Binary to floating point representation using IEEE-754

I am given a Character array of length 32 made of only zeros and ones and i want to convert in floating point representation using IEEE-754. Can any one help???? My string is '10101110101101011010010110101001'. and my code uptill now is

sign=binary(1);

exp=binary(2:9);

mantissa=binary(10:32)

subt=bin2dec(exp);

e=-127+subt;

num=0;

for i=1:length(mantissa)

num=mantissa(i)*2^(-i)+num;

end

[Arslan Ahmad]

i did it at last

function [result] = mySingle2Decimal(binary)

sign=str2num(binary(1));

exp=binary(2:9);

mant=binary(10:32);

subt=bin2dec(exp);

e=-127+subt;

num=0;

for i=1:length(mant)

num=str2num(mant(i))*2^(-i)+num;

end

result=(-1)^sign*(1+num)*2^(e);

Answer 1

[Stephen23]

It is simple to avoid the loop:

S = '10101110101101011010010110101001';
%S = '00111110001000000000000000000000'; % Wikipedia example
V = S-'0'; % convert to numeric
frc = 1+sum(V(10:32).*2.^(-1:-1:-23))
pow = sum(V(2:9).*2.^(7:-1:0))-127
sgn = (-1)^V(1)
val = sgn * frc * 2^pow

See also https://en.wikipedia.org/wiki/Single-precision_floating-point_format

[Arslan Ahmad]

Thanks a lot

Answer 2

[James Tursa]

Any method you choose is going to have to make assumptions about bit/byte ordering and the handling of the special inf, nan, and denormalized bit patterns. The following method simply assumes that the char array bit pattern you have matches the machine you are currently using and accounts for these special patterns:

result = typecast(uint32(bin2dec(S)),'single');

The other methods shown in this thread that work with the sign, exponent, and mantissa bits directly do not account for these special bit patterns.

[Anton Shishkin]

Thanks a lot