CODEWITHSUNDEEP
clinuxfopenfile-permissions
stdcall
stdcall

Reputation: 28860

fopen fails when file exists

I have the following snippet of code running inside a massive Linux daemon. I'm trying to debug to a log file, but when the log file exists, the fopen fails

if ( ( debugFILE = fopen( "/home/lala/debug.log", "a" ) ) == NULL )
{
    perror("error: ");
}

The error I get is: "Permission denied".

This is a the output of ls of the specific file:

----rw---- 1 lala lala 0 Mar 11 18:26 debug.log

First, why the file was created in the firstplace with these permissions. Second, why the fopen succeeds when creating, but not when opening ?

Upvotes: 3

Views: 4915

Answers (2)

mariux
mariux

Reputation: 3117

fopen creates files with permission 0666 modified by the process's umask.

So if you do not manually change the files permissions within your program or after your program has finished.

The process will most likely have a wrong umask.

Do you set the umask within your programm or within the context of the calling process? Your umask -S output actually looks fine (looks like umask 002).

Upvotes: 4

teppic
teppic

Reputation: 8195

The "a" option will always create a file if it doesn't exist, and if successful return a valid pointer. It's created according to the umask setting of the process - and in your case the process is creating a file without proper permissions, hence the next time fopen fails. If you don't want to mess with the umask, just call this before and after the fopen:

chmod("/home/lala/debug.log", 0644);

It's fine to call chmod in this way if the file doesn't exist, it'll just do nothing (except for setting errno appropriately).

Upvotes: 4

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