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open a hex file

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Ronaldo
Ronaldo 2013-10-16
评论： Ronaldo 2013-10-17
I use the following code to open a hex file (please look at the attachment).
fid = fopen('FileName');
A = fread(fid);
My problem is instead of getting A as a cell containg n*1 (n is the number of rows in the hex file) I get one row of numbers. I would appreciate it if you help me get a result like below:
00 00 00 00 49 40 40 0D
03 00 30 43 C6 02 00 00
A3 6B 74 23 90 47 E4 40
and so on
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Azzi Abdelmalek
Azzi Abdelmalek 2013-10-16
Can you post the result you've got? maybe with simple use of reshape function you can get the expected result

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Jan
Jan 2013-10-16
编辑：Jan 2013-10-16
There is not format you can call "hex file". You can interpret a file as stream of bytes and display them by hex numbers. But from this point of view, any file is a hex file.
Please try this:
fid = fopen('FileName');
A = fread(fid, Inf, 'uint8');
fclose(fid);
Fmt = repmat('%02X ', 1, 16); % Create format string
Fmt(end) = '*'; % Trailing star
S = sprintf(Fmt, A); % One long string
C = regexp(S, '*', 'split'); % Split at stars
Perhaps a string with line breaks is sufficient:
Fmt = repmat('%02X ', 1, 16); % Create format string
Fmt(end:end+1) = '\n';
S = sprintf(Fmt, A);
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Jan
Jan 2013-10-16
@Ronaldo: Please specify "large". Some people in the forum use this for 1000 elements, some for 1e15 elements.
When importing the file exhausts too much memory, install more memory, use a 64 bit version of Matlab and OS, free arrays, which are not used anymore. You can import the data by fread(fid, Inf, '*uint8') also, which occupies just an 8.th of the memory. But for SPRINTF a modern Matlab version is required then, because older ones did not accept UINT8 as input.
Ronaldo
Ronaldo 2013-10-17
The file that I want to open is 5 GB. I am using a 64 bit version of MATLAB and OS. Available physical memory of my computer is 31.2 GB. I was wondering whether there is anyway that I can prevent OUT of MEMORY problem if I want to open the 5 GB file.

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Azzi Abdelmalek
Azzi Abdelmalek 2013-10-16
编辑：Azzi Abdelmalek 2013-10-16
fid = fopen('file.txt');
A = textscan(fid,'%s %s %s %s %s %s %s %s')
fclose(fid)
A=[A{:}]
arrayfun(@(x) strjoin(A(1,:)),(1:3)','un',0)
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Azzi Abdelmalek
Azzi Abdelmalek 2013-10-16
编辑：Azzi Abdelmalek 2013-10-16
Decimal numbers? please edit your question and make it as clear as possible
Jan
Jan 2013-10-16
@Ronaldo: A memory leak? I assume, you mean something completely different.

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