问题描述
IEEE 754浮点除法的可逆性是什么?我的意思是，该标准是否保证如果double y = 1.0 / x然后x == 1.0 / y，即x可以精确地一点一点恢复?
What is the invertability of the IEEE 754 floating-point division? I mean is it guaranteed by the standard that if double y = 1.0 / x then x == 1.0 / y, i.e. x can be restored precisely bit by bit?
y是infinity或NaN的情况是明显的例外.
The cases when y is infinity or NaN are obvious exceptions.
推荐答案
是的，IEEE 754双精度(*)值x就是x != 1.0 / (1.0 / x).
Yes, there are IEEE 754 double-precision(*) values x that are such x != 1.0 / (1.0 / x).
使用此属性可以轻松构建一个正常值的示例:在0x1.fffffffffffffp0编写的示例/> C99的浮点值十六进制表示法就是1.0 / (1.0 / 0x1.fffffffffffffp0) == 0x1.ffffffffffffep0.很自然地期望0x1.fffffffffffffp0是一个反例，因为1.0 / 0x1.fffffffffffffp0落在binade的开头，在那里浮点数的密度较小，因此最里面的除法必须发生较大的相对误差.更准确地说，1.0 / 0x1.fffffffffffffp0恰好位于0.5及其双精度后继字符之间的中点上方，因此1.0 / 0x1.fffffffffffffp0舍入为0.5的后继字符，具有较大的相对误差.
It is easy to build an example of a normal value with this property by hand: the one that's written 0x1.fffffffffffffp0 in C99's hexadecimal notation for floating-point values is such that 1.0 / (1.0 / 0x1.fffffffffffffp0) == 0x1.ffffffffffffep0. It was natural to expect 0x1.fffffffffffffp0 to be a counter-example because 1.0 / 0x1.fffffffffffffp0 falls at the beginning of a binade, where floating-point numbers are less dense, so a larger relative error had to happen on the innermost division. More precisely, 1.0 / 0x1.fffffffffffffp0 falls just above the midpoint between 0.5 and its double-precision successor, so that 1.0 / 0x1.fffffffffffffp0 is rounded up to the successor of 0.5, with a large relative error.
以十进制%.16e格式，0x1.fffffffffffffp0是1.9999999999999998e+00，而0x1.ffffffffffffep0是1.9999999999999996e+00.
In decimal %.16e format, 0x1.fffffffffffffp0 is 1.9999999999999998e+00 and 0x1.ffffffffffffep0 is 1.9999999999999996e+00.
(*)对于任何IEEE 754格式，没有理由使反函数具有该属性.
(*) there is no reason for the inverse function to have the property in the question for any of the IEEE 754 format
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