计算C中的浮点的epsilon [英] Calculate epsilon for a float in C

问题描述

我必须使用位/整数运算来找出C中给定值的epsilon.根据位模式，我知道计算下一个邻居需要增加尾数，如果溢出，则要增加指数-但我不确定从哪里开始计算ε.

I have to use bit/integer ops to figure out the epsilon for a given value in C. I know in terms of bit patterns that computing the next neighbor involves incrementing the mantissa, and if that overflows, to increment the exponent - but I'm not sure where to begin in terms of calculating the epsilon.

我无法使用浮点数学运算-因此，我需要直接生成位模式.这使操作变得棘手，因为我无法减去.

这是我的理解(基于我在SO上所做的一些研究):范围随着​​数字的增加而明显变化，但我不确定如何使用 FLT_EPSILON生成正确的数字.对于2 ^ 2x数字，也许是(FLT_EPSILON-1)*数字?

Here's my understanding (based on some research I've done here on SO): the range changes obviously as the numbers get bigger, but I'm not sure how to use FLT_EPSILON to generate the right numbers. For 2^2x numbers, perhaps it is (FLT_EPSILON - 1) * number?

推荐答案

如果epsilon只是尾数增加一个LSB​​的值，则该值应为2 ^(exp-23).因此，如果您输入的是 [s] [exp] [mantissa] ，则epsilon应该是 [0] [exp-23] [0] .当然，当输入为1.0( [0] [127] [0] )时，结果为 [0] [104] [0] = 2 ^-23 = FLT_EPSILON .

If your epsilon is simply the value of one LSB increment to the mantissa, the value should be 2^(exp-23). So if your input is [s][exp][mantissa] your epsilon should be [0][exp-23][0]. Sure enough, when given an input of 1.0 ([0][127][0]), the result is [0][104][0] = 2^-23 = FLT_EPSILON.

这篇关于计算C中的浮点的epsilon的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！