ieee-754 | 易学教程

Why does frexp() not yield scientific notation?

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问题 Scientific notation is the common way to express a number with an explicit order of magnitude. First a nonzero digit, then a radix point, then a fractional part, and the exponent. In binary, there is only one possible nonzero digit. Floating-point math involves an implicit first digit equal to one, then the mantissa bits "follow the radix point." So why does frexp() put the radix point to the left of the implicit bit, and return a number in [0.5, 1) instead of scientific-notation-like [1, 2)?

Are all integer values perfectly represented as doubles? [duplicate]

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This question already has an answer here: Representing integers in doubles 5 answers My question is whether all integer values are guaranteed to have a perfect double representation. Consider the following code sample that prints "Same": // Example program #include <iostream> #include <string> int main() { int a = 3; int b = 4; double d_a(a); double d_b(b); double int_sum = a + b; double d_sum = d_a + d_b; if (double(int_sum) == d_sum) { std::cout << "Same" << std::endl; } } Is this guaranteed to be true for any architecture, any compiler, any values of a and b ? Will any integer i converted

Computing a correctly rounded / an almost correctly rounded floating-point cubic root

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问题 Suppose that correctly rounded standard library functions such as found in CRlibm are available. Then how would one compute the correctly rounded cubic root of a double-precision input? This question is not an “actual problem that [I] face”, to quote the FAQ. It is a little bit like homework this way. But the cubic root is a frequently found operation and one could imagine this question being an actual problem that someone faces. Since “best Stack Overflow questions have a bit of source code

Properties of 80-bit extended precision computations starting from double precision arguments

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问题 Here are two implementations of interpolation functions. Argument u1 is always between 0. and 1. . #include <stdio.h> double interpol_64(double u1, double u2, double u3) { return u2 * (1.0 - u1) + u1 * u3; } double interpol_80(double u1, double u2, double u3) { return u2 * (1.0 - (long double)u1) + u1 * (long double)u3; } int main() { double y64,y80,u1,u2,u3; u1 = 0.025; u2 = 0.195; u3 = 0.195; y64 = interpol_64(u1, u2, u3); y80 = interpol_80(u1, u2, u3); printf("u2: %a\ny64:%a\ny80:%a\n", u2

How does javascript print 0.1 with such accuracy?

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问题 I've heard that javascript Numbers are IEEE 754 floating points, which explains why > 0.3 - 0.2 0.09999999999999998 but I don't understand > 0.1 0.1 I thought 0.1 couldn't be accurately stored as a base 2 floating point, but it prints right back out, like it's been 0.1 all along. What gives? Is the interpreter doing some rounding before it prints? It's not helping me that there are at least 2 versions of IEEE 754: 1984 edition and 2008. It sounds like the latter added full support for decimal

What are the other NaN values?

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问题 The documentation for java.lang.Double.NaN says that it is A constant holding a Not-a-Number (NaN) value of type double . It is equivalent to the value returned by Double.longBitsToDouble(0x7ff8000000000000L) . This seems to imply there are others. If so, how do I get hold of them, and can this be done portably? To be clear, I would like to find the double values x such that Double.doubleToRawLongBits(x) != Double.doubleToRawLongBits(Double.NaN) and Double.isNaN(x) are both true. 回答1: You

What uses do floating point NaN payloads have?

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问题 I know that IEEE 754 defines NaNs to have the following bitwise representation: The sign bit can be either 0 or 1 The exponent field contains all 1 bits Some bits of the mantissa are used to specify whether it's a quiet NaN or signalling NaN The mantissa cannot be all 0 bits because that bit pattern is reserved for representing infinity The remaining bits of the mantissa form a payload The payload is propagated (as is the NaN as a whole) to the result of a floating point calculation when the

What are the applications/benefits of an 80-bit extended precision data type?

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问题 Yeah, I meant to say 80-bit . That's not a typo... My experience with floating point variables has always involved 4-byte multiples, like singles (32 bit), doubles (64 bit), and long doubles (which I've seen refered to as either 96-bit or 128-bit). That's why I was a bit confused when I came across an 80-bit extended precision data type while I was working on some code to read and write to AIFF (Audio Interchange File Format) files: an extended precision variable was chosen to store the

Extreme numerical values in floating-point precision in R

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Can somebody please explain me the following output. I know that it has something to do with floating point precision, but the order of magnitue (difference 1e308) surprises me. 0: high precision > 1e-324==0 [1] TRUE > 1e-323==0 [1] FALSE 1: very unprecise > 1 - 1e-16 == 1 [1] FALSE > 1 - 1e-17 == 1 [1] TRUE R uses IEEE 754 double-precision floating-point numbers. Floating-point numbers are more dense near zero. This is a result of their being designed to compute accurately (the equivalent of about 16 significant decimal digits, as you have noticed) over a very wide range. Perhaps you expected

Is SSE floating-point arithmetic reproducible?

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The x87 FPU is notable for using an internal 80-bit precision mode, which often leads to unexpected and unreproducible results across compilers and machines. In my search for reproducible floating-point math on .NET, I discovered that both major implementations of .NET (Microsoft's and Mono) emit SSE instructions rather than x87 in 64-bit mode. SSE(2) uses strictly 32-bit registers for 32-bit floats, and strictly 64-bit registers for 64-bit floats. Denormals can optionally be flushed to zero by setting the appropriate control word . It would therefore appear that SSE does not suffer from the