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

1 个评论

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);

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 采纳的回答

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

更多回答(1 个)

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.

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