MATLAB: Using filepath input in fopen

Question

Hi all, I can't seem to use fopen to open a file using a variable that stores the name and path of the file. For example, the variable [pathvariable] stores /imper/codec/data.txt. I am doing fid = fopen([pathvariable],'r') but it is not working.

I've tried the following too:

fid = fopen(pathvariable,'r')
fid = fopen('pathvariable','r')

In all cases, I get a fid of -1.

Could someone please help. Thanks.

Answer 1

The proper syntax is indeed your first example:

fid = fopen(pathvariable, 'r');

There could be many reasons for why it does not work: invalid path(file not found), file locked, permission denied, etc. First thing to do would be to look at the errmsg output of:

[fid, errmsg] = fopen(pathvariable, 'r')

which should give you more info.

If you're on Windows, /imper/codec/data.txt does not look like a absolute path which may be the problem. I would always use absolute paths with fopen to avoid any issue:

fid = fopen(fullfile('C:\some\where', relativepath), 'r');