I am being asked in a homework to represent the decimal 0.1 in IEEE 754 representation. Here are the steps I made:
However online converters, and this answer on stack exchange suggests otherwise. They put this solution:
s eeeeeeee mmmmmmmmmmmmmmmmmmmmmmm
0 01111011 10011001100110011001101
The difference is the number 1 at the right. Why isn't it 1100, why is it 1101?
As njuffa said in a comment, rounding is the explanation for the difference you see. Converters usually produce the nearest floating-point value to the decimal number you put in. The IEEE 754 standard recommends that the rounding mode be taken into account for conversions from one base to another (such as from decimal to binary), and the default rounding mode is “to nearest”.
The two closest single-precision floating-point values to 1/10 are 1.10011001100110011001100×2-4 and 1.10011001100110011001101×2-4 (below and above 1/10).
The digits that are cut off are “11001100…”, indicating that the real 1/10 is closer to the upper bound than to the lower bound(if the remaining digits had been “100000000…”, the real number would have been exactly in-between the two). For this reason, the upper value 1.10011001100110011001101×2-4 is chosen as the conversion of 1/10 to binary32 when converting in round-to-nearest mode.
Related
I'm currently learning about IEEE754 standard and rounding, and I have an exercise which is the following:
Add -325.875 to 0.546875 in IEEE754, but with 3 bits dedicated to the mantissa instead of 23.
I'm having a lot of trouble doing this, especially representing the intermediary values, and the guard/round/sticky bits. Can someone give me a step-by-step solution, to the problem?
My biggest problem is that obviously I can't represent 0.546875 as 0.100011 as that would have more precision than the system has. So how would that be represented?
Apologies if the wording is confusing.
Preliminaries
The preferred term for the fraction portion of a floating-point number is “significand,” not “mantissa.” “Mantissa” is an old word for the fraction portion of a logarithm. Mantissas are logarithmic; adding to the mantissa multiplies the number represented. Significands are linear; adding to the significand adds to the number represented (as scaled by the exponent).
When working with a significand, use its mathematical precision, not the number of bits in the storage format. The IEEE-754 binary32 format has 23 bits in its primary field for the encoding of a significand, but another bit is encoded via the exponent field. Mathematically, numbers in the binary32 format behave as if they have 24 bits in their significands.
So, the task is to work with numbers with four bits in their significands, not three.
Work
In binary, −325.875 is −101000101.1112•2. In scientific notation, that is −1.010001011112•28. Rounding it to four bits in the significand gives −1.0102•28.
In binary, 0.546875 is .1000112. In scientific notation, that is 1.000112•2−1. Rounding it to four bits in the significand gives 1.0012•2−1. Note that the first four bits are 1000, but they are immediately followed by 11, so we round up. 1.00011 is closer to 1.001 than it is to 1.000.
So, in a floating-point format with four-bit significands, we want to add −1.0102•28 and 1.0012•2−1. If we adjust the latter number to have the same exponent as the former, we have −1.0102•28 and 0.0000000010012•28. To add those, we note the signs are different, so we want to subtract the magnitudes. It may help to line up the digits as we were taught in elementary school:
1.010000000000
0.000000001001
——————————————
1.001111110111
Thus, the mathematical result would be −1.0011111101112•28. However, we need to round the significand to four bits. The first four bits are 1001, but they are followed by 11, so we round up, producing 1010. So the final result is −1.0102•28.
−1.0102•28 is −1.25•28 = −320.
I'm reading this article about exponent bias in floating point numbers and it says the following:
n IEEE 754 floating point numbers, the exponent is biased in the
engineering sense of the word – the value stored is offset from the
actual value by the exponent bias. Biasing is done because exponents
have to be signed values in order to be able to represent both tiny
and huge values, but two's complement, the usual representation for
signed values, would make comparison harder. To solve this problem the
exponent is biased before being stored, by adjusting its value to put
it within an unsigned range suitable for comparison. By arranging the
fields so that the sign bit is in the most significant bit position,
the biased exponent in the middle, then the mantissa in the least
significant bits, the resulting value will be ordered properly,
whether it's interpreted as a floating point or integer value. This
allows high speed comparisons of floating point numbers using fixed
point hardware.
I've also found this explanation from wikipedia's article about offset binary:
This has the consequence that the "zero" value is represented by a 1
in the most significant bit and zero in all other bits, and in general
the effect is conveniently the same as using two's complement except
that the most significant bit is inverted. It also has the consequence
that in a logical comparison operation, one gets the same result as
with a two's complement numerical comparison operation, whereas, in
two's complement notation a logical comparison will agree with two's
complement numerical comparison operation if and only if the numbers
being compared have the same sign. Otherwise the sense of the
comparison will be inverted, with all negative values being taken as
being larger than all positive values.
I don't really understand what kind of comparison they are talking about here. Can someone please explain using a simple example?
'Comparison' here refers to the usual comparison of numbers by size: 5 > 4, etc. Suppose floating-point numbers were stored with as
[sign bit] [unbiased exponent] [mantissa]
For example, if the exponent is a 2's complement 3-bit binary number and the mantissa is a 4-bit unsigned binary number, you'd have
1 010 1001 = 4.5
1 110 0111 = 0.21875
You can see that the first is bigger than the second, but to figure this out, the computer would have to calculate 1.001 x 2^2 and 0.111 x 2^(-2) and then compare the resulting floating-point numbers. This is already complex with floating-point hardware, and if there is no such hardware for this computer, then...
So the number is stored as
[sign bit] [biased exponent] [mantissa]
Using the same 3-bit binary number for the exponent (this time biased; see a related question) and unsigned 4-bit mantissa, we have
1 101 1001 = 4.5
1 001 0111 = 0.21875
But now comparison is very easy! You can treat the two numbers as integers 11011001 and 10010111 and see that the first is obviously bigger: obvious even to a computer, as integer comparisons are easy. This is why biased exponents are used.
In my computer architecture course we use a 14 bit binary model;(1 bit for sign,5 bits for exponent, and 8 bits for our mantissa). When inputting the Exponent my instructor has us add 16 to offset it.(bias 16) Why are we using 16 bias? Is it because 5 bits can only represent up to 31 numbers? If so please elaborate and compare to IEEE single precision that uses a 127 bias when using the exponent. Lastly if someone can give me a clear definition of bias used in this context and in binary I would greatly appreciate it. Please comment if anything I said was unclear.
The IEEE 754 binary float formats follow a simple pattern for the exponent bias. When the exponent has p bits the bias is . With this the exponent has an equal number of positive and negative exponents.
For single precision floats p is 8 and therefore the bias is 127. For your format p is 5 and the bias is 15. Maybe your instructor changed the bias to 16 because the format don't support denorm, infinity and NaN.
There are several ways of representing a range of numbers including both positive and negative. Adding a bias is particularly flexible. The range [-n, m) can be represented by adding n to each number, mapping it to the range [0, m+n).
That system is used for the exponent in all the floating point systems I have used. It simplifies some comparisons, because larger unsigned binary value of the non-sign bits represents larger absolute magnitude of the float, except for special values such as NaNs.
For float exponents, the bias is around half the exponent range, so that approximately half the values are on each side of zero. Exact balance is impossible because there are an even number of bit patterns, and one is used for zero.
As discussed in another answer, the IEEE 754 standard would use a bias of 15 for a 5 bit exponent.
There are several possible reasons for choosing 16:
There is some actual technical reason, such as the suggested one of not treating 31 as special.
Bias 16 makes the representation of 1.0 particularly simple, with a single non-zero bit.
Being subtly different from IEEE 754 helps convince students that floating point does not imply IEEE 754. There are other floating point formats.
Being subtly different from IEEE 754 may discourage use of existing tools to get the results for exercises without understanding how the representation works.
It is an arbitrary choice of one of the reasonable values for the exponent bias, without reference to IEEE 754.
Which type (Float or decimal) is best used to store prices in a mysql database?
Floats are not exact and can introduce cumulative rounding errors. Decimal is the best format for financial information that must be exact.
Prices are decimal values, and calculations on them are expected to behave like decimal fractions when it comes to rounding, literals, etc.
That's exactly what decimal types do.
Floats are stored as binary fractions, and they do not behave like decimal fractions - their behaviour is frequently not what people used to decimal math expect. Read The Floating-Point Guide for detailed explanations.
For money values, never never use binary float types - especially when you have a perfectly good decimal type available!
For Financial calculations use Decimal
According to IEEE 754 Floats were always binary, only the new standard IEEE 754R defined decimal formats. Many of the fractional binary parts can never equal the exact decimal representation. Any binary number can be written as m/2^n (m, n positive integers), any decimal number as m/(2^n*5^n). As binarys lack the prime factor 5, all binary numbers can be exactly represented by decimals, but not vice versa.
0.3 = 3/(2^1 * 5^1) = 0.3
0.3 = [0.25/0.5] [0.25/0.375] [0.25/3.125] [0.2825/3.125]
1/4 1/8 1/16 1/32
So for Financial calculations use Decimal not FLOAT
When we store a number in float we don't save the exact number,it is an approximation. The integer part gets the priority and fractional part is as close as the type size. So if you have calculations and need accurate result use Decimal.
Please Use BigDecimal , as it is the best for the prices , since pennies are rounded properly to dollar.
Joshua Bloch recommends BigDecimal.
Table 2 in section 5.6 on the bottom of page 11 of the IEEE 754 specification lists the ranges of decimal values for which decimal to binary floating-point conversion must be performed. The ranges of exponents don't make sense to me. For example, for double-precision, the table says the maximum decimal value eligible to be converted is (1017-1)*10999. That's way bigger then DBL_MAX, which is approximately 1.8*10308. Obviously I'm missing something -- can someone explain this table to me? Thanks.
[Side note: technically, the document you link to is no longer a standard; "IEEE 754" should really only be used to refer to the updated edition of the standard published in 2008.]
My understanding is that, as you say, the left-hand column of that table describes the range of valid inputs to any decimal string to binary float conversion that's provided. So for example, a decimal string that's something like '1.234e+879' represents the value 1234*10^876 (M = 1234, N = 876), so is within the table limits, and is required to be accepted by the conversion functionality. Though note that exactly what form the decimal strings are allowed to take is outside the scope of IEEE 754; it's only the value that's relevant here.
I don't think it's a problem that some of the allowed inputs can be outside the representable range of a double; the usual rules for overflow should be followed in this case; see section 7.3 of the document. That is, the overflow exception should be signaled, and assuming that it's not trapped the result of the conversion (for a positive out-of-range value, say) is positive infinity if the rounding mode is round to nearest or round towards positive infinity, and the largest finite representable value if the rounding mode is round towards negative infinity or round towards zero.
Slightly more subtly, from my reading of this document, a decimal string like '1e+1000' should also be accepted by the conversion function, since the value it represents is expressible in the form 10 * 10^999, or even 10000000000000000 * 10^984. See the sentence that starts 'On input, trailing zeros shall be appended to or stripped from M ...' in section 5.6.
The current version of IEEE 754 seems to be a bit different in this respect, judging by the publicly available draft version (version 1.2.5): it just requires each implementation to specify bounds [-η, η] on the exponent of the decimal string, with η large enough to accomodate the decimal strings corresponding to finite binary values in the largest supported binary format; so if the binary64 format is the largest supported format, it looks to me as though η = 400 would be plenty big enough, for example.