Question about epsilone()!

I'm reading the code that someone wrote epsilone() epsilon() appears in the associated document as "double type minus 1 from the smallest number greater than 1."

So epsilone() is the smallest number greater than zero? Or is there a value that can be expressed as double within the (0,0+epsilone) range?

double someValue = ... //assign something
if (someValue <  std::numeric_limits<double>::epsilon() && 
    someValue > -std::numeric_limits<double>::epsilon()) {
  someValue = 0.0;
}

c++ double

2022-09-21 23:09

1 Answers

64-bit IEEE double consists of 1-bit sign, 52-bit mantissa, and 11-bit extent.

The first one is 1.000000000000000000000000000000000000000000000000×2^0

The smallest number in water greater than 1 is 1.0000 00000000 00000000 00000000 00000000 00000000 00000001 × 2^0 = 1 + 2^-52 This is.

So epsilon = (1+2^-52) - 1 = 2^-52 It is.

There are many values within the (0, epilon) range that can be expressed as double. the smallest positive factor 1.0000 00000000 00000000 00000000 00000000 00000000 00000000 × 2^-1022 = 2^-1022 Starting with

There are about (1022 - 52 + 1)×2^52 = 4372995238176751616.

2022-09-21 23:09

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