How to truncate a binary number and else?

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Marcela Matos 2020-3-16

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评论： Walter Roberson 2020-3-19

Having a binary number:

101.010101000111101011100001

101.01010100011110101110000
101.01010100011110101110001

Truncate de binary number (IEEE 754) That my binary number only have 23 digits on the right
That the binary number have only 23 digits on the right but not with truncation but to round. If the 24th digit is 1 then it must add it to the 23th digit etc)

How do I do that in matlab?

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Walter Roberson 2020-3-16

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N = '10101010100011110101110000';
sum((N-'0').*2.^(3-(1:length(N))))

Walter Roberson 2020-3-16

This code does not require that all of N be processed.

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Answer 1

Aghamarsh Varanasi 2020-3-19

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Hi,

I understand that you want to implement truncate and round-off operations on binary numbers. Considering that the binary numbers are stored as strings in MATLAB, we can implement the truncate and round-off operations as,

Truncate

N = '101.010101000111101011100001'; 
%index at which the binary number ends -- 23 digits to right of decimal 
index = strfind(N,'.')+23; 
 
% truncate 
N = N(1:min(index,length(N))); 

Round off

N = '101.010101000111101011100001'; 
%index at which the binary number ends -- 23 digits to right of decimal 
index = strfind(N,'.')+23; 
if index < length(N) 
    roundOffPart = N(index:end); 
 
    %truncate N 
    N = N(1:min(index,length(N))); 
 
    % If there is 1 in the string after 23 digits 
    % carry it farward to N 
    if contains(roundOffPart,'1') 
        N(end) = '1';  
    end 
end 

Hope this helps

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Walter Roberson 2020-3-19

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    % If there is 1 in the string after 23 digits 
    % carry it farward to N

That would be rounding up, but the user only asked for rounding, so only the location immediately after 23 needs to be checked. Unless the user did not explain properly, as rounding-up is a valid thing to want to do and maybe was intended.

    if contains(roundOffPart,'1') 
        N(end) = '1';  
    end
    

But suppose that N(end) is already '1'. Which is the case. The number has been carefully constructed so that rounding up would require carrying the '1' by several places. Possibly past the decimal point.

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