Using filepath input in fopen

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.

[KSSV]

How did you provide pathvariable? Check whether the file is present or not with exist.

[Stephen23]

@Suha: in 99 percent of cases the filepath is incorrect or there is a spelling mistake somewhere. Check the filename carefully.

[Suha]

Thank you KSSV and Stephen for your prompt reply ! My pathvariable was indeed the issue, which I have now fixed. :)

Answer 1

[Guillaume]

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');

[Suha]

Thank you Guillaume for taking time out to write such a clear and detailed response. Much appreciated. Using the errmsg output I was able to spot that my pathvariable was incorrect, hence the file could not be detected. All fixed and working now. Many thanks !!