IEEE 754中表示的最大浮点数 [英] max float represented in IEEE 754
问题描述
$(b)(1.11111111111111111111111)_b * 2 ^ [(11111111)_b- 127]
这里 有人知道为什么吗? 最大值以单精度表示,约为3.4028235×1038,实际上是1.11111111111111111111111_b2×11111110_b -127 。 另请参阅 http:/ /en.wikipedia.org/wiki/Single-precision_floating-point_format I am wondering if the max float represented in IEEE 754 is: Here Does anybody know why? Thank you. The exponent 11111111b is reserved for infinities and NaNs, so your number cannot be represented. The greatest value that can be represented in single precision, approximately 3.4028235×1038, is actually 1.11111111111111111111111b×211111110b-127. See also http://en.wikipedia.org/wiki/Single-precision_floating-point_format 这篇关于IEEE 754中表示的最大浮点数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! _b
表示二进制表示。但是这个值是 3.403201383 * 10 ^ 38
,它与 3.402823669 * 10 ^ 38
不同,它是(1.0)_b * 2 ^ [(11111111)_b-127]
,例如<限定>
。是不是
(1.11111111111111111111111)_b * 2 ^ [(11111111)_b-127]
在框架中可表示和更大?
(1.11111111111111111111111)_b*2^[(11111111)_b-127]
_b
means binary representation. But that value is 3.403201383*10^38
, which is different from 3.402823669*10^38
, which is (1.0)_b*2^[(11111111)_b-127]
and given by for example c++
<limits>
. Isn't
(1.11111111111111111111111)_b*2^[(11111111)_b-127]
representable and larger in the framework?