ieee-754

So I'm trying to learn more about Denormalized numbers as defined in the IEEE 754 standard for Floating Point numbers. I've already read several articles thanks to Google search results, and I've gone through several StackOverFlow posts. However I still have some questions unanswered. First off, just to review my understanding of what a Denormalized float is: Numbers which have fewer bits of precision, and are smaller (in magnitude) than normalized numbers Essentially, a denormalized float has the ability to represent the SMALLEST (in magnitude) number that is possible to be represented with

It seems that the IEEE 754 standard defines 16,777,214 32-bit floating point values as NaNs, or 0.4% of all possible values. I wonder what is the rationale for reserving so many useful values, while only 2 ones essentially needed: one for signaling and one for quiet NaN. Sorry if this question is trivial, I couldn't find any explanation on the internet. Robert Harvey The IEEE-754 standard defines a NaN as a number with all ones in the exponent, and a non-zero significand. The highest-order bit in the significand specifies whether the NaN is a signaling or quiet one. The remaining bits of the

Consider the following C++ code: double someZero = 0; std::cout << 0 - someZero << '\n'; // prints 0 std::cout << -someZero << std::endl; // prints -0 The question arises: what is negative zero good for, and should it be defensively avoided (i.e. use subtraction instead of smacking a minus onto a variable)? NPE From Wikipedia: It is claimed that the inclusion of signed zero in IEEE 754 makes it much easier to achieve numerical accuracy in some critical problems[1], in particular when computing with complex elementary functions[2]. The first reference is "Branch Cuts for Complex Elementary

How do I subtract IEEE 754 numbers? For example: 0,546875 - 32.875... -> 0,546875 is 0 01111110 10001100000000000000000 in IEEE-754 -> -32.875 is 1 10000111 01000101111000000000000 in IEEE-754 So how do I do the subtraction? I know I have to to make both exponents equal but what do I do after that? 2'Complement of -32.875 mantissa and add with 0.546875 mantissa? old_timer Really not any different than you do it with pencil and paper. Okay a little different 123400 - 5432 = 1.234*10^5 - 5.432*10^3 the bigger number dominates, shift the smaller number's mantissa off into the bit bucket until the

This topic has come up many times on StackOverflow, but I believe this is a new take. Yes, I have read Bruce Dawson's articles and What Every Computer Scientist Should Know About Floating-Point Arithmetic and this nice answer . As I understand it, on a typical system there are four basic problems when comparing floating-point numbers for equality: Floating point calculations are not exact Whether a-b is "small" depends on the scale of a and b Whether a-b is "small" depends on the type of a and b (e.g. float, double, long double) Floating point typically has +-infinity, NaN, and denormalized

I know that most decimals don't have an exact floating point representation ( Is floating point math broken? ). But I don't see why 4*0.1 is printed nicely as 0.4 , but 3*0.1 isn't, when both values actually have ugly decimal representations: >>> 3*0.1 0.30000000000000004 >>> 4*0.1 0.4 >>> from decimal import Decimal >>> Decimal(3*0.1) Decimal('0.3000000000000000444089209850062616169452667236328125') >>> Decimal(4*0.1) Decimal('0.40000000000000002220446049250313080847263336181640625') The simple answer is because 3*0.1 != 0.3 due to quantization (roundoff) error (whereas 4*0.1 == 0.4 because

With GCC 5.3 the following code compield with -O3 -fma float mul_add(float a, float b, float c) { return a*b + c; } produces the following assembly vfmadd132ss %xmm1, %xmm2, %xmm0 ret I noticed GCC doing this with -O3 already in GCC 4.8 . Clang 3.7 with -O3 -mfma produces vmulss %xmm1, %xmm0, %xmm0 vaddss %xmm2, %xmm0, %xmm0 retq but Clang 3.7 with -Ofast -mfma produces the same code as GCC with -O3 fast . I am surprised that GCC does with -O3 because from this answer it says The compiler is not allowed to fuse a separated add and multiply unless you allow for a relaxed floating-point model.