An IEEE double has 53 significant bits (that’s the value of DBL_MANT_DIG in <cfloat>). That’s approximately 15.95 decimal digits (log10(253)); the implementation sets DBL_DIG to 15, not 16, because it has to round down. So you have nearly an extra decimal digit of precision (beyond what’s implied by DBL_DIG==15) because of that. The nextafter() function … Read more
The MSDN documentation is nonsensical and wrong. Bad concept. Binary-floating-point format does not have any precision in decimal digits because it has no decimal digits at all. It represents numbers with a sign, a fixed number of binary digits (bits), and an exponent for a power of two. Wrong on the high end. The floating-point … Read more
Just what is this “hint” to the runtime? As you conjecture, the compiler tracks whether a conversion to double or float was actually present in the source code, and if it was, it always inserts the appropriate conv opcode. Does the C# spec stipulate that an explicit cast to float causes the insertion of a … Read more
The Wikipedia page on it is a good place to start. To sum up: float is represented in 32 bits, with 1 sign bit, 8 bits of exponent, and 23 bits of the significand (or what follows from a scientific-notation number: 2.33728*1012; 33728 is the significand). double is represented in 64 bits, with 1 sign … Read more
I know of no way to way to make normal floating points deterministic in .net. The JITter is allowed to create code that behaves differently on different platforms(or between different versions of .net). So using normal floats in deterministic .net code is not possible. The workarounds I considered: Implement FixedPoint32 in C#. While this is … Read more
For a given IEEE-754 floating point number X, if 2^E <= abs(X) < 2^(E+1) then the distance from X to the next largest representable floating point number (epsilon) is: epsilon = 2^(E-52) % For a 64-bit float (double precision) epsilon = 2^(E-23) % For a 32-bit float (single precision) epsilon = 2^(E-10) % For a … Read more
Huge difference. As the name implies, a double has 2x the precision of float[1]. In general a double has 15 decimal digits of precision, while float has 7. Here’s how the number of digits are calculated: double has 52 mantissa bits + 1 hidden bit: log(253)÷log(10) = 15.95 digits float has 23 mantissa bits + … Read more
The accepted answer is 100% without question WRONG. Not halfway wrong or even slightly wrong. I fear this issue is going to confuse and mislead programmers for a long time to come when this question pops up in searches. NaN is designed to propagate through all calculations, infecting them like a virus, so if somewhere … Read more
2mantissa bits + 1 + 1 The +1 in the exponent (mantissa bits + 1) is because, if the mantissa contains abcdef… the number it represents is actually 1.abcdef… × 2^e, providing an extra implicit bit of precision. Therefore, the first integer that cannot be accurately represented and will be rounded is: For float, 16,777,217 … Read more
The biggest/largest integer that can be stored in a double without losing precision is the same as the largest possible value of a double. That is, DBL_MAX or approximately 1.8 × 10308 (if your double is an IEEE 754 64-bit double). It’s an integer. It’s represented exactly. What more do you want? Go on, ask me what … Read more