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

1 Commento

Azzi Abdelmalek
Azzi Abdelmalek il 16 Ott 2013
Can you post the result you've got? maybe with simple use of reshape function you can get the expected result

Accedi per commentare.

 Risposta accettata

Jan
Jan il 16 Ott 2013
Modificato: Jan il 16 Ott 2013

5 voti

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);

3 Commenti

Ronaldo
Ronaldo il 16 Ott 2013
When the file is large, I get the following error for line 2:
OUT of MEMORY
Jan
Jan il 16 Ott 2013
@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 il 17 Ott 2013
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.

Accedi per commentare.

Più risposte (1)

Azzi Abdelmalek
Azzi Abdelmalek il 16 Ott 2013
Modificato: Azzi Abdelmalek il 16 Ott 2013

1 voto

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)

8 Commenti

Walter Roberson
Walter Roberson il 16 Ott 2013
Use '%x%x%x%x%x' instead of %s if the numeric form is desired.
Ronaldo
Ronaldo il 16 Ott 2013
Modificato: Ronaldo il 18 Ott 2013
The extension of the original file is not txt.
Ronaldo
Ronaldo il 16 Ott 2013
Modificato: Ronaldo il 16 Ott 2013
If I use the following code for the problem, I get the result that I want
fid = fopen('FIleName');
a = fread(fid);
for i=1:size(a,1)
sprintf('%02X %02X %02X %02X %02X %02X %02X %02X %02X %02X %02X %02X %02X %02X %02X %02X',a(8*i-7:8*i,1))
end
BUT if the file size is large then this code is not working. This is the reason why I want "a" to be (n*1 cell which each cell contains one row of the image attached to the question) instead of a complete row or column. of all the elements in the hex file.
Azzi Abdelmalek
Azzi Abdelmalek il 16 Ott 2013
Modificato: Azzi Abdelmalek il 16 Ott 2013
Post the first samples you've got with
A = fread(fid);
If the data you've got are like:
A={'00' ;'00' ;'00';'00'; '49'; '40' ;'40' ;'0D'; '03'; '00' ;'30' ;'43'; 'C6'; '02'; '00' ;'00'}
You can make some transformations
out=reshape(A,8,[])'
result=arrayfun(@(x) strjoin(out(x,:)),(1:size(out,1))','un',0)
Ronaldo
Ronaldo il 16 Ott 2013
Modificato: Ronaldo il 16 Ott 2013
It is a column of double numbers. If I use dec2hex for each element of the column, I get the equivalent value in the original hex file. As you can see in the attachment the hex file contains different rows, I like "a" also contains several rows to avoid memory leak!
Azzi Abdelmalek
Azzi Abdelmalek il 16 Ott 2013
Modificato: Azzi Abdelmalek il 16 Ott 2013
Decimal numbers? please edit your question and make it as clear as possible
Jan
Jan il 16 Ott 2013
@Ronaldo: A memory leak? I assume, you mean something completely different.

Accedi per commentare.

Categorie

Richiesto:

il 16 Ott 2013

Commentato:

il 17 Ott 2013

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by