问题描述
朋友&我正在辩论今天午餐时如何储存Inf和NaN。以Fortran 90为例。 4字节的实数可以获得Inf或NaN的值。这是如何存储在内部的?据推测,一个4字节的实数是由一个32位二进制数字内部表示的数字。 Inf和NaN是以33位二进制数存储的吗?
pbs.edu/general/software/packages/ieee/ieee.phprel =noreferrer> link :
IEEE单精度浮点标准表示需要一个32位字,它可以表示为从0到31的编号,从左到右。第一位是符号位, S
,接下来的八位是指数位,' E
',最后的23位是小数点' F
':
EEEEEEEE FFFFFFFFFFFFFFFFFFFFFFF
0 1 8 9 31
由该词表示的 V
可以如下确定:
E = 255
和 F
非零,则 V = NaN
(不是数字)
F
为零且 S
为 1
,然后 V = -Infinity
E = 255
和 F
为零且 S
为 0
,则 V = Infinity
0 然后
V =( - 1)** S * 2 **(E-127)*(1.F)
其中 1.F
旨在表示通过用隐式前导1和二进制前缀F创建的二进制数
点。如果 E = 0
且 F
不为零,则 V =( - 1)** S * 2 **(-126)*(0.F)
这些
是非标准化值。
E = 0
且 F
为零且 S
是 1
,然后 V = -0
E = 0
且 F
为零且 S
0
,然后 V = 0
A friend & I were debating how Inf's and NaN's are stored during lunch today.
Take Fortran 90 for example. 4-byte reals can obtain the value of Inf or NaN. How is this stored internally? Presumably, a 4-byte real is a number represented internally by a 32 digit binary number. Are Inf's and NaN's stored as 33 bit binary numbers?
Specifically from Pesto's link:
The IEEE single precision floating point standard representation requires a 32 bit word, which may be represented as numbered from 0 to 31, left to right. The first bit is the sign bit, S
, the next eight bits are the exponent bits, 'E
', and the final 23 bits are the fraction 'F
':
S EEEEEEEE FFFFFFFFFFFFFFFFFFFFFFF 0 1 8 9 31
The value V
represented by the word may be determined as follows:
- If
E=255
andF
is nonzero, thenV=NaN
("Not a number") - If
E=255
andF
is zero andS
is1
, thenV=-Infinity
- If
E=255
andF
is zero andS
is0
, thenV=Infinity
- If
0<E<255
thenV=(-1)**S * 2 ** (E-127) * (1.F)
where "1.F
" is intended to represent the binary number created by prefixing F with an implicit leading 1 and a binarypoint. - If
E=0
andF
is nonzero, thenV=(-1)**S * 2 ** (-126) * (0.F)
Theseare "unnormalized" values. - If
E=0
andF
is zero andS
is1
, thenV=-0
- If
E=0
andF
is zero andS
is0
, thenV=0
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