UNIX for Advanced & Expert Users

I have two files,like follows:

$ls -l foo bar
-rw------- bar
-rw-rw-rw- foo

Then I execute follow code:

Code:

chmod("foo",(statbuf.st_mode & ~S_IXGRP) | S_ISGID)
chmod("bar",S_IRUSR | S_IWUSR | S_IRGRP | S_IROTH)

Then I view the result

$ls -l foo bar
-rw-r--r-- bar
-rw-rwlrw- foo

How to get above result? Could you explain the code in detail?

Code:

chmod("foo",(statbuf.st_mode & ~S_IXGRP) | S_ISGID)
chmod("bar",S_IRUSR | S_IWUSR | S_IRGRP | S_IROTH)

By the way,what's the meaning of 'l' in result?

Thanks

do you understand bit manipulation in C ? It won't make sense unless you understand that. Basically the chmod "foo" code is taking the existing file's status bits, zero-ing out the execute bit while setting the set-gid bit. Normally, if you tell a file that it is set-gid, you see something like this:

Code:

-rwxrwsr-x  .... foo

However, if the execute bit is not set, then setgid doesn't make sense. To the best of my knowledge, this behavior is undefined. The "l" might be reflecting this undefined status.