Reading 64 bit words

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Francesc Massanes Basi
Francesc Massanes Basi on 29 Sep 2020
I have tried reading a binary file 64 bits at a time, and is not behaving as expecting.
I have a binary file with the word = 0xFAFAFAFA00010001, if I read it like this:
word = fread(fid,1,'uint64'), and look at hex2dec(word), the word is 0xFAFAFAFA00010000 (yes, one bit has been changed).
but if I read:
word_l = fread(fid,1,'uint64'); word_h = fread(fid,1,'uint64'), I get 0x00010001 as hex2dec(word_l) and 0xFAFAFAFA as hex2dec(word_h).
I do not know if the problem is on the read (don't think so), or the 64bit management (I think is here), but I tried doing this too:
a=hex2dec('FAFAFAFA');
b = hex2dec('00010001');
dec2hex(bitshift(a,32))
% Output is 0xFAFAFAFA00000000
dec2hex(b)
% Output is 0x00010001
dec2hex(bitor(bitshift(a,32),b))
% Output is 0xFAFAFAFA00010000
Is there an issue with matlab and numbers larger than 53 bits? 60 bits?

Accepted Answer

James Tursa
James Tursa on 29 Sep 2020
Try reading and keeping the type as uint64 (using the *) instead of converting to double:
word = fread(fid,1,'*uint64');
  3 Comments
James Tursa
James Tursa on 29 Sep 2020
Edited: James Tursa on 29 Sep 2020
I'm not sure what the issue is now. If you use the '*uint64' input format, the 64-bits are read into a uint64 type and all bits are retained. If you use the 'uint64' input format (without the asterisk) the bits are read and then converted to a double, which will lose trailing bits because of the mantissa limitation. Why isn't the '*uint64' format doing what you want?
The hex2dec('FAFAFAFA00010001') function also converts the value to double, losing trailing bits. So that is not a good comparison.
Francesc Massanes Basi
Francesc Massanes Basi on 2 Oct 2020
You are right, I had to go to another hex to decimal convertion and then compare to the word read with '*uint64'.

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