[Patch, Fortran] Set umask(0600) before calling mkstemp

[Tobias Burnus]

Depending on the system mkstemp might create the scratch files with 0666 
permission (e.g. glibc <= 2.06);  for security reasons, it should use 
0600. Thus, one is supposed to set a umask before calling the function 
(see, e.g., the Linux man page for mkstemp).

Build and regtested on x86-64-linux.
OK for the trunk? Is this something one should back port to the branches?

Tobias
-------------- next part --------------
2012-10-10  Tobias Burnus  <burnus@net-b.de>

	PR fortran/54878
	* io/unix.c (tempfile_open): Set umask before calling mkstemp.

diff --git a/libgfortran/io/unix.c b/libgfortran/io/unix.c
index 805d4bb..713d54c 100644
--- a/libgfortran/io/unix.c
+++ b/libgfortran/io/unix.c
@@ -1051,6 +1051,9 @@ tempfile_open (const char *tempdir, char **fname)
 {
   int fd;
   const char *slash = "/";
+#if defined(HAVE_UMASK) && defined(HAVE_MKSTEMP)
+  mode_t mode_mask;
+#endif
 
   if (!tempdir)
     return -1;
@@ -1072,8 +1075,17 @@ tempfile_open (const char *tempdir, char **fname)
   snprintf (template, tempdirlen + 23, "%s%sgfortrantmpXXXXXX", 
 	    tempdir, slash);
 
+#ifdef HAVE_UMASK
+  /* Temporarily set the umask such that the file has 0600 permissions.  */
+  mode_mask = umask (S_IXUSR | S_IRWXG | S_IRWXO);
+#endif
+
   fd = mkstemp (template);
 
+#ifdef HAVE_UMASK
+  (void) umask (mode_mask);
+#endif
+
 #else /* HAVE_MKSTEMP */
   fd = -1;
   int count = 0;