float 和 double 精度相关的概念

为什么精度float小数点后最多 6 位，精度double小数点后最多15位？

任何人都可以给一个数学解释 of it?

说一下精度float or double是一些小数位数是草率的术语。float and double通常使用 IEEE-754 二进制 32 和二进制 64 格式实现，它们分别使用 24 位和 53 位的有效数。 （有效数是浮点表示形式的小数部分。它伴随着符号位和指数。）这些格式的精度为 24 位和 53 位。

精度足够，对于float，任何具有六位有效数字的十进制数字都可以转换为float足够准确，转换回六位十进制数字会产生相同的数字。 （本文讨论的所有转换都使用舍入到最接近的值。）六是限制；有一些七位十进制数字无法在往返转换后继续存在。为了double，限制为 15 位有效数字。

Here is how we find the limits. A 24-bit significand can represent 2²⁴ values. That is 16,777,216, so it is certainly capable of representing all seven-digit decimal numerals from 0 to 9,999,999. However, if we want to represent some seven-digit decimal numeral a.bcdefg•10ⁿ, the last decimal numeral in this interval, 9.999999•10ⁿ, has some float representation f•2^e. Because the powers of 10 are irregular relative to the powers of two, that f might be small, meaning that we are not representing the numbers from 0.000000•10ⁿ to 9.999999•10ⁿ with significands entirely in a single sequence 0 to 16,777,215. (I am glossing over some details here; numbers which would have significands under 2²² are rescaled to another exponent in the normal range. But being in a finer exponent range gives more precision, so that is not an issue.) Rather, there was a rollover at some point, and the exponent increased. So we lose one bit of precision. So the number of decimal digits we can guarantee survive a round-trip conversion to a floating-point format with p digits in base b is floor((p−1)•log₁₀b). (That formula is given in C 2018 5.2.4.2.2 12.)