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Update number.md #270
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IEEE 754 中，双精度浮点数精度的计算我个人认为是这样的 ，因为最大的尾数是1.{52个1}（默认隐藏一个固定的1）, 指数能让小数点往前移动52位，所以最大整数是1....一共53个1，然后就是2∧52+2∧51+…+2∧1+2∧0 一个等比数列求和的计算~结果为2∧53-1，所以精度范围是 -(253 – 1) 和253 – 1 之间的整数
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其实范围是 -2^^53 + 1 到 2^^53 - 1，我来改掉。 |
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IEEE 754 中，双精度浮点数精度的计算我个人认为是这样的 ，因为最大的尾数是1.{52个1}（默认隐藏一个固定的1）, 指数能让小数点往前移动52位，所以最大整数是1....一共53个1，然后就是2∧52+2∧51+…+2∧1+2∧0 一个等比数列求和的计算~结果为2∧53-1，所以精度范围是 -(2的53次方 – 1) 和2的53次方 – 1 之间的整数