octal to binary, flip the binary, convert to hex

2014 6 5
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更新时间：2014 6 5
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I have a unique situation where I must take an octal label (e.g., '52') and ultimately arrive at a hex value that happens to be the flipped version of the binary of the octal. Poor wording... I know. Here is an example:
Octal label = 52
Binary of octal 52 = 00101010
Flipped binary of octal 52 = 01010100
Hex of flipped binary of octal 52 = 54
I have 78 different octal numbers I may have to do this for and am looking for a simple solution. The complicated part for me is the fact that MATLAB drops off the leading zeroes, so using fliplr() doesn't help too much when trying to flip the binary number. Here is what I have so far:
for i = 1:length(chosenLabels)
chosenLables(i) = bin2hex(fliplr(dec2bin(base2dec(chosenLabels(i), 8))));
end
This is how I found out bin2hex isn't a function :) And as I said, I know the fliplr isn't going to work. Ideally, one line of code would be awesome but I'm guessing I may need to use a for loop and create a binary vector that includes the leading zeroes.

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Joseph Cheng
Joseph Cheng 2014-6-5
编辑：Joseph Cheng 2014-6-5

1 个投票

You can try something like this
octstr = ['52';'51']; %your oct inputs
bin_str = dec2bin(base2dec(octstr,8),8) %convert to decimal using base 8 and 8 indexes long.
bin_str=bin_str(:,end:-1:1) %reverse the binary
hexstr = dec2hex(bin2dec(bin_str)) %convert bin->dec->hex
and you don't even need a for loop!

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Adam Kaas
Adam Kaas 2014-6-5
This is incredibly close to what I need. The difference is my chosenLables array is a 1x9 double and base2dec() doesn't like it. I did a num2str(chosenLabels) and it liked it, but the answer isn't readable as my binary string is one very long string of binary with no separation. Suggestions? I apologize... I'm not the most well versed MATLAB user.
Adam Kaas
Adam Kaas 2014-6-5
Here's what I ended up doing.
bin_str = dec2bin(oct2dec(chosenLabels),8);
bin_str = bin_str(:,end:-1:1);
hexstr = dec2hex(bin2dec(bin_str));
Thanks for the pointers in the right direction Joseph!
Joseph Cheng
Joseph Cheng 2014-6-5
编辑：Joseph Cheng 2014-6-5
So you're saying the octal number is not a string but an octal number written as a decimal number?
then you can do this.
clc,clear all;
octdouble = [46 46 23 35 5 4 32 47 57];
octstr = num2str(octdouble')
octstr(octstr==' ')='0'
bin_str = dec2bin(base2dec(octstr,8),8)
bin_str=bin_str(:,end:-1:1)
hexstr = dec2hex(bin2dec(bin_str))
Joseph Cheng
Joseph Cheng 2014-6-5
编辑：Joseph Cheng 2014-6-5
you don't need to ' ' to '0' substitution. got a random error but that was because i was using a random integer generator to make the octals and didn't notice i had 8 and 9 in in there which wouldn't be a valid octal number, duh.

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