Is this defined by the language? Is there a defined maximum? Is it different in different browsers?
JavaScript has two number types: Number and BigInt.
The most frequently-used number type, Number, is a 64-bit floating point IEEE 754 number.
The largest exact integral value of this type is Number.MAX_SAFE_INTEGER, which is:
253-1, or
+/- 9,007,199,254,740,991, or
nine quadrillion seven trillion one hundred ninety-nine billion two hundred fifty-four million seven hundred forty thousand nine hundred ninety-one
To put this in perspective: one quadrillion bytes is a petabyte (or one thousand terabytes).
"Safe" in this context refers to the ability to represent integers exactly and to correctly compare them.
From the spec:
Note that all the positive and negative integers whose magnitude is no
greater than 253 are representable in the Number type (indeed, the
integer 0 has two representations, +0 and -0).
To safely use integers larger than this, you need to use BigInt, which has no upper bound.
Note that the bitwise operators and shift operators operate on 32-bit integers, so in that case, the max safe integer is 231-1, or 2,147,483,647.
const log = console.log
var x = 9007199254740992
var y = -x
log(x == x + 1) // true !
log(y == y - 1) // also true !
// Arithmetic operators work, but bitwise/shifts only operate on int32:
log(x / 2) // 4503599627370496
log(x >> 1) // 0
log(x | 1) // 1
Technical note on the subject of the number 9,007,199,254,740,992: There is an exact IEEE-754 representation of this value, and you can assign and read this value from a variable, so for very carefully chosen applications in the domain of integers less than or equal to this value, you could treat this as a maximum value.
In the general case, you must treat this IEEE-754 value as inexact, because it is ambiguous whether it is encoding the logical value 9,007,199,254,740,992 or 9,007,199,254,740,993.
>= ES6:
Number.MIN_SAFE_INTEGER;
Number.MAX_SAFE_INTEGER;
<= ES5
From the reference:
Number.MAX_VALUE;
Number.MIN_VALUE;
console.log('MIN_VALUE', Number.MIN_VALUE);
console.log('MAX_VALUE', Number.MAX_VALUE);
console.log('MIN_SAFE_INTEGER', Number.MIN_SAFE_INTEGER); //ES6
console.log('MAX_SAFE_INTEGER', Number.MAX_SAFE_INTEGER); //ES6
It is 253 == 9 007 199 254 740 992. This is because Numbers are stored as floating-point in a 52-bit mantissa.
The min value is -253.
This makes some fun things happening
Math.pow(2, 53) == Math.pow(2, 53) + 1
>> true
And can also be dangerous :)
var MAX_INT = Math.pow(2, 53); // 9 007 199 254 740 992
for (var i = MAX_INT; i < MAX_INT + 2; ++i) {
// infinite loop
}
Further reading: http://blog.vjeux.com/2010/javascript/javascript-max_int-number-limits.html
In JavaScript, there is a number called Infinity.
Examples:
(Infinity>100)
=> true
// Also worth noting
Infinity - 1 == Infinity
=> true
Math.pow(2,1024) === Infinity
=> true
This may be sufficient for some questions regarding this topic.
Jimmy's answer correctly represents the continuous JavaScript integer spectrum as -9007199254740992 to 9007199254740992 inclusive (sorry 9007199254740993, you might think you are 9007199254740993, but you are wrong!
Demonstration below or in jsfiddle).
console.log(9007199254740993);
However, there is no answer that finds/proves this programatically (other than the one CoolAJ86 alluded to in his answer that would finish in 28.56 years ;), so here's a slightly more efficient way to do that (to be precise, it's more efficient by about 28.559999999968312 years :), along with a test fiddle:
/**
* Checks if adding/subtracting one to/from a number yields the correct result.
*
* #param number The number to test
* #return true if you can add/subtract 1, false otherwise.
*/
var canAddSubtractOneFromNumber = function(number) {
var numMinusOne = number - 1;
var numPlusOne = number + 1;
return ((number - numMinusOne) === 1) && ((number - numPlusOne) === -1);
}
//Find the highest number
var highestNumber = 3; //Start with an integer 1 or higher
//Get a number higher than the valid integer range
while (canAddSubtractOneFromNumber(highestNumber)) {
highestNumber *= 2;
}
//Find the lowest number you can't add/subtract 1 from
var numToSubtract = highestNumber / 4;
while (numToSubtract >= 1) {
while (!canAddSubtractOneFromNumber(highestNumber - numToSubtract)) {
highestNumber = highestNumber - numToSubtract;
}
numToSubtract /= 2;
}
//And there was much rejoicing. Yay.
console.log('HighestNumber = ' + highestNumber);
Many earlier answers have shown 9007199254740992 === 9007199254740992 + 1 is true to verify that 9,007,199,254,740,991 is the maximum and safe integer.
But what if we keep doing accumulation:
input: 9007199254740992 + 1 output: 9007199254740992 // expected: 9007199254740993
input: 9007199254740992 + 2 output: 9007199254740994 // expected: 9007199254740994
input: 9007199254740992 + 3 output: 9007199254740996 // expected: 9007199254740995
input: 9007199254740992 + 4 output: 9007199254740996 // expected: 9007199254740996
We can see that among numbers greater than 9,007,199,254,740,992, only even numbers are representable.
It's an entry to explain how the double-precision 64-bit binary format works. Let's see how 9,007,199,254,740,992 be held (represented) by using this binary format.
Using a brief version to demonstrate it from 4,503,599,627,370,496:
1 . 0000 ---- 0000 * 2^52 => 1 0000 ---- 0000.
|-- 52 bits --| |exponent part| |-- 52 bits --|
On the left side of the arrow, we have bit value 1, and an adjacent radix point. By consuming the exponent part on the left, the radix point is moved 52 steps to the right. The radix point ends up at the end, and we get 4503599627370496 in pure binary.
Now let's keep incrementing the fraction part with 1 until all the bits are set to 1, which equals 9,007,199,254,740,991 in decimal.
1 . 0000 ---- 0000 * 2^52 => 1 0000 ---- 0000.
(+1)
1 . 0000 ---- 0001 * 2^52 => 1 0000 ---- 0001.
(+1)
1 . 0000 ---- 0010 * 2^52 => 1 0000 ---- 0010.
(+1)
.
.
.
1 . 1111 ---- 1111 * 2^52 => 1 1111 ---- 1111.
Because the 64-bit double-precision format strictly allots 52 bits for the fraction part, no more bits are available if we add another 1, so what we can do is setting all bits back to 0, and manipulate the exponent part:
┏━━▶ This bit is implicit and persistent.
┃
1 . 1111 ---- 1111 * 2^52 => 1 1111 ---- 1111.
|-- 52 bits --| |-- 52 bits --|
(+1)
1 . 0000 ---- 0000 * 2^52 * 2 => 1 0000 ---- 0000. * 2
|-- 52 bits --| |-- 52 bits --|
(By consuming the 2^52, radix
point has no way to go, but
there is still one 2 left in
exponent part)
=> 1 . 0000 ---- 0000 * 2^53
|-- 52 bits --|
Now we get the 9,007,199,254,740,992, and for the numbers greater than it, the format can only handle increments of 2 because every increment of 1 on the fraction part ends up being multiplied by the left 2 in the exponent part. That's why double-precision 64-bit binary format cannot hold odd numbers when the number is greater than 9,007,199,254,740,992:
(consume 2^52 to move radix point to the end)
1 . 0000 ---- 0001 * 2^53 => 1 0000 ---- 0001. * 2
|-- 52 bits --| |-- 52 bits --|
Following this pattern, when the number gets greater than 9,007,199,254,740,992 * 2 = 18,014,398,509,481,984 only 4 times the fraction can be held:
input: 18014398509481984 + 1 output: 18014398509481984 // expected: 18014398509481985
input: 18014398509481984 + 2 output: 18014398509481984 // expected: 18014398509481986
input: 18014398509481984 + 3 output: 18014398509481984 // expected: 18014398509481987
input: 18014398509481984 + 4 output: 18014398509481988 // expected: 18014398509481988
How about numbers between [ 2 251 799 813 685 248, 4 503 599 627 370 496 )?
1 . 0000 ---- 0001 * 2^51 => 1 0000 ---- 000.1
|-- 52 bits --| |-- 52 bits --|
The value 0.1 in binary is exactly 2^-1 (=1/2) (=0.5)
So when the number is less than 4,503,599,627,370,496 (2^52), there is one bit available to represent the 1/2 times of the integer:
input: 4503599627370495.5 output: 4503599627370495.5
input: 4503599627370495.75 output: 4503599627370495.5
Less than 2,251,799,813,685,248 (2^51)
input: 2251799813685246.75 output: 2251799813685246.8 // expected: 2251799813685246.75
input: 2251799813685246.25 output: 2251799813685246.2 // expected: 2251799813685246.25
input: 2251799813685246.5 output: 2251799813685246.5
/**
Please note that if you try this yourself and, say, log
these numbers to the console, they will get rounded. JavaScript
rounds if the number of digits exceed 17. The value
is internally held correctly:
*/
input: 2251799813685246.25.toString(2)
output: "111111111111111111111111111111111111111111111111110.01"
input: 2251799813685246.75.toString(2)
output: "111111111111111111111111111111111111111111111111110.11"
input: 2251799813685246.78.toString(2)
output: "111111111111111111111111111111111111111111111111110.11"
And what is the available range of exponent part? 11 bits allotted for it by the format.
From Wikipedia (for more details, go there)
So to make the exponent part be 2^52, we exactly need to set e = 1075.
To be safe
var MAX_INT = 4294967295;
Reasoning
I thought I'd be clever and find the value at which x + 1 === x with a more pragmatic approach.
My machine can only count 10 million per second or so... so I'll post back with the definitive answer in 28.56 years.
If you can't wait that long, I'm willing to bet that
Most of your loops don't run for 28.56 years
9007199254740992 === Math.pow(2, 53) + 1 is proof enough
You should stick to 4294967295 which is Math.pow(2,32) - 1 as to avoid expected issues with bit-shifting
Finding x + 1 === x:
(function () {
"use strict";
var x = 0
, start = new Date().valueOf()
;
while (x + 1 != x) {
if (!(x % 10000000)) {
console.log(x);
}
x += 1
}
console.log(x, new Date().valueOf() - start);
}());
The short answer is “it depends.”
If you’re using bitwise operators anywhere (or if you’re referring to the length of an Array), the ranges are:
Unsigned: 0…(-1>>>0)
Signed: (-(-1>>>1)-1)…(-1>>>1)
(It so happens that the bitwise operators and the maximum length of an array are restricted to 32-bit integers.)
If you’re not using bitwise operators or working with array lengths:
Signed: (-Math.pow(2,53))…(+Math.pow(2,53))
These limitations are imposed by the internal representation of the “Number” type, which generally corresponds to IEEE 754 double-precision floating-point representation. (Note that unlike typical signed integers, the magnitude of the negative limit is the same as the magnitude of the positive limit, due to characteristics of the internal representation, which actually includes a negative 0!)
ECMAScript 6:
Number.MAX_SAFE_INTEGER = Math.pow(2, 53)-1;
Number.MIN_SAFE_INTEGER = -Number.MAX_SAFE_INTEGER;
Other may have already given the generic answer, but I thought it would be a good idea to give a fast way of determining it :
for (var x = 2; x + 1 !== x; x *= 2);
console.log(x);
Which gives me 9007199254740992 within less than a millisecond in Chrome 30.
It will test powers of 2 to find which one, when 'added' 1, equals himself.
Anything you want to use for bitwise operations must be between 0x80000000 (-2147483648 or -2^31) and 0x7fffffff (2147483647 or 2^31 - 1).
The console will tell you that 0x80000000 equals +2147483648, but 0x80000000 & 0x80000000 equals -2147483648.
JavaScript has received a new data type in ECMAScript 2020: BigInt. It introduced numerical literals having an "n" suffix and allows for arbitrary precision:
var a = 123456789012345678901012345678901n;
Precision will still be lost, of course, when such big integer is (maybe unintentionally) coerced to a number data type.
And, obviously, there will always be precision limitations due to finite memory, and a cost in terms of time in order to allocate the necessary memory and to perform arithmetic on such large numbers.
For instance, the generation of a number with a hundred thousand decimal digits, will take a noticeable delay before completion:
console.log(BigInt("1".padEnd(100000,"0")) + 1n)
...but it works.
Try:
maxInt = -1 >>> 1
In Firefox 3.6 it's 2^31 - 1.
I did a simple test with a formula, X-(X+1)=-1, and the largest value of X I can get to work on Safari, Opera and Firefox (tested on OS X) is 9e15. Here is the code I used for testing:
javascript: alert(9e15-(9e15+1));
I write it like this:
var max_int = 0x20000000000000;
var min_int = -0x20000000000000;
(max_int + 1) === 0x20000000000000; //true
(max_int - 1) < 0x20000000000000; //true
Same for int32
var max_int32 = 0x80000000;
var min_int32 = -0x80000000;
Let's get to the sources
Description
The MAX_SAFE_INTEGER constant has a value of 9007199254740991 (9,007,199,254,740,991 or ~9 quadrillion). The reasoning behind that number is that JavaScript uses double-precision floating-point format numbers as specified in IEEE 754 and can only safely represent numbers between -(2^53 - 1) and 2^53 - 1.
Safe in this context refers to the ability to represent integers exactly and to correctly compare them. For example, Number.MAX_SAFE_INTEGER + 1 === Number.MAX_SAFE_INTEGER + 2 will evaluate to true, which is mathematically incorrect. See Number.isSafeInteger() for more information.
Because MAX_SAFE_INTEGER is a static property of Number, you always use it as Number.MAX_SAFE_INTEGER, rather than as a property of a Number object you created.
Browser compatibility
In JavaScript the representation of numbers is 2^53 - 1.
However, Bitwise operation are calculated on 32 bits ( 4 bytes ), meaning if you exceed 32bits shifts you will start loosing bits.
In the Google Chrome built-in javascript, you can go to approximately 2^1024 before the number is called infinity.
Scato wrotes:
anything you want to use for bitwise operations must be between
0x80000000 (-2147483648 or -2^31) and 0x7fffffff (2147483647 or 2^31 -
1).
the console will tell you that 0x80000000 equals +2147483648, but
0x80000000 & 0x80000000 equals -2147483648
Hex-Decimals are unsigned positive values, so 0x80000000 = 2147483648 - thats mathematically correct. If you want to make it a signed value you have to right shift: 0x80000000 >> 0 = -2147483648. You can write 1 << 31 instead, too.
Firefox 3 doesn't seem to have a problem with huge numbers.
1e+200 * 1e+100 will calculate fine to 1e+300.
Safari seem to have no problem with it as well. (For the record, this is on a Mac if anyone else decides to test this.)
Unless I lost my brain at this time of day, this is way bigger than a 64-bit integer.
Node.js and Google Chrome seem to both be using 1024 bit floating point values so:
Number.MAX_VALUE = 1.7976931348623157e+308
In MSACCESS VBA, I convert a HEX string to decimal by prefixing the string with "&h"
?CLng("&h1234")
4660
?CLng("&h80000000")
-2147483648
What should I do to convert it to an unsigned integer?
Using CDbl doesn't work either:
?CDbl("&h80000000")
-2147483648
Your version seems like the best answer, but can be shortened a bit:
Function Hex2Dbl(h As String) As Double
Hex2Dbl = CDbl("&h0" & h) ' Overflow Error if more than 2 ^ 64
If Hex2Dbl < 0 Then Hex2Dbl = Hex2Dbl + 4294967296# ' 16 ^ 8 = 4294967296
End Function
Double will have rounding precision error for most values above 2 ^ 53 - 1 (about 16 decimal digits), but Decimal can be used for values up to 16 ^ 12 - 1 (Decimal uses 16 bytes, but only 12 of them for the number)
Function Hex2Dec(h)
Dim L As Long: L = Len(h)
If L < 16 Then ' CDec results in Overflow error for hex numbers above 16 ^ 8
Hex2Dec = CDec("&h0" & h)
If Hex2Dec < 0 Then Hex2Dec = Hex2Dec + 4294967296# ' 2 ^ 32
ElseIf L < 25 Then
Hex2Dec = Hex2Dec(Left$(h, L - 9)) * 68719476736# + CDec("&h" & Right$(h, 9)) ' 16 ^ 9 = 68719476736
End If
End Function
If you want to go higher than 2^31 you could use Decimal or LongLong. LongLong and CLngLngonly work on 64bit platforms though. Since I only have 32 bit office at the moment, this is for Decimal and CDec.
There seems to be an issue when converting 8-digit Hex numbers because apparently signed 32-bit is used somewhere in the process which results in the sign mistake even though Decimal could handle the number.
'only for positive numbers
Function myHex2Dec(hexString As String) As Variant
'cut off "&h" if present
If Left(hexString, 2) = "&h" Or Left(hexString, 2) = "&H" Then hexString = Mid(hexString, 3)
'cut off leading zeros
While Left(hexString, 1) = "0"
hexString = Mid(hexString, 2)
Wend
myHex2Dec = CDec("&h" & hexString)
'correct value for 8 digits onle
If myHex2Dec < 0 And Len(hexString) = 8 Then
myHex2Dec = CDec("&h1" & hexString) - 4294967296#
'cause overflow for 16 digits
ElseIf myHex2Dec < 0 Then
Error (6) 'overflow
End If
End Function
Test:
Sub test()
Dim v As Variant
v = CDec("&H80000000") '-2147483648
v = myHex2Dec("&H80000000") '2147483648
v = CDec("&H7FFFFFFFFFFFFFFF") '9223372036854775807
v = myHex2Dec("&H7FFFFFFFFFFFFFFF") '9223372036854775807
v = CDec("&H8000000000000000") '-9223372036854775808
v = myHex2Dec("&H8000000000000000") 'overflow
End Sub
With remark of #arcadeprecinct I was able to create a function for it:
Function Hex2UInt(h As String) As Double
Dim dbl As Double: dbl = CDbl("&h" & h)
If dbl < 0 Then
dbl = CDbl("&h1" & h) - 4294967296#
End If
Hex2UInt = dbl
End Function
Some example output:
?Hex2UInt("1234")
4660
?Hex2UInt("80000000")
2147483648
?Hex2UInt("FFFFFFFFFFFF")
281474976710655
Maximum value to represent as an integer is 0x38D7EA4C67FFF
?Hex2UInt("38D7EA4C67FFF")
999999999999999
?Hex2UInt("38D7EA4C68000")
1E+15
a proposal, result in h
sh = "&H80000000"
h = CDbl(sh)
If h < 0 Then
fd = Hex$(CDbl(Left(sh, 3)) - 8)
sh = "&h" & fd & Mid(sh, 4)
h = CDbl(sh) + 2 ^ 31
End If
I found my way here looking for a Word VBA solution, but what I've discovered might also apply to other Office apps. I realise that this is a very old question and that there are some ingenious solutions to it, but I'm surprised that nobody has explained what it is that seems to be the root cause of the problem, and hence what might possibly be a one-line solution in many cases. When I was an assembly language programmer in the 1970s, working more in binary and octal than anything else, this was a very common issue, known as "2s complement".
I'll explain it in its simplest form, from first principles, by the way it works on a byte, so that it's understandable even by absolute beginners.
Normally, the most significant bit is bit-7 at the left which has a value of 128, the least significant bit is bit-0 at the right which has a value of 1. Therefore, the highest possible value if all bits are set is 255. However in 2s complement, bit-7 is the "sign bit". This only leaves the seven bits from 0 to 6 to hold the actual value, giving them a maximum value of 127. The sign bit has a value of -128. If all 8 bits are set, the byte value becomes (-128 + 127) which gives the negative decimal value of -1. The 2s complement range of values for 8 bits is from -128 (with only bit-7 set) to +127 (with only bits 0 to 6 set). If the sign bit is set, the value of the byte is -128 plus the positive value of whatever is stored in bits 0 to 6. E.g. binary 11111101 = hex FD = decimal (-128 + 125) = -3, 10110100 = hex B4 = decimal (-128 + 52) = -76.
2s complement applies the same effect at each increasing 8-bit boundary, thus for 16 bits, the sign bit is bit-15 (with a value of -32,768) and the positive value is in bits 0 to 14, giving a 16-bit range of values from -32768 to 32767. Similarly, the 24-bit range is from -8388608 to 8388607, and so on.
I recently encountered this conversion problem in some code that was converting hexadecimal RGB colour values which originated as a 6-character text string in a Word document. Having successfully processed tens of thousands of these I was suddeny presented with an "out of range" error pop-up. The string that had caused the problem was "008080". The command ... = Val("&H" + variable) had converted this to -32896, an invalid value to pass as a colour property. The Val() function had removed the leading zeros and treated 8080 as a signed 2s complement 16-bit value.
In my case the solution was simple. Because I know that I'll always be dealing with 24-bit, 6-character hex values. I just added an extra "1" text character to the front of the hex code (thus making it longer than 16 bits), then, in effect, subtracted the same value. So, with the original 6-character hex RGB code held in the variable HexCode, I get the right decimal result using the command
DecCode = Val("&H" + "1" + HexCode) - Val("&H" + "1000000")
Problem solved, by just adding a little extra code to an existing line. I hope that my explanation of the cause of the problem helps others to devise their own solutions where it's appropriate.
Question
Let's say I have a string or array which represents a number in base N, N>1, where N is a power of 2. Assume the number being represented is larger than the system can handle as an actual number (an int or a double etc).
How can I convert that to a decimal string?
I'm open to a solution for any base N which satisfies the above criteria (binary, hex, ...). That is if you have a solution which works for at least one base N, I'm interested :)
Example:
Input: "10101010110101"
-
Output: "10933"
It depends on the particular language. Some have native support for arbitrary-length integers, and others can use libraries such as GMP. After that it's just a matter of doing the lookup in a table for the digit value, then multiplying as appropriate.
This is from a Python-based computer science course I took last semester that's designed to handle up to base-16.
import string
def baseNTodecimal():
# get the number as a string
number = raw_input("Please type a number: ")
# convert it to all uppercase to match hexDigits (below)
number = string.upper(number)
# get the base as an integer
base = input("Please give me the base: ")
# the number of values that we have to change to base10
digits = len(number)
base10 = 0
# first position of any baseN number is 1's
position = 1
# set up a string so that the position of
# each character matches the decimal
# value of that character
hexDigits = "0123456789ABCDEF"
# for each 'digit' in the string
for i in range(1, digits+1):
# find where it occurs in the string hexDigits
digit = string.find(hexDigits, number[-i])
# multiply the value by the base position
# and add it to the base10 total
base10 = base10 + (position * digit)
print number[-i], "is in the " + str(position) + "'s position"
# increase the position by the base (e.g., 8's position * 2 = 16's position)
position = position * base
print "And in base10 it is", base10
Basically, it takes input as a string and then goes through and adds up each "digit" multiplied by the base-10 position. Each digit is actually checked for its index-position in the string hexDigits which is used as the numerical value.
Assuming the number that it returns is actually larger than the programming language supports, you could build up an array of Ints that represent the entire number:
[214748364, 8]
would represent 2147483648 (a number that a Java int couldn't handle).
That's some php code I've just written:
function to_base10($input, $base)
{
$result = 0;
$length = strlen($input);
for ($x=$length-1; $x>=0; $x--)
$result += (int)$input[$x] * pow($base, ($length-1)-$x);
return $result;
}
It's dead simple: just a loop through every char of the input string
This works with any base <10 but it can be easily extended to support higher bases (A->11, B->12, etc)
edit: oh didn't see the python code :)
yeah, that's cooler
I would choose a language which more or less supports natively math representation like 'lisp'. I know it seems less and less people use it, but it still has its value.
I don't know if this is large enough for your usage, but the largest integer number I could represent in my common lisp environment (CLISP) was 2^(2^20)
>> (expt 2 (expt 2 20)
In lisp you can easily represent hex, dec, oct and bin as follows
>> \#b1010
10
>> \#o12
10
>> 10
10
>> \#x0A
10
You can write rationals in other bases from 2 to 36 with #nR
>> #36rABCDEFGHIJKLMNOPQRSTUVWXYZ
8337503854730415241050377135811259267835
For more information on numbers in lisp see: Practical Common Lisp Book
In erlang, there are bitwise operations to operate on integers, for example:
1> 127 bsl 1.
254
there is also the ability to pack integers into a sequence of bytes
<< 16#7F, 16#FF >>
is it possible, or are there any operators or BIFs that can perform bitwise operations (eg AND, OR, XOR, SHL, SHR) on binary packed data?
for example (if bsl worked on binary packages - which it does not):
1> << 16#7F, 16#FF >> bsl 1.
<< 255, 254 >>
Try out this way:
bbsl(Bin,Shift) -> <<_:Shift,Rest/bits>> = Bin, <<Rest/bits,0:Shift>>.
Using Erlang's unbounded integer sizes we can accomplish this:
1> Bits = <<16#0FFFFFFF:(4*8)>>.
<<15,255,255,255>>
2> size(Bits).
4
3> Size=size(Bits)*8.
32
4> <<Num:Size>> = Bits.
<<15,255,255,255>>
5> Num.
268435455
6> Num2 = Num bsl 4.
4294967280
7> Bits2 = <<Num2:Size>>.
<<"ÿÿÿð">>
8> <<A:8,B:8,C:8,D:8>>=Bits2.
<<"ÿÿÿð">>
9> A.
255
10> D.
240
as we expected.
Note that in my solution I anticipated how many shifts I would need (4) by adding 4 '0's to the initial string stored in the binary (16#0F... the first 4 positions are 0)
Not sure how I would handle it if I had to shift beyond the 'boundary' of the binary container, I guess you would just AND with 2^Size-1.