floating-point,ieee-754,underflow

The IEEE754 2008 standard (§7.5) defines that the underflow exception shall be signalled when the result is non-zero, and strictly between -MinNorm and +MinNorm: it leaves it up to implementation as to whether this is before or after rounding, so you could have values just below minNorm that get rounded...

Per the comments: The approach with nextafter is exactly what you should be doing. However, it has some complications that may lead to unexpected results. Quoting cppreference std::nextafter: float nextafter( float from, float to ); (1) (since C++11) double nextafter( double from, double to ); (2) (since C++11) long double...

c#,double,type-conversion,precision,ieee-754

Being able to store the maximum value of UInt32 without losing information doesn't necessarily mean you'll be able to store all values in UInt32 without losing information - after all, there are plenty of long values which can be stored in a double even though some smaller values can't. (262...

c++,floating-point,double,decimal,ieee-754

There is no contradiction. As you can see, the value of x is incorrect at the first 7 in its decimal expansion; I count 16 correct digits before that. std::setprecision doesn't control the precision of the inputs to std::cout, it simply displays as many digits as you request. Perhaps std::setprecision...

python,ieee-754,floating-point-conversion

As you can see in Java they are using structure to do the trick. /* * Find the float corresponding to a given bit pattern */ JNIEXPORT jfloat JNICALL Java_java_lang_Float_intBitsToFloat(JNIEnv *env, jclass unused, jint v) { union { int i; float f; } u; u.i = (long)v; return (jfloat)u.f; }...

No need to test, the second snippet works just fine. (provided circularIndex is not 0).

There are 2139095039 positive floats. There are as many negative floats. Do you want to include +0.0 and -0.0 as two items or as one? Depending on the answer the total is 2 * 2139095039 + 2 or 2 * 2139095039 + 1, that is, respectively, 4278190080 or 4278190079. Source...

javascript,floating-point,ieee-754

JavaScript uses IEEE-754 double-precision numbers ("binary64" as the 2008 spec puts it; that is, as you suspected, it's the base 2 version, not the 2008 base 10 version). The reason you get the string "0.1" for the number value 0.1, even though 0.1 can't be perfectly represented in binary64, is...

language-agnostic,floating-point,floating-point-precision,ieee-754

From the Wikipedia entry on IEEE-754: The number representations described above are called normalized, meaning that the implicit leading binary digit is a 1. To reduce the loss of precision when an underflow occurs, IEEE 754 includes the ability to represent fractions smaller than are possible in the normalized representation,...

I wasn't aware that IEEE 754 defines an 8-bit format. (In fact I'm still not convinced it does.) But we can extrapolate from the formats it does define. You don't mention a hidden bit, but the 16, 32, 64, and 128 bit IEEE 754 formats all use a hidden bit,...

1) No, not guaranteed to produce identical answers. Even with IEEE, subtle rounding effects may result in a different of a 1 or 2 ULP by using a/x or a*(1/x). 2) If x is extremely small (that is a bit smaller than DBL_MIN (minimum normalized positive floating-point number) as in...

floating-point,numbers,ieee-754,scientific-notation

Since float has about 7 significant digits, you should switch to scientific notation if log10(abs(x)) > 7 or log10(abs(x)) < -7. Update: As the float still has binary format, it's better to focus on binary values. It has 23 significant binary digits, so you can check abs(x) > 223 and...

python,numpy,nan,ieee-754,multidimensional-array

On newer versions of numpy you get this warning: FutureWarning: numpy equal will not check object identity in the future. The comparison did not return the same result as suggested by the identity (`is`)) and will change. my guess is that numpy is using id test as a shortcut, for...

assembly,floating-point,masm,ieee-754

Did you mean something like this: ; Conversion of an ASCII-coded decimal rational number (DecStr) ; to an ASCII-coded decimal binary number (BinStr) as 32-bit single (IEEE 754) include \masm32\include\masm32rt.inc ; MASM32 headers, mainly for printf .data DecStr db "-1.75",0 BinStr db 40 dup (0) result REAL4 ? number dd...

c#,floating-point,double,ieee-754

It doesn't add one more decimal digit - just a single binary digit. So instead of 23 bits, you have 24 bits. This is handy, because the only number you can't represent as starting with a one is zero, and that's a special value. In short, you're not looking at...

c,math,floating-point,ieee-754

In the project you linked, they included a Mathematica notebook with an explanation of their algorithms, which includes the "mysterious" -126.94269 value. If you need a viewer, you can get one from the Mathematica website for free. Edit: Since I'm feeling generous, here's the relevant section in screenshot form. Simply...

When using fixed formatting (%f) you get a format with a decimal point and up to 6 digits. Since the value you used rounds to a value smaller than 0.000001 it seems reasonable to have 0.000000 printed. You can either use more digits (I think using %.10f but I'm not...

I assume you have some sort of textbook or spec on whatever floating point spec you intend to simulate. Look in there for definitions. For something more general, you can read: http://en.wikipedia.org/wiki/NaN Here is what wikipedia says on sNaNs: Signaling NaN[edit] Signaling NaNs, or sNaNs, are special forms of a...

Something like that: uint fb = Convert.ToUInt32(f); return BitConverter.ToSingle(BitConverter.GetBytes((int) fb), 0); ...

c++,matlab,floating-point,ieee-754

Firstly, if your numerical method depends on the accuracy of sin to the last bit, then you probably need to use an arbitrary precision library, such as MPFR. The IEEE754 2008 standard doesn't require that the functions be correctly rounded (it does "recommend" it though). Some C libms do provide...

There are 8388609. Of those 2^23 are in 1 <= x < 2, all having the same sign and exponent, with 23 bits of fraction. Add one for including 2, which is the same bit pattern as 1 except for adding 1 to the exponent.

java,binary,decimal,converter,ieee-754

Yes, there are ways to read from a binary representation. But you don't have a representation in an IEEE format. I would ignore the period and read as a BigInteger base2, then create a value to divide by also using BigInteger: private static double binaryStringToDouble(String s) { return stringToDouble(s, 2);...

Assuming that none of the operations would yield an overflow or an underflow, and your input values have uniformly distributed significands, then this is equivalent. Well, I suppose that to have a rigorous proof, one should do an exhaustive test (probably not possible in practice for double precision since there...

numbers,ieee-754,representation

Using a handy online IEEE-754 conversion utility: 2.0 = 0x40000000 3.0 = 0x40400000 So: 0x40400000 - 0x40000000 = 0x400000 = 4194304 Answer: 4 million or so....

c,math,floating-point,precision,ieee-754

-1.#IND is Microsoft's way of outputting an indeterminate value, specifically NaN. One of the ways this can happen is with 0 / 0 but I would check all operations to see where the issue lies: double secant_method(double(*f)(double), double a, double b){ double c; printf("DBG =====\n"); for (int i = 0;...

This behavior is due to the /fp:fast MSVC compiler option, which (among other things) permits the compiler to perform comparisons without regard to proper NaN behavior in an effort to generate faster code. Using /fp:precise or /fp:strict instead causes these comparisons to behave as expected when presented with NaN arguments.

Convert the hex codepoints to a byte string first; binascii.unhexlify() can do this for you, provided you remove the whitespace: import binascii import struct fdata = struct.unpack('<f', binascii.unhexlify(data.replace(' ', '')))[0] Demo: >>> import binascii >>> import struct >>> data = '38 1A A3 44' >>> struct.unpack('<f', binascii.unhexlify(data.replace(' ', ''))) (1304.8193359375,)...

c++,floating-point,double,ieee-754

Bit format is well-defined, but not all machines are little-endian. The IEEE standard does not require floating-point numbers to be a certain endian, either. You can run the following program to see the byte pattern of the double 42.0: #include <stdio.h> #include <numeric> #include <limits> using namespace std; int main()...

Maybe. It depends on how the compiler decide to evaluate floating-point expressions (read about FLT_EVAL_METHOD, invented by C99 but now part of the C++ standard, if you want the gory details.) As soon as a can be greater than 4, the product a*b expressed as a float will round up...

ieee-754,floating-point-precision

The standard double has 52 bit mantissa, so yes, it is capable to hold and exactly reproduce a 32 bit integer. Another problem is the requirement that they have to be beetween 0 and 1. The reciprocal is not the way to do that! Counterexample: 1/3 is not exactly representable...

The significand is, by definition, a sequence of p “digits” in base β, where a digit in base β is one of β possible values, from the digit representing 0 to the digit representing β-1. How many choices are there for sequences of p digits where each digit has β...

The IEEE 754 rules of arithmetic for signed zeros state that +0.0 + -0.0 depends on the rounding mode. In the default rounding mode, it will be +0.0. When rounding towards -∞, it will be -0.0. You can check this in C++ like so: #include <iostream> int main() { std::cout...

floating-point,double,floating-accuracy,ieee-754

In Java, a - b is never equal to 0 if a != b. This is because Java mandates IEEE 754 floating point operations which support denormalized numbers. From the spec: In particular, the Java programming language requires support of IEEE 754 denormalized floating-point numbers and gradual underflow, which make...

optimization,filter,average,ieee-754,multiplying

I don't think that multiplying a floating-point number by a power of two is faster in practice than a generic multiplication (though I agree that in theory it should be faster, assuming no overflow/underflow). Said otherwise, I don't think that there is a hardware optimization. Now, I can assume that...

c#,.net,floating-point,ieee-754

The sequence of bits representing the number 3333 is not found in the IEEE 754 representation of 2.3333, nor 0.3333, because IEEE 754 uses a binary exponent, not decimal. That is, you are looking for the numerator in 3333 / 10000 but the internal representation is (when converted to decimal)...

Note: The question was edited to add "that is a divisor of" long after this answer was posted, see below the fold for an update. What is the risk of precision loss... It's virtually guaranteed, depending on the integer and floating point number involved, because of the mismatch between what...

c++,algorithm,floating-point,ieee-754

For normal numbers like 3.14159, the procedure is as follows: separate the number into sign, biased exponent, and significand add the difference in the exponent biases for long double and double (0x3fff - 0x3ff) to the exponent. assemble the sign, new exponent, and significand (remembering to make the leading bit...

binary,numbers,32-bit,ieee-754,ieee

Since i will assume you already know the Standard we can convert as following , Convert your number to base 2 1011.01000 Shift this binary number 1.01101000 2**3 (shifted by 3) add exponent 127+3=130 convert 130 to binary format 10000010 And so we have sign * 2^exponent * mantissa Sign...

hash,floating-point,64bit,ieee-754,floating-point-conversion

Apart from +0/-0, are there denormalized floats, which really translate to the same discrete value, or can I just hash their binary representation as is, without fear of generating different hashes for identical values? There is no translation from the representation to something else that would be the “real”...

math,floating-point,division,ieee-754

One practical way of correctly rounding the result of iterative division is to produce a preliminary quotient to within one ulp of the mathematical result, then use the exactly-computed residual to compute the final result. The tool of choice for the exact computation of residuals is the fused-multiply add (FMA)...

floating-point,double,ieee-754,floating-point-conversion,significant-digits

First, for this question it is better to use the total significand sizes 24 and 53. The fact that the leading bit is not represented is just an aspect of the encoding. If you are interested only in a vague explanation, one decimal digits contains exactly log2(10) (about 3.32) bits...

c,floating-point,floating-accuracy,ieee-754

I am going to assume IEEE 754 binary floating point arithmetic, with float 32 bit and double 64 bit. In general, there is no advantage to doing the calculation in double, and in some cases it may make things worse through doing two rounding steps. Conversion from float to double...

This does not guarantee the correct answer if the underlying machine is something esoteric, however: float f = 3.14; uint32_t u; memcpy(&u, &f, sizeof u); for (int i = 31; i >= 0; i--) putchar('0' + ((u >> i) & 1)); ...