiar with Informix, but I think you may be able to use the standard to_date function to convert the string value to a date and then use an addMilliseconds function to add the milliseconds.
http://pic.dhe.ibm.com/infocenter/informix/v121/index.jsp?topic=%2Fcom.ibm.netpr.doc%2Fids_net_093.htm