I'm learning about the representation of floating-point IEEE 754 numbers, and my textbook says:
To pack even more bits into the significand, IEEE 754 makes the leading 1-bit of normalized binary numbers implicit. Hence, the number is actually 24 bits long in single precision (implied 1 and 23-bit fraction), and 53 bits long in double precision (1 + 52).
I don't get what "implicit" means here... what's the difference between an explicit bit and an implicit bit? Don't all numbers have the bit, regardless of their sign?
Yes, all normalised numbers (other than the zeroes) have that bit set to one (a), so they make it implicit to prevent wasting space storing it.
In other words, they save that bit totally, and reuse it so that it can be used to increase the precision of your numbers.
Keep in mind that this is the first bit of the fraction, not the first bit of the binary pattern. The first bit of the binary pattern is the sign, followed by a few bits of exponent, followed by the fraction itself.
For example, a single precision number is (sign, exponent, fraction):
<1> <--8---> <---------23----------> <- bit widths
s eeeeeeee fffffffffffffffffffffff
If you look at the way the number is calculated, it's:
(-1)sign x 1.fraction x 2exponent-bias
So the fractional part used for calculating that value is 1.fffff...fff (in binary).
(a) There is actually a class of numbers (the denormalised ones and the zeroes) for which that property does not hold true. These numbers all have a biased exponent of zero but the vast majority of numbers follow the rule.
Here is what they are saying. The first non-zero bit is always going to be 1. So there is no need for the binary representation to include that bit, since you know what it is. So they don't. They tell you where that first 1 is, and then they give the bits after it. So there is a 1 that is not explicitly in the binary representation, whose location is implicit from the fact that they told you where it was.
It may also be helpful to note that we are dealing in binary representations of a number. The reason that the first digit of a normalized binary number (that is, no leading zeroes) has to be 1 is that 1 is the only non-zero value available to us in this representation. So, the same would not be true for, say, base-three representations.
Related
I was reading about the gradual underflow concept and how it is op important in the music industry Gradual overflow Application in Music
I well understand the problem of an overflow buffer, but this i don't know how to represent an underflow.
Can you please give me an example(a program preferably in c or c++) as in how a computer handles gradual underflow?
Gradual underflow is related to what IEEE 754 calls "subnormal" numbers.
Consider using scientific notation in which you have (say) 10 digits of precision and exponents that can range from -99 through 99.
Under normal circumstances, you treat everything as scientific notation, so if you want to represent 1000, you represent it as 1e3 -- that is, 1 * 103.
Now, consider a number like 1.234e-102. The smallest exponent you can represent is -99. So, if you do your job the simplest possible way, you simply that since it has an exponent smaller than that, it's just 0. That would be "fast underflow".
In IEEE 754 (and related standards) you can store that as (essentially) 0.001234 * 10-99. In doing so, you may lose some precision compared to a normal number that has an exponent in the -99...99 range. On the other hand, you lose less than if you just rounded it to zero because it has an exponent smaller than -99. In fact, in this case it started with 4 significant digits, and as represented it retains all 4 significant digits.
On a computer, the numbers are represented in binary, so the numbers of significant digits and/or maximum range of exponents aren't round numbers when converted to decimal, but the same basic idea applies--when we have a number that's too small to represent in the normal format, we can still store it with the smallest exponent that can be represented, but also includes some leading zeros.
This does lead to one difficulty: numbers are normally stored in what's called normalized form. The "significand" part is normalized by shifting it left until the first digit is a 1 (keep in mind that since it's binary it can only be 0 or 1). Since we know it's a 1, we cheat a little: we don't normally store that 1 in the number as it's stored. So, a double precision floating point number normally has 53 bits of precision, but only actually stores 52 bits of significand.
With a subnormal number, that's no longer the case. That's not terribly difficult to deal with, but it still introduces a special case--and one that's only rarely used, so CPU designers (and such) rarely try to optimize for it. As a result, the exact same code can suddenly run a lot slower when executing on data that contains subnormals.
In this wiki article it shows 23 bits for precision, 8 for exponent, and 1 for sign
Where is the hidden 24th bit in float type that makes (23+1) for 7 significand digits?
Floating point numbers are usually normalized. Consider, for example, scientific notation as most of us learned it in school. You always scale the exponent so there's exactly one digit before the decimal point. For example, instead of 123.456, you write 1.23456x102.
Floating point on a computer is normally handled (almost1) the same way: numbers are normalized so there's exactly one digit before the binary point (binary point since most work in binary instead of decimal). There's one difference though: in the case of binary, that means the digit before the decimal point must be a 1. Since it's always a 1, there's no real need to store that bit. To save a bit of storage in each floating point number, that 1 bit is implicit instead of being stored.
As usual, there's just a bit more to the situation than that though. The main difference is denormalized numbers. Consider, for example, if you were doing scientific notation but you could only use exponents from -99 to +99. If you wanted to store a number like, say, 1.234*10-102, you wouldn't be able to do that directly, so it would probably just get rounded down to 0.
Denormalized numbers give you a way to deal with that. Using a denormalized number, you'd store that as 0.001234*10-99. Assuming (as is normally the case on a computer) that the number of digits for the mantissa and exponent are each limited, this loses some precision, but still avoids throwing away all the precision and just calling it 0.
1 Technically, there are differences, but they make no difference to the basic understanding involved.
http://en.wikipedia.org/wiki/Single_precision_floating-point_format#IEEE_754_single_precision_binary_floating-point_format:_binary32
The true significand includes 23
fraction bits to the right of the
binary point and an implicit leading
bit (to the left of the binary point)
with value 1 unless the exponent is
stored with all zeros
Explains it pretty well, it is by convention/design that last bit is not stored explicitly but rather stated by specification that it is there unless everything is 0'os.
As you write, the single-precision floating-point format has a sign bit, eight exponent bits, and 23 significand bits. Let s be the sign bit, e be the exponent bits, and f be the significand bits. Here is what various combinations of bits stand for:
If e and f are zero, the object is +0 or -0, according to whether s is 0 or 1.
If e is zero and f is not, the object is (-1)s * 21-127 * 0.f. "0.f" means to write 0, period, and the 23 bits of f, then interpret that as a binary numeral. E.g., 0.011000... is 3/8. These are the "subnormal" numbers.
If 0 < e < 255, the object is (-1)s * 2e-127 * 1.f. "1.f" is similar to "0.f" above, except you start with 1 instead of 0. This is the implicit bit. Most of the floating-point numbers are in this format; these are the "normal" numbers.
If e is 255 and f is zero, the object is +infinity or -infinity, according to whether s is 0 or 1.
If e is 255 and f is not zero, the object is a NaN (Not a Number). The meaning of the f field of a NaN is implementation dependent; it is not fully specified by the standard. Commonly, if the first bit is zero, it is a signaling NaN; otherwise it is a quiet NaN.
I found myself today doing some bit manipulation and I decided to refresh my floating-point knowledge a little!
Things were going great until I saw this:
... 23 fraction bits of the significand appear in the memory format but the total precision is 24 bits
I read it again and again but I still can't figure out where the 24th bit is, I noticed something about a binary point so I assumed that it's a point in the middle between the mantissa and the exponent.
I'm not really sure but I believe he author was talking about this bit:
Binary point?
|
s------e-----|-------------m----------
0 - 01111100 - 01000000000000000000000
^ this
The 24th bit is implicit due to normalization.
The significand is shifted left (and one subtracted from the exponent for each bit shift) until the leading bit of the significand is a 1.
Then, since the leading bit is a 1, only the other 23 bits are actually stored.
There is also the possibility of a denormal number. The exponent is stored as a "bias" format signed number, meaning that it's an unsigned number where the middle of the range is defined to mean 01. So, with 8 bits, it's stored as a number from 0..255, but 0 is interpreted to mean -128, 128 is interpreted to mean 0, and 255 is interpreted as 127 (I may have a fencepost error there, but you get the idea).
If, in the process of normalization, this is decremented to 0 (meaning an actual exponent value of -128), then normalization stops, and the significand is stored as-is. In this case, the implicit bit from normalization it taken to be a 0 instead of a 1.
Most floating point hardware is designed to basically assume numbers will be normalized, so they assume that implicit bit is a 1. During the computation, they check for the possibility of a denormal number, and in that case they do roughly the equivalent of throwing an exception, and re-start the calculation with that taken into account. This is why computation with denormals often gets drastically slower than otherwise.
In case you wonder why it uses this strange format: IEEE floating point (like many others) is designed to ensure that if you treat its bit pattern as an integer of the same size, you can compare them as signed, 2's complement integers and they'll still sort into the correct order as floating point numbers. Since the sign of the number is in the most significant bit (where it is for a 2's complement integer) that's treated as the sign bit. The bits of the exponent are stored as the next most significant bits -- but if we used 2's complement for them, an exponent less than 0 would set the second most significant bit of the number, which would result in what looked like a big number as an integer. By using bias format, a smaller exponent leaves that bit clear, and a larger exponent sets it, so the order as an integer reflects the order as a floating point.
Normally (pardon the pun), the leading bit of a floating point number is always 1; thus, it doesn't need to be stored anywhere. The reason is that, if it weren't 1, that would mean you had chosen the wrong exponent to represent it; you could get more precision by shifting the mantissa bits left and using a smaller exponent.
The one exception is denormal/subnormal numbers, which are represented by all zero bits in the exponent field (the lowest possible exponent). In this case, there is no implicit leading 1 in the mantissa, and you have diminishing precision as the value approaches zero.
For normal floating point numbers, the number stored in the floating point variable is (ignoring sign) 1. mantissa * 2exponent-offset. The leading 1 is not stored in the variable.
c++ pow(2,1000) is normaly to big for double, but it's working. why?
So I've been learning C++ for couple weeks but the datatypes are still confusing me.
One small minor thing first: the code that 0xbadc0de posted in the other thread is not working for me.
First of all pow(2,1000) gives me this more than once instance of overloaded function "pow" matches the argument list.
I fixed it by changing pow(2,1000) -> pow(2.0,1000)
Seems fine, i run it and get this:
http://i.stack.imgur.com/bbRat.png
Instead of
10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376
it is missing a lot of the values, what might be cause that?
But now for the real problem.
I'm wondering how can 302 digits long number fit a double (8 bytes)?
0xFFFFFFFFFFFFFFFF = 18446744073709551616 so how can the number be larger than that?
I think it has something to do with the floating point number encoding stuff.
Also what is the largest number that can possibly be stored in 8 bytes if it's not 0xFFFFFFFFFFFFFFFF?
Eight bytes contain 64 bits of information, so you can store 2^64 ~ 10^20 unique items using those bits. Those items can easily be interpreted as the integers from 0 to 2^64 - 1. So you cannot store 302 decimal digits in 8 bytes; most numbers between 0 and 10^303 - 1 cannot be so represented.
Floating point numbers can hold approximations to numbers with 302 decimal digits; this is because they store the mantissa and exponent separately. Numbers in this representation store a certain number of significant digits (15-16 for doubles, if I recall correctly) and an exponent (which can go into the hundreds, of memory serves). However, if a decimal is X bytes long, then it can only distinguish between 2^(8X) different values... unlikely enough for exactly representing integers with 302 decimal digits.
To represent such numbers, you must use many more bits: about 1000, actually, or 125 bytes.
It's called 'floating point' for a reason. The datatype contains a number in the standard sense, and an exponent which says where the decimal point belongs. That's why pow(2.0, 1000) works, and it's why you see a lot of zeroes. A floating point (or double, which is just a bigger floating point) number contains a fixed number of digits of precision. All the remaining digits end up being zero. Try pow(2.0, -1000) and you'll see the same situation in reverse.
The number of decimal digits of precision in a float (32 bits) is about 7, and for a double (64 bits) it's about 16 decimal digits.
Most systems nowadays use IEEE floating point, and I just linked to a really good description of it. Also, the article on the specific standard IEEE 754-1985 gives a detailed description of the bit layouts of various sizes of floating point number.
2.0 ^ 1000 mathematically will have a decimal (non-floating) output. IEEE floating point numbers, and in your case doubles (as the pow function takes in doubles and outputs a double) have 52 bits of the 64 bit representation allocated to the mantissa. If you do the math, 2^52 = 4,503,599,627,370,496. Because a floating point number can represent positive and negative numbers, really the integer representation will be ~ 2^51 = 2,251,799,813,685,248. Notice there are 16 digits. there are 16 quality (non-zero) digits in the output you see.
Essentially the pow function is going to perform the exponentiation, but once the exponentiation moves past ~2^51, it is going to begin losing precision. Ultimately it will hold precision for the top ~16 decimal digits, but all other digits right will be un-guaranteed.
Thus it is a floating point precision / rounding problem.
If you were strictly in unsigned integer land, the number would overflow after (2^64 - 1) = 18,446,744,073,709,551,616. What overflowing means, is that you would never actually see the number go ANY HIGHER than the one provided, infact I beleive the answer would be 0 from this operation. Once the answer goes beyond 2^64, the result register would be zero, and any multiply afterwords would be 0 * 2, which would always result in 0. I would have to try it.
The exact answer (as you show) can be obtained using a standard computer using a multi-precision libary. What these do is to emulate a larger bit computer by concatenating multiple of the smaller data types, and use algorithms to convert and print on the fly. Mathematica is one example of a math engine that implements an arbitrary precision math calculation library.
Floating point types can cover a much larger range than integer types of the same size, but with less precision.
They represent a number as:
a sign bit s to indicate positive or negative;
a mantissa m, a value between 1 and 2, giving a certain number of bits of precision;
an exponent e to indicate the scale of the number.
The value itself is calculated as m * pow(2,e), negated if the sign bit is set.
A standard double has a 53-bit mantissa, which gives about 16 decimal digits of precision.
So, if you need to represent an integer with more than (say) 64 bits of precision, then neither a 64-bit integer nor a 64-bit floating-point type will work. You will need either a large integer type, with as many bits as necessary to represent the values you're using, or (depending on the problem you're solving) some other representation such as a prime factorisation. No such type is available in standard C++, so you'll need to make your own.
If you want to calculate the range of the digits that can be hold by some bytes, it should be (2^(64bits - 1bit)) to (2^(64bits - 1bit) - 1).
Because the left most digit of the variable is for representing sign (+ and -).
So the range for negative side of the number should be : (2^(64bits - 1bit))
and the range for positive side of the number should be : (2^(64bits - 1bit) - 1)
there is -1 for the positive range because of 0(to avoid reputation of counting 0 for each side).
For example if we are calculating 64bits, the range should be ==> approximately [-9.223372e+18] to [9.223372e+18]
How would i go about finding the value of the two-byte two’s complement value 0xFF72 is"?
Would i start by converting 0xFF72 to binary?
reverse the bits.
add 1 in binary notation. // lost here.
write decimal.
I just dont know..
Also,
What about an 8 byte double that has the value: 0x7FF8000000000000. Its value as a floating point?
I would think that this was homework, but for the particular double that is listed. 0x7FF8000000000000 is a quiet NaN per the IEEE-754 spec, not a very interesting value to put on a homework assignment:
The sign bit is clear.
The exponent field is 0x7ff, the largest possible exponent, which means that the number is either an infinity or a NaN.
The significand field is 0x8000000000000. Since it isn't zero, the number is not an infinity, and must be a NaN. Since the leading bit is set, it is a quiet NaN, not a so-called "signaling NaN".
Step 3 just means add 1 to the value. It's really as simple as it sounds. :-)
Example with 0xFF72 (assumed 16-bits here):
First, invert it: 0x008D (each digit is simply 0xF minus the original value)
Then add 1: 0x008E
This sounds like homework and for openness you should tag it as such if it is.
As for interpreting an 8 byte (double) floating point number, take a look at this Wikipedia article.