Can anyone help me to understand the concept of this code ( IEEE 754 REPRESENTATION

1 visualización (últimos 30 días)
clc
clear all
close all % Converting input of base 10
format long g
x=randi([1 50],1,1)
%Converting of mantissa (x_1) to between 1 and 2 of base-2 scientific notation
n_exponent=0;
x_1=x;
if abs(x)<1
while abs(x_1)<1
n_exponent=n_exponent-1;
x_1=x/(2.^n_exponent);
end
else
while abs(x_1)>=2
n_exponent=n_exponent+1;
x_1=x/(2.^n_exponent);
end
end
%Exponent to 8-bit binary (single precision 127)
exp_127=n_exponent+127;
bit8=0;
order=1;
for n=1:8
bit8=rem(exp_127,2)*order+bit8;
exp_127=fix(exp_127/2);
order=order*10;
end
% Converting fraction to 23-bit mantissa (leading 1 hidden)
fraction=x_1-1;
bit23=0;
order1=1;
for n=1:23
fraction=fraction*2;
bit23=order1*fix(fraction)+bit23;
fraction=rem(fraction,1);
order1=order1*10;
end
%Determining sign of x
sign=0;
if x<0
sign=1;
end
%Full IEEE 754 binary floating point
N=32;
len=length(sign)+length(num2str(bit8))+length(num2str(bit23))
fprintf('%d', [sign, fliplr(bit8), fliplr(bit23),zeros(1,N-len)])
  3 comentarios
David Goodmanson
David Goodmanson el 10 de Ag. de 2018
Editada: David Goodmanson el 10 de Ag. de 2018
Hi Rajashree, take a look at https://en.wikipedia.org/wiki/Double-precision_floating-point_format
Rajashree Annapillai
Rajashree Annapillai el 12 de Ag. de 2018
Thank you sir.Can you please explain how the exponent is calculated by this code.

Iniciar sesión para comentar.

Respuestas (0)

Categorías

Más información sobre Numbers and Precision en Help Center y File Exchange.

Productos


Versión

R2016a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by