I have two cards which contains a hex value, I am struggling to find out what kind of algorithm is used ti get the decimal value.
8HEX from chip: 0b98c44a Printed on card: 3491308370
8HEX from chip: 0c96425c Printed on card: 812204602
does any one of you number experts in here find out how it is done, you helped me with this before :)
The relation between the numbers seems to be that the bits of every byte has been reversed.
When you print first pair in binary they are:
11010000 00011001 00100011 01010010 = 3491308370
00001011 10011000 11000100 01001010 = 0x0b98c44a
The second pair is:
00110000 01101001 01000010 00111010 = 812204602
00001100 10010110 01000010 01011100 = 0x0c96425c
If you want to know how to convert one number to the other, you should mention which programming language you are using.
Related
MQL4 documentation states that the value limits for double type variables is:
"Minimal Positive Value" = 2.2250738585072014e-308
"Maximum Value" = 1.7976931348623158e+308
See https://docs.mql4.com/basis/types/double
Why does StringToDouble() alter the value converted?
Am I doing one thing while expecting a different result?
void OnStart() {
string s1 = "5554535251504900090807060504030201";
double d1 = StringToDouble(s1);
string s2 = DoubleToString(d1);
Print("s2<",s2,">");
printf("%099.8f",d1);
Print("s1<",s1,">");
return;<br>
}
Here's what I get when I run that code:
s1<5554535251504900090807060504030201>
d1<000000000000000000000000000000000000000000000000000000005554535251504899684469244159852544.00000000>
s2<5554535251504899684469244159852544>
5554535251504900090807060504030201 amounts to5.55454E+33.
Obviously, that doesn't even come remotely close to the 1.7976931348623158e+308 limit.
What am I missing here?
Q : "What am I missing here?"
The documented facts.
MQL4 uses no more than 4-bytes to store int.
MQL4 uses no more than 8-bytes to store double.
IEEE-754 standard defines the rest - how many bits from those 64 are reserved for: exponent ( -308, 0, +308 )
sign ( +, - ) and
the rest, for normalised form of the mantissa : 0.???????...????
Argument, that an actual number is far from either "edge" of < DBL_MIN, DBL_MAX > does explain nothing about the shallow-ness of the exact number reduced-precision representation ( see DBL_EPSILON ~ 2E-16 or DBL_DIG ~ 15-significant digits, or DBL_MANT_DIG ~ 53-bits, left from a 64-bit ( 8-Byte ) storage-cell for mantissa ).
There are many numbers, that simply cannot be stored exactly, using IEEE-754 floating point number representation.
Tons of literature explain this, so feel free to dig deeper, or may use another tools, that rely on infinite-(unlimited)-precision number representation, should your use-case requires that.
How do I get a random number in Lua to the eighth decimal?
Example : 0.00000001
I have tried the following and several variations of this but can not get the format i need.
math.randomseed( os.time() )
x = math.random(10000000,20000000) * 0.00000001
print(x)
i would like to put in say 200 and get this 0.00000200
Just grab a random number from 0-9, and slide it down 6 places. You can use format specifiers to create the string representation of the number that you desire. For floats we use %f, and indicate how many decimal places we want to have with an intermediate .n, where n is a number.
math.randomseed(os.time())
-- random(9) to exclude 0
print(('%.8f'):format(math.random(0, 9) * 1e-6))
--> '0.00000400'
string.format("%.8f",math.random())
to help anyone else. my question should have been worded a bit better. i wanted to be able to get random numbers and get it to the 8th decimal place.
but i wanted to be able to have those numbers from 1-10,000 so he is updated how i wanted it and the help of Oka got me to this
math.randomseed(os.time())
lowest = 1
highest = 7000
rand=('%.8f'):format(math.random(lowest, highest) / 100000000)
print(rand)
Hope this helps someone else or if it can be cleaned up please let me know
I'm writing a Wireshark dissector in lua and trying to decode a time-based protocol field.
I've two components 1)
local ref_time = os.time{year=2000, month=1, day=1, hour=0, sec=0}
and 2)
local offset_time = tvbuffer(0:5):bytes()
A 5-Byte (larger than uint32 range) ByteArray() containing the number of milliseconds (in network byte order) since ref_time. Now I'm looking for a human readable date. I didn't know this would be so hard, but 1st it seems I cannot simple add an offset to an os.time value and 2nd the offset exceeds Int32 range ...and most function I tested seem to truncate the exceeding input value.
Any ideas on how I get the date from ref_time and offset_time?
Thank you very much!
Since ref_time is in seconds and offset_time is in milliseconds, just try:
os.date("%c",ref_time+offset_time/1000)
I assume that offset_time is a number. If not, just reconstruct it using arithmetic. Keep in mind that Lua uses doubles for numbers and so a 5-byte integer fits just fine.
I am generating a unique and random alphanumeric string segment to represent certain links that will be generated by the users. For doing that I was approaching with "uuid" number to ensure it's uniqueness and randomness, but, as per my requirements the string shouldn't be more than 5 characters long. So I dropped that idea.
Then I decided to generate such a string using random function of ruby and current time stamp.
The code for my random string goes like this:-
temp=DateTime.now
temp=temp + rand(DateTime.now.to_i)
temp= hash.abs.to_s(36)
What I did is that I stored the current DateTime in a temp variable and then I generated a random number passing the current datetime as parameter. Then in the second line actually added current datetime and random number together to make a unique and random string.
Soon I found,while I was testing my application in two different machines and send the request at the same time, it generated the same string(Though it's rare) once after more than 100 trials.
Now I'm thinking that I should add one more parameter like mac address or client ip address before passing to_s(36) on temp variable. But can't figure out how to do it and even then whether it will be unique or nor...
Thanks....
SecureRandom in ruby uses process id (if available) and current time. You can use the urlsafe_base64(n= 16) class method to generate the sequence you need. According to your requirements I think this is your best bet.
Edit: After a bit of testing, I still think that this approach will generate non-unique keys. The way I solved this problem for barcode generation was:
barcode= barcode_sql_id_hash("#{sql_id}#{keyword}")
Here, your keyword can be time + pid.
If you are certain that you will never need more than a given M amount of unique values, and you don't need more than rudimentary protection against guessing the next generated id, you can use a Linear Congruentual Generator to generate your identificators. All you have to do is remember the last id generated, and use that to generate a new one using the following formula:
newid = (A * oldid + B) mod M
If 2³² distinct id values are enough to suit your needs, try:
def generate_id
if #lcg
#lcg = (1664525 * #lcg + 1013904223) % (2**32)
else
#lcg = rand(2**32) # Random seed
end
end
Now just pick a suitable set of characters to represent the id in as little as 6 character. Uppercase and lowercase letters should do the trick, since (26+26)^6 > 2^32:
ENCODE_CHARS = [*?a..?z, *?A..?Z]
def encode(n)
6.times.map { |i|
n, mod = n.divmod(ENCODE_CHARS.size)
ENCODE_CHARS[mod]
}.join
end
Example:
> 10.times { n = generate_id ; puts "%10d = %s" % [n, encode(n)] }
2574974483 = dyhjOg
3636751446 = QxyuDj
368621501 = bBGvYa
1689949688 = yuTgxe
1457610999 = NqzsRd
3936504298 = MPpusk
133820481 = PQLpsa
2956135596 = yvXpOh
3269402651 = VFUhFi
724653758 = knLfVb
Due to the nature of the LCG, the generated id will not repeat until all 2³² values have been used exactly once each.
There is no way you can generate a unique UUID with only five chars, with chars and numbers you have a basic space of around 56 chars, so there is a max of 56^5 combinations , aprox 551 million (Around 2^29).
If with this scheme you were about to generate 10.000 UUIDs (A very low number of UUIDs) you would have a probability of 1/5.000 of generating a collision.
When using crypto, the standard definition of a big enough space to avert collisions is around 2^80.
To put this into perspective, your algorithm would be better off if it generated just a random integer (a 32 bit uint is 2^32, 8 times the size you are proposing) which is clearly a bad idea.
Hi
I want to convert "qrysth.Fields[i].AsFloat" to a string so I use the following code:
FormatFloat('0.###############',qrysth.Fields[i].AsFloat)
but I find the result string is 12.000000000000001 while qrysth.Fields[i].AsFloat is 12.00. I know FormatFloat actually not use 12.00 to do the convert, but use an infinite number of binary to do the convert. (like 0.1 in decimal system is 0.1, but it is an infinite number in binary system 0.00011001100...)
Is there other way I could get 12.00 in the case above? or 12.000000000000000 at least?
If you really get 12.000000000000001, then your field didn't hold exactly 12, so the output is correct. You asked for high precision by putting so many # characters in the format. If you don't want it so precise, then use a less precise format string.
FormatFloat('0.00',qrysth.Fields[i].AsFloat) will give '12.00'.
To be able to get '12.000000000000000' you should do the rounding yourself, as there's no loss of precision.
I want to convert
"qrysth.Fields[i].AsFloat" to a string
Then why not use AsString?
qrysth.Fields[i].AsString
This will give you the best representation, as long as you're not concerned about the exact width. If you are, use FormatFloat with the exact number of digits you need - in other words, if you're looking for 12.00, use FormatFloat('##.##', qrysth.Fields[i].AsFloat), or even better CurrToStrand AsCurrency, as they automatically uses two digits after the decimal point.
function MyFormatFloat(V: Double): String;
const
DesiredMinPrec = '0.000000000000000';
AssumedMaxPrec = '0.#####';
begin
Result := FormatFloat(DesiredMinPrec, StrToFloat(FormatFloat(AssumedMaxPrec, V)));
end;