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VBA rounding problem
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Closed 9 months ago.
I am trying to compare two cells in a table:
The column "MR" is calculated using the formula =ABS([#Value]-A1) to determine the moving range of the column "Value". The values in the "Value" column are not rounded. The highlighted cells in the "MR" column (B3 and B4) are equal. I can enter the formula =B3=B4 into a cell and Excel says that B3 is equal to B4.
But when I compare them in VBA, VBA says that B4 is greater than B3. I can select cell B3 and enter the following into the Immediate Window ? selection.value = selection.offset(1).value. That statement evaluates to false.
I tried removing the absolute value from the formula thinking that might have had something to do with it, but VBA still says they aren't equal.
I tried adding another row where Value=1.78 so MR=0.18. Interestingly, the MR in the new row (B5) is equal to B3, but is not equal to B4.
I then tried increasing the decimal of A4 to match the other values, and now VBA says they are equal. But when I added the absolute value back into the formula, VBA again says they are not equal. I removed the absolute value again and now VBA is saying they are not equal.
Why is VBA telling me the cells are not equal when Excel says they are? How can I reliably handle this situation through VBA going forward?
The problem is that the IEEE 754 Standard for Floating-Point Arithmetic is imprecise by design. Virtually every programming language suffers because of this.
IEEE 754 is an extremely complex topic and when you study it for months and you believe you understand fully, you are simply fooling yourself!
Accurate floating point value comparisons are difficult and error prone. Think long and hard before attempting to compare floating point numbers!
The Excel program gets around the issue by cheating on the application side. VBA on the other hand follows the IEEE 754 spec for Double Precision (binary64) faithfully.
A Double value is represented in memory using 64 bits. These 64 bits are split into three distinct fields that are used in binary scientific notation:
The SIGN bit (1 bit to represent the sign of the value: pos/neg)
The EXPONENT (11 bits, biased in value by +1023)
The MANTISSA (53 bits, 52 bits stored + 1 bit implied)
The mantissa in this system leverages the fact that all binary numbers begin with a digit of 1 and so that 1 is not stored in the bit-pattern. It is implied, increasing the mantissa precision to 53-bits for normal values.
The math works like this: Stored Value = SIGN VALUE * 2^UNBIASED EXPONENT * MANTISSA
Note that a stored value of 1 for the sign bit denotes a negative SIGN VALUE (-1) while a 0 denotes a positive SIGN VALUE (+1). The formula is SIGN VALUE = (-1) ^ (sign bit).
The problem always boils down to the same thing.
The vast majority of real numbers cannot be expressed precisely
within this system which introduces small rounding errors that propagate
like weeds.
It may help to think of this system as a grid of regularly spaced points. The system can represent ONLY the point-values and NONE of the real numbers between the points. All values assigned to a float will be rounded to one of the point-values (usually the closest point, but there are modes that enforce rounding upwards to the next highest point, or rounding downwards). Conducting any calculation on a floating-point value virtually guarantees the resulting value will require rounding.
To accent the obvious, there are an infinite number of real numbers between adjacent representable point-values on this grid; and all of them are rounded to the discreet grid-points.
To make matters worse, the gap size doubles at every Power-of-Two as the grid expands away from true zero (in both directions). For example, the gap length between grid points for values in the range of 2 to 4 is twice as large as it is for values in the range of 1 to 2. When representing values with large enough magnitudes, the grid gap length becomes massive, but closer to true zero, it is miniscule.
With your example numbers...
1.24 is represented with the following binary:
Sign bit = 0
Exponent = 01111111111
Mantissa = 0011110101110000101000111101011100001010001111010111
The Hex pattern over the full 64 bits is precisely: 3FF3D70A3D70A3D7.
The precision is derived exclusively from the 53-bit mantissa and the exact decimal value from the binary is:
0.2399999999999999911182158029987476766109466552734375
In this instance a leading integer of 1 is implied by the hidden bit associated with the mantissa and so the complete decimal value is:
1.2399999999999999911182158029987476766109466552734375
Now notice that this is not precisely 1.24 and that is the entire problem.
Let's examine 1.42:
Sign bit = 0
Exponent = 01111111111
Mantissa = 0110101110000101000111101011100001010001111010111000
The Hex pattern over the full 64 bits is precisely: 3FF6B851EB851EB8.
With the implied 1 the complete decimal value is stored as:
1.4199999999999999289457264239899814128875732421875000
And again, not precisely 1.42.
Now, let's examine 1.6:
Sign bit = 0
Exponent = 01111111111
Mantissa = 1001100110011001100110011001100110011001100110011010
The Hex pattern over the full 64 bits is precisely: 3FF999999999999A.
Notice the repeating binary fraction in this case that is truncated
and rounded when the mantissa bits run out? Obviously 1.6 when
represented in binary base2 can never be precisely accurate in the
same way as 1/3 can never be accurately represented in decimal base10
(0.33333333333333333333333... ≠ 1/3).
With the implied 1 the complete decimal value is stored as:
1.6000000000000000888178419700125232338905334472656250
Not exactly 1.6 but closer than the others!
Now let's subtract the full stored double precision representations:
1.60 - 1.42 = 0.18000000000000015987
1.42 - 1.24 = 0.17999999999999993782
So as you can see, they are not equal at all.
The usual way to work around this is threshold testing, basically an inspection to see if two values are close enough... and that depends on you and your requirements. Be forewarned, effective threshold testing is way harder than it appears at first glance.
Here is a function to help you get started comparing two Double Precision numbers. It handles many situations well but not all because no function can.
Function Roughly(a#, b#, Optional within# = 0.00001) As Boolean
Dim d#, x#, y#, z#
Const TINY# = 1.17549435E-38 'SINGLE_MIN
If a = b Then Roughly = True: Exit Function
x = Abs(a): y = Abs(b): d = Abs(a - b)
If a <> 0# Then
If b <> 0# Then
z = x + y
If z > TINY Then
Roughly = d / z < within
Exit Function
End If
End If
End If
Roughly = d < within * TINY
End Function
The idea here is to have the function return True if the two Doubles are Roughly the same Within a certain margin:
MsgBox Roughly(3.14159, 3.141591) '<---dispays True
The Within margin defaults to 0.00001, but you can pass whatever margin you need.
And while we know that:
MsgBox 1.60 - 1.42 = 1.42 - 1.24 '<---dispays False
Consider the utility of this:
MsgBox Roughly(1.60 - 1.42, 1.42 - 1.24) '<---dispays True
#chris neilsen linked to an interesting Microsoft page about Excel and IEEE 754.
And please read David Goldberg's seminal What Every Computer Scientist Should Know About Floating-Point Arithmetic. It changed the way I understood floating point numbers.
Consider this code:
x = 1 # 0001
x << 2 # Shift left 2 bits: 0100
# Result: 4
x | 2 # Bitwise OR: 0011
# Result: 3
x & 1 # Bitwise AND: 0001
# Result: 1
I can understand the arithmetic operators in Python (and other languages), but I never understood 'bitwise' operators quite well. In the above example (from a Python book), I understand the left-shift but not the other two.
Also, what are bitwise operators actually used for? I'd appreciate some examples.
Bitwise operators are operators that work on multi-bit values, but conceptually one bit at a time.
AND is 1 only if both of its inputs are 1, otherwise it's 0.
OR is 1 if one or both of its inputs are 1, otherwise it's 0.
XOR is 1 only if exactly one of its inputs are 1, otherwise it's 0.
NOT is 1 only if its input is 0, otherwise it's 0.
These can often be best shown as truth tables. Input possibilities are on the top and left, the resultant bit is one of the four (two in the case of NOT since it only has one input) values shown at the intersection of the inputs.
AND | 0 1 OR | 0 1 XOR | 0 1 NOT | 0 1
----+----- ---+---- ----+---- ----+----
0 | 0 0 0 | 0 1 0 | 0 1 | 1 0
1 | 0 1 1 | 1 1 1 | 1 0
One example is if you only want the lower 4 bits of an integer, you AND it with 15 (binary 1111) so:
201: 1100 1001
AND 15: 0000 1111
------------------
IS 9 0000 1001
The zero bits in 15 in that case effectively act as a filter, forcing the bits in the result to be zero as well.
In addition, >> and << are often included as bitwise operators, and they "shift" a value respectively right and left by a certain number of bits, throwing away bits that roll of the end you're shifting towards, and feeding in zero bits at the other end.
So, for example:
1001 0101 >> 2 gives 0010 0101
1111 1111 << 4 gives 1111 0000
Note that the left shift in Python is unusual in that it's not using a fixed width where bits are discarded - while many languages use a fixed width based on the data type, Python simply expands the width to cater for extra bits. In order to get the discarding behaviour in Python, you can follow a left shift with a bitwise and such as in an 8-bit value shifting left four bits:
bits8 = (bits8 << 4) & 255
With that in mind, another example of bitwise operators is if you have two 4-bit values that you want to pack into an 8-bit one, you can use all three of your operators (left-shift, and and or):
packed_val = ((val1 & 15) << 4) | (val2 & 15)
The & 15 operation will make sure that both values only have the lower 4 bits.
The << 4 is a 4-bit shift left to move val1 into the top 4 bits of an 8-bit value.
The | simply combines these two together.
If val1 is 7 and val2 is 4:
val1 val2
==== ====
& 15 (and) xxxx-0111 xxxx-0100 & 15
<< 4 (left) 0111-0000 |
| |
+-------+-------+
|
| (or) 0111-0100
One typical usage:
| is used to set a certain bit to 1
& is used to test or clear a certain bit
Set a bit (where n is the bit number, and 0 is the least significant bit):
unsigned char a |= (1 << n);
Clear a bit:
unsigned char b &= ~(1 << n);
Toggle a bit:
unsigned char c ^= (1 << n);
Test a bit:
unsigned char e = d & (1 << n);
Take the case of your list for example:
x | 2 is used to set bit 1 of x to 1
x & 1 is used to test if bit 0 of x is 1 or 0
what are bitwise operators actually used for? I'd appreciate some examples.
One of the most common uses of bitwise operations is for parsing hexadecimal colours.
For example, here's a Python function that accepts a String like #FF09BE and returns a tuple of its Red, Green and Blue values.
def hexToRgb(value):
# Convert string to hexadecimal number (base 16)
num = (int(value.lstrip("#"), 16))
# Shift 16 bits to the right, and then binary AND to obtain 8 bits representing red
r = ((num >> 16) & 0xFF)
# Shift 8 bits to the right, and then binary AND to obtain 8 bits representing green
g = ((num >> 8) & 0xFF)
# Simply binary AND to obtain 8 bits representing blue
b = (num & 0xFF)
return (r, g, b)
I know that there are more efficient ways to acheive this, but I believe that this is a really concise example illustrating both shifts and bitwise boolean operations.
I think that the second part of the question:
Also, what are bitwise operators actually used for? I'd appreciate some examples.
Has been only partially addressed. These are my two cents on that matter.
Bitwise operations in programming languages play a fundamental role when dealing with a lot of applications. Almost all low-level computing must be done using this kind of operations.
In all applications that need to send data between two nodes, such as:
computer networks;
telecommunication applications (cellular phones, satellite communications, etc).
In the lower level layer of communication, the data is usually sent in what is called frames. Frames are just strings of bytes that are sent through a physical channel. This frames usually contain the actual data plus some other fields (coded in bytes) that are part of what is called the header. The header usually contains bytes that encode some information related to the status of the communication (e.g, with flags (bits)), frame counters, correction and error detection codes, etc. To get the transmitted data in a frame, and to build the frames to send data, you will need for sure bitwise operations.
In general, when dealing with that kind of applications, an API is available so you don't have to deal with all those details. For example, all modern programming languages provide libraries for socket connections, so you don't actually need to build the TCP/IP communication frames. But think about the good people that programmed those APIs for you, they had to deal with frame construction for sure; using all kinds of bitwise operations to go back and forth from the low-level to the higher-level communication.
As a concrete example, imagine some one gives you a file that contains raw data that was captured directly by telecommunication hardware. In this case, in order to find the frames, you will need to read the raw bytes in the file and try to find some kind of synchronization words, by scanning the data bit by bit. After identifying the synchronization words, you will need to get the actual frames, and SHIFT them if necessary (and that is just the start of the story) to get the actual data that is being transmitted.
Another very different low level family of application is when you need to control hardware using some (kind of ancient) ports, such as parallel and serial ports. This ports are controlled by setting some bytes, and each bit of that bytes has a specific meaning, in terms of instructions, for that port (see for instance http://en.wikipedia.org/wiki/Parallel_port). If you want to build software that does something with that hardware you will need bitwise operations to translate the instructions you want to execute to the bytes that the port understand.
For example, if you have some physical buttons connected to the parallel port to control some other device, this is a line of code that you can find in the soft application:
read = ((read ^ 0x80) >> 4) & 0x0f;
Hope this contributes.
I didn't see it mentioned above but you will also see some people use left and right shift for arithmetic operations. A left shift by x is equivalent to multiplying by 2^x (as long as it doesn't overflow) and a right shift is equivalent to dividing by 2^x.
Recently I've seen people using x << 1 and x >> 1 for doubling and halving, although I'm not sure if they are just trying to be clever or if there really is a distinct advantage over the normal operators.
I hope this clarifies those two:
x | 2
0001 //x
0010 //2
0011 //result = 3
x & 1
0001 //x
0001 //1
0001 //result = 1
Think of 0 as false and 1 as true. Then bitwise and(&) and or(|) work just like regular and and or except they do all of the bits in the value at once. Typically you will see them used for flags if you have 30 options that can be set (say as draw styles on a window) you don't want to have to pass in 30 separate boolean values to set or unset each one so you use | to combine options into a single value and then you use & to check if each option is set. This style of flag passing is heavily used by OpenGL. Since each bit is a separate flag you get flag values on powers of two(aka numbers that have only one bit set) 1(2^0) 2(2^1) 4(2^2) 8(2^3) the power of two tells you which bit is set if the flag is on.
Also note 2 = 10 so x|2 is 110(6) not 111(7) If none of the bits overlap(which is true in this case) | acts like addition.
Sets
Sets can be combined using mathematical operations.
The union operator | combines two sets to form a new one containing items in either.
The intersection operator & gets items only in both.
The difference operator - gets items in the first set but not in the second.
The symmetric difference operator ^ gets items in either set, but not both.
Try It Yourself:
first = {1, 2, 3, 4, 5, 6}
second = {4, 5, 6, 7, 8, 9}
print(first | second)
print(first & second)
print(first - second)
print(second - first)
print(first ^ second)
Result:
{1, 2, 3, 4, 5, 6, 7, 8, 9}
{4, 5, 6}
{1, 2, 3}
{8, 9, 7}
{1, 2, 3, 7, 8, 9}
This example will show you the operations for all four 2 bit values:
10 | 12
1010 #decimal 10
1100 #decimal 12
1110 #result = 14
10 & 12
1010 #decimal 10
1100 #decimal 12
1000 #result = 8
Here is one example of usage:
x = raw_input('Enter a number:')
print 'x is %s.' % ('even', 'odd')[x&1]
Another common use-case is manipulating/testing file permissions. See the Python stat module: http://docs.python.org/library/stat.html.
For example, to compare a file's permissions to a desired permission set, you could do something like:
import os
import stat
#Get the actual mode of a file
mode = os.stat('file.txt').st_mode
#File should be a regular file, readable and writable by its owner
#Each permission value has a single 'on' bit. Use bitwise or to combine
#them.
desired_mode = stat.S_IFREG|stat.S_IRUSR|stat.S_IWUSR
#check for exact match:
mode == desired_mode
#check for at least one bit matching:
bool(mode & desired_mode)
#check for at least one bit 'on' in one, and not in the other:
bool(mode ^ desired_mode)
#check that all bits from desired_mode are set in mode, but I don't care about
# other bits.
not bool((mode^desired_mode)&desired_mode)
I cast the results as booleans, because I only care about the truth or falsehood, but it would be a worthwhile exercise to print out the bin() values for each one.
Bit representations of integers are often used in scientific computing to represent arrays of true-false information because a bitwise operation is much faster than iterating through an array of booleans. (Higher level languages may use the idea of a bit array.)
A nice and fairly simple example of this is the general solution to the game of Nim. Take a look at the Python code on the Wikipedia page. It makes heavy use of bitwise exclusive or, ^.
There may be a better way to find where an array element is between two values, but as this example shows, the & works here, whereas and does not.
import numpy as np
a=np.array([1.2, 2.3, 3.4])
np.where((a>2) and (a<3))
#Result: Value Error
np.where((a>2) & (a<3))
#Result: (array([1]),)
i didnt see it mentioned, This example will show you the (-) decimal operation for 2 bit values: A-B (only if A contains B)
this operation is needed when we hold an verb in our program that represent bits. sometimes we need to add bits (like above) and sometimes we need to remove bits (if the verb contains then)
111 #decimal 7
-
100 #decimal 4
--------------
011 #decimal 3
with python:
7 & ~4 = 3 (remove from 7 the bits that represent 4)
001 #decimal 1
-
100 #decimal 4
--------------
001 #decimal 1
with python:
1 & ~4 = 1 (remove from 1 the bits that represent 4 - in this case 1 is not 'contains' 4)..
Whilst manipulating bits of an integer is useful, often for network protocols, which may be specified down to the bit, one can require manipulation of longer byte sequences (which aren't easily converted into one integer). In this case it is useful to employ the bitstring library which allows for bitwise operations on data - e.g. one can import the string 'ABCDEFGHIJKLMNOPQ' as a string or as hex and bit shift it (or perform other bitwise operations):
>>> import bitstring
>>> bitstring.BitArray(bytes='ABCDEFGHIJKLMNOPQ') << 4
BitArray('0x142434445464748494a4b4c4d4e4f50510')
>>> bitstring.BitArray(hex='0x4142434445464748494a4b4c4d4e4f5051') << 4
BitArray('0x142434445464748494a4b4c4d4e4f50510')
the following bitwise operators: &, |, ^, and ~ return values (based on their input) in the same way logic gates affect signals. You could use them to emulate circuits.
To flip bits (i.e. 1's complement/invert) you can do the following:
Since value ExORed with all 1s results into inversion,
for a given bit width you can use ExOR to invert them.
In Binary
a=1010 --> this is 0xA or decimal 10
then
c = 1111 ^ a = 0101 --> this is 0xF or decimal 15
-----------------
In Python
a=10
b=15
c = a ^ b --> 0101
print(bin(c)) # gives '0b101'
You can use bit masking to convert binary to decimal;
int a = 1 << 7;
int c = 55;
for(int i = 0; i < 8; i++){
System.out.print((a & c) >> 7);
c = c << 1;
}
this is for 8 digits you can also do for further more.
How many inversions are there in a binary string of length n ?
For example , n = 3
000->0
001->0
010->1
011->0
100->2
101->1
110->2
111->0
So total inversions are 6
The question looks like a homework, that's why let me omit the details. You can:
Solve the problem as a recurrency (see Толя's answer)
Make up and solve the characteristic equation, get the solution as a close formula with some arbitrary constants (c1, c2, ..., cn); as the matter of fact you'll get just one unknown constant.
Put some known solutions (e.g. f(1) = 0, f(3) = 6) into the formula and find out all the unknown coefficients
The final answer you should get is
f(n) = n*(n-1)*2**(n-3)
where ** means raising into power (2**(n-3) is 2 in n-3 power). In case you don't want to deal with recurrency and the like stuff, you can just prove the formula by induction.
It is easy recurrent function.
Assume that we know answer for n-1.
And after ato all previous sequences we add 0 or 1 as first character.
if we adding 0 as first character that mean that count of inversions will not be changed: hence answer will be same as for n-1.
if we adding 1 as first character that mean count of inversions will be same as before and will be added extra inversion equals to count of 0 into all previous sequences.
Count of zeros ans ones in sequences of length n-1 will be:
(n-1)*2^(n-1)
Half of them is zeros it will give following result
(n-1)*2^(n-2)
It means that we have following formula:
f(1) = 0
f(n) = 2*f(n-1) + (n-1)*2^(n-2)
I have some large numbers in an Excel sheet and I want to convert them to binary.
e.g.
12345678
965321458
-12457896
If we are talking positive number between 0 and 2^32-1 you can use this formula:
=DEC2BIN(MOD(QUOTIENT($A$1,256^3),256),8)&DEC2BIN(MOD(QUOTIENT($A$1,256^2),256),8)&DEC2BIN(MOD(QUOTIENT($A$1,256^1),256),8)&DEC2BIN(MOD(QUOTIENT($A$1,256^0),256),8)
NOTE: =DEC2BIN() function cannot handle numbers larger than 511 so as you see my formula breaks your number into four 8-bit chunks, converts them to binary format and then concatenates the results.
Well, theoretically you can extend this formula up to six 8-bit chunks. Maximum precision you can get in Excel is 15 (fifteen) decimal digits. When exceeded, only the most significant 15 digits remain, the rest is rounded. I.e. if you type 12345678901234567 Excel will store it as 12345678901234500. So since 2^48-1 is 15 decimal digits long the number won't get rounded.
Perhaps a simpler option:
For positive numbers only, just use BASE (as in BASE2) for numbers between 0 to 2^53 in Excel. Here are some examples:
=BASE(3,2) # returns 11
=BASE(11,2) # returns 1011
Credit for answer goes here:
https://ask.libreoffice.org/en/question/69797/why-is-dec2bin-limited-to-82bits-in-an-32-and-64-bits-world/
Negative numbers: Come to think of it, negative numbers could be handled as well by building upon howy61's answer. He shifts everything by a power of two (2^31 in his case) to use the 2's complement:
=BASE(2^31+MyNum, 2)
so (using 2^8 for only 8 bits):
=BASE(2^8+(-1),2) # returns 11111111
=BASE(2^8+(-3),2) # returns 11111101
The numbers given by the OP requires more bits, so I'll use 2^31 (could go up to 2^53):
=BASE(2^31+(-12457896),2) # returns 11111111010000011110100001011000
For either positive or negative, both formulas could be coupled in a single IF formula. Here are two ways you could do it that give the same answer, where MyNum is the decimal number you start with:
=IF(MyNum<0, BASE(2^31+MyNum,2), BASE(MyNum, 2))
or
=BASE(IF(MyNum<0, MyNum+2^32, MyNum), 2)
See VBA posted here
' The DecimalIn argument is limited to 79228162514264337593543950245
' (approximately 96-bits) - large numerical values must be entered
' as a String value to prevent conversion to scientific notation. Then
' optional NumberOfBits allows you to zero-fill the front of smaller
' values in order to return values up to a desired bit level.
Function DecToBin(ByVal DecimalIn As Variant, Optional NumberOfBits As Variant) As String
DecToBin = ""
DecimalIn = CDec(DecimalIn)
Do While DecimalIn <> 0
DecToBin = Trim$(Str$(DecimalIn - 2 * Int(DecimalIn / 2))) & DecToBin
DecimalIn = Int(DecimalIn / 2)
Loop
If Not IsMissing(NumberOfBits) Then
If Len(DecToBin) > NumberOfBits Then
DecToBin = "Error - Number too large for bit size"
Else
DecToBin = Right$(String$(NumberOfBits, "0") & _
DecToBin, NumberOfBits)
End If
End If
End Function
I just tried the formula above, and found that Microsoft screwed up the DEC2BIN function in another way that keeps the formula from working correctly with negative numbers. Internally, DEC2BIN uses a ten bit result; leading zeroes are dropped from the text result, unless the optional length parameter is used, in which case the required number of leading zeroes are left in the string. But here's the rub: a negative number always starts with a one, so there are no leading zeroes to drop, so DEC2BIN will always show all ten bits! Thus, DEC2BIN(-1,8), which should show 11111111 (eight ones) will instead show 1111111111 (ten ones.)
To fix this, use RIGHT to trim each eight bit chunk to eight bits, dumb as that sounds.
=RIGHT(DEC2BIN(QUOTIENT(A1,256^3),8),8) & RIGHT(...
(I read through the VBA, and it does not have the same problem, but it doesn't look like it will handle negatives at all.)
To add easier to read formatting to Taosique's great answer, you can also break it up into chunks of 4 bits with spaces in between, although the formula grows to be a monster:
=DEC2BIN(MOD(QUOTIENT($A$1,16^7),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^6),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^5),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^4),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^3),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^2),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^1),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^0),16),4)
1101 0100 1111 0110 0011 0001 0000 0001
Of course, you can just use the right half of it, if you're just interested in 16 bit numbers:
=DEC2BIN(MOD(QUOTIENT($A$1,16^3),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^2),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^1),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^0),16),4)
0011 0001 0000 0001
While I didn't write this for negatives or decimals, it should be relatively easy to modify. This VBA will convert any super large (or not so large if you want, but that wasn't the point) decimal up to the converted binary result containing up to 32767 digits (maximum string length in VBA).
Enter decimal in cell "A1" as a string, result will be in "B1" as a string.
Dim NBN As String
Dim Bin As String
5 Big = Range("A1")
AA = Len(Big)
For XX = 1 To AA
L1 = Mid(Big, XX, 1) + CRY
CRY = 0
If L1 = 0 Then
FN = "0"
GoTo 10
End If
If Int(L1 / 2) = L1 / 2 Then
FN = L1 / 2
GoTo 10
End If
If Int(L1 / 2) <> L1 / 2 Then
FN = Int(L1 / 2)
CRY = 10
GoTo 10
End If
10 NBN = NBN & FN
Next XX
If Left(NBN, 1) = "0" Then
NBN = Right(NBN, (Len(NBN) - 1))
End If
If CRY = 10 Then Bin = "1" & Bin Else Bin = "0" & Bin
Range("A1") = NBN
Range("A2") = Bin
If Len(NBN) > 0 Then
NBN = ""
CRY = 0
GoTo 5
End If
Someone can find binary shift operations more clear and relevant here
=DEC2BIN(BITRSHIFT($A$1,24),8) & DEC2BIN(MOD(BITRSHIFT($A$1,16),256),8) & DEC2BIN(MOD(BITRSHIFT($A$1,8),256),8) & DEC2BIN(MOD($A$1,256),8)
This formula is for 32-bit values
This vba function solves the problem of binary conversion of numbers greater than 511 that can not be done with WorksheetFunction.dec2bin.
The code takes advantage of the WorksheetFunction.dec2bin function by applying it in pieces.
Function decimal2binary(ByVal decimal2convert As Long) As String
Dim rest As Long
If decimal2convert = 0 Then
decimal2binary = "0"
Exit Function
End If
Do While decimal2convert > 0
rest = decimal2convert Mod 512
decimal2binary = Right("000000000" + WorksheetFunction.Dec2Bin(rest), 9) + decimal2binary
decimal2convert = (decimal2convert - rest) / 512
Loop
decimal2binary = Abs(decimal2binary)
End Function
=IF(Decimal>-1,BASE(Decimal,2,32),BASE(2^32+(Decimal),2))
Does both positive and negative numbers.
Took a bit LOL. Tech pun.
You're welcome.
Here's another way. It's not with a single formula, but I have tried and converted up to the number 2,099,999,999,999. My first intention was to build a 51 bit counter, but somehow it does not work with numbers beyond the one I mentioned. Download from
http://www.excelexperto.com/content/macros-production/contador-binario-de-51-bits/
I hope it's useful. Regards.
Without VBA and working with negative numbers as well (here: sint16), however, taking much more space:
You can download the excel file here: (sorry, didn't know where to put the file)
int16 bits to decimal.xlsx
or alternatively follow these steps (if your Excel is not in English, use Excel Translator to "translate" the formula into your MS Office language):
Enter the binary number in 4-bit nibbles (A4 = most significant to D4 = least significant) like shown in the screenshot. Enter all 4 digits (even if starting with 0) and format them as "text"
Enter formula in F4:
=IF(NUMBERVALUE(A4)>=1000,TRUE,FALSE)
Enter the letter "A" in G2-J2, "B" in K2-N2, "C" in O2-R2, "D" in S2-V2
Enter "1" in G3, K3, O3 and S3; "2" in H3, L3, P3 and T3; "3" in I3, M3, Q3 and U3; "4" in J3, N3, R3 and V3
In G4, enter:
=MID(INDIRECT(G$2&ROW()),G$3,1)
Copy the formula to H4-V4
In X4, enter:
=IF(G4="1",0,1)
Copy X4 to Y4-AM4
In BD3 enter "1"
In BC4, enter:
=IF((AM$4+BD3)=2,1,0)
IN BD4, enter:
=IF((AM$4+BD3)=2,0,IF((AM$4+BD3)=1,1,0))
Copy BD4 and BD4 and insert it 15 times diagonally one row further down and one column further left (like in the screenshot), i.e. insert it to BB5 and BC5, then BA6 and BB6, ..., AN19 and AO19.
In AO20, enter "=AO19"; in AP20, enter "=AP18" and so on until BD20 ("=BD4") - i.e. bring down the numbers into one line as seen in the screenshot
In BE20, enter (this is your result):
=IF(F4=FALSE,BIN2DEC(A4&B4)*2^8+BIN2DEC(C4&D4),-1*(BIN2DEC(AO20&AP20&AQ20&AR20&AS20&AT20&AU20&AV20)*2^8+BIN2DEC(AW20&AX20&AY20&AZ20&BA20&BB20&BC20&BD20)))
There maybe a simple solution. I have several 4.2 billion cells that are actually a negative Two's Complement and this works to get the correct value:
=SUM(2^31-(A1-2^31))
In Commodore 64 BASIC V2, PRINT'ing a true boolean expression outputs -1:
READY.
A=(5=5)
READY.
PRINT A
-1
Why -1 and not 1?
Commodore Basic doesn't have a boolean data type. A boolean expression evaluates to a number, where 0 means False and -1 means true.
As there is no boolean data type there are no boolean type expressions. You can use any numeric expression in an IF statement, and it will interpret any non-zero value as True.
Some languages where a boolean value is numeric or where it can be converted to a numeric value uses -1 to represent a true value. For an integer value 0 all bits are cleared, and for -1 all bits are set, so they can be seen as natural complements for each other.
Eventhough Commodore Basic doesn't use integer numbers but floating point numbers, the value -1 was supposedly chosen because some other languages uses it.
Short answer
Why -1 and not 1?
It's just a convention
Some speculation...
The reason underlying the choice for this convention may be related to the internal representation of integers numbers on most platforms:
Suppose a boolean value is stored in a 16 bits wide integer; false is 0 (every bit unset).
0 => 00000000 00000000
It makes sense to agree upon another convention where true is all bits set:
11111111 11111111
(a very reasonable choice since all 1 is the bitwise NOT of all 0)
The decimal representation of a signed integer whose bits are all set is -1
-1 => 11111111 11111111
While the binary representation of 1 is
1 => 00000001 00000000 (on a little endian platform).
So that's why -1 and not 1: is just a convention; but if you look at the binary representation of the value you may agree that the convention makes sense.
However this convention is far from being universally adopted despite all the above considerations. On many languages true casted to a numeric value is 1.
An important note about the code you posted:
You wrote
A=(5=5)
A is a real variable whose value is represented by 5 bytes (1 for exponent, 4 for mantissa).
The internal C64 representation of the real value 0 is all bits 0, but the representation of -1 is far from being all bits 1.
(all bits 1 would lead to the value -1.70141183e+38)
So again C64 Basic just sticks to a convention.
Finally let's explain the behaviuour of the IF statements in your code.
Any value different from 0 is evaluated as true in an IF statement (this happens on most languages, maybe all).
Bottom note:
If you want to take a look at the internal representation of a C64 variable (integer or real only, not strings) you may use the following code:
READY.
10 REM THE VARIABLE TO INSPECT
20 A=(5=5)
30 REM THE ADDR. OF THE FIRST VARIABLE
40 B=PEEK(45)+256*PEEK(46)
50 REM DISPLAY THE VAR'S 7 BYTES
60 FOR C=B TO B+6
70 PRINT C;": ";PEEK(C)
80 NEXT C
RUN
2227 : 65
2228 : 0
2229 : 129
2230 : 128
2231 : 0
2232 : 0
2233 : 0
Note the first two addresses (2227, 2228) store the variable's name (65, 0 for A)
You may try declaring the variable as integer and see the result with
20 A%=(5=5)
There is a way that this may be used, by customising a for loop, i.e., if you wanted something to happen until a key is pressed, one might:
0 for i=-1 to 0
1 rem logic here ...
10 get a$: i=(a$=""): next i
This is the same sort of logic as a do...while loop.
Edit - If you specifically wanted 0 to be false and 1 to be true, you could define a function as follows (I forgot about the ABS keyword as I've not used it in probably 20 years :-|) :
0 def fn b(x) = abs(x)
1 i = 7
2 a$ = "hello"
3 if fn b (i=6) then print "i is 6"
4 if fn b (i<10) then print "i is less than 10"
5 if fn b (a$="hi") then print "hey there!"
6 if fn b (a$="hello") then print a$