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Flip last 3 bits of vector

Asked by Jorge Zapata on 14 Mar 2013

I have a uint16 vector which I need to flip the last 3 bits of every number.

I already have done this but I think there must be an easier solution to do this. Here is my code.

%Turn the vector to binary
V_bin = dec2bin(V,16);
for i=1:length(V)
  %Get the last 3 bits
  tmp = V_bin(14:end);
  %Convert the string to decimal
  tmpdec = bin2dec(tmp);
  %Do the flip
  tmpflip = bitcmp(uint8(tmpdec));
%Flipped to binary
tmpbin = dec2bin(tmpflip);
%Replace the flipped bits in the original string
V_bin(14:end) = tmpbin(6:end);
end
V = bin2dec(V_bin);

As you can see there are a lot of lines for a simple operation, I wonder if there is a more efficient method to do the same. Thanks

0 Comments

Jorge Zapata

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1 Answer

Answer by Teja Muppirala on 14 Mar 2013
Accepted answer

You can use BITXOR,

V = uint16( round(65535*rand(5,1)) );
V2 = bitxor(V,1+2+4); % 1+2+4 = 7 = 0000000000000111
dec2bin(V)
dec2bin(V2)

2 Comments

Jan Simon on 14 Mar 2013

+1, I cannot stress enough how much better this is compared to the dec2bin conversion tricks. Converting numerical data to strings for a manipulation is too indirect to be efficient.

Jorge Zapata on 14 Mar 2013

In fact strings operations affect performance so your answer is what I was looking for, thanks

Teja Muppirala

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