If I assign a 23 index greater than 16777215 how to undo the decimal number? More than 23 at the back of the tail 1 s and 0 s how to choose?

CodePudding user response:

https://www.cnblogs.com/fengliu-/p/7455246.html
Suggest to understand the ~

CodePudding user response:

See my question in reply, I do not want you to tell me how single precision is stored, my question is if the mantissa is beyond how to convert decimal after 23 for example, I define a binary is 1.00000000000000000000011 * 24 float016777219 mantissa has 24 beyond mantissa said, the result is 16777220 is why my question is if after more than 23 index such as index of 25 how to restore a decimal

CodePudding user response:

The custom want to data types, overloading need to calculation,

CodePudding user response:

Large range of floating-point Numbers is said, but the precision is not high, with 23 binary precision decimal precision is 6-7, more than the precision will be lost

F=16777219.0

E=log (f)/log (2)
E=0 + (e | 0)

M=(1 & lt; <23)
N=(2, e) - (f/pow 1.0)
N=n * m
N=((n + 0.5) | 0)//round here, precision loss

H=0 | ((e + 0 x7f) & lt; <23) | n//here is the hexadecimal stored data

V=h& (m - 1)
V=(v + 0.0)/m
F=(1.0 + v) * pow (2 e)//the end result here into 16777220



The following is a calculation steps:

 
F=16777219.00000000 
16777219.00000000 
E=log (f)/log (2) 
E=16.63553251225261/log (2) 
E=16.63553251225261/0.6931471805599453 
E=24.00000025797396 
24.00000025797396 
E=0 + (e | 0) 
E=0 + 0 x0000000000000018 
E=24 
24 
M=(1 & lt; <23) 
M=0 x0000000000800000 
0 x0000000000800000 
N=(2, e) - (f/pow 1.000000000000000) 
N=(f/16777216.00000000-1.000000000000000) 
N=(1.000000178813934 1.000000000000000) 
N=1.788139343261719 e-007 
1.788139343261719 e-007 
N=n * m 
N=1.500000000000000 
1.500000000000000 
N=((n + 0.5000000000000000) | 0) 
N=(2.000000000000000 | 0) 
N=0 x0000000000000002 
0 x0000000000000002 
H=0 | ((e + 0 x000000000000007f) & lt; <23) | n 
H=0 | (151 & lt; <23) | n 
H=0 | 0 x000000004b800000 | n 
H=0 x000000004b800000 | n 
H=0 x000000004b800002 
0 x000000004b800002 
V=h & amp; (m - 1) 
V=h & amp; 0 x00000000007fffff 
V=0 x0000000000000002 
0 x0000000000000002 
V=(v + 0.0000000000000000)/m 
V=2.000000000000000/m 
V=2.384185791015625 e-007 
2.384185791015625 e-007 
F=(1.000000000000000 + v) * pow (2 e) 
(f=1.000000238418579 * pow, 2 e) 
F=1.000000238418579 * 16777216.00000000 
F=16777220.00000000 
16777220.00000000