common: check is_pseudofs or has_fsck more reliably

There might be filesystems we can't cover in common,
e.g. out-of-tree ones. Use findmnt / check if the
fsck command exists to make it more reliable.

common

@@ -436,11 +436,12 @@ dm_name_for_devnode() {
 }
 
 fstype_is_pseudofs() {
-  (( pseudofs_types["$1"] ))
+  (( pseudofs_types["$1"] )) || findmnt --pseudo "$1" &>/dev/null
 }
 
 fstype_has_fsck() {
-  (( fsck_types["$1"] ))
+  (( fsck_types["$1"] == 0 )) && return 1
+  (( fsck_types["$1"] )) || command -v "fsck.$1" &>/dev/null
 }
 
 # vim: et ts=2 sw=2 ft=sh:

genfstab.in

@@ -53,7 +53,7 @@ optstring_apply_quirks() {
   # Prune 'relatime' option for any pseudofs. This seems to be a rampant
   # default which the kernel often exports even if the underlying filesystem
   # doesn't support it. Example: https://bugs.archlinux.org/task/54554.
-  if awk -v fstype="$fstype" '$1 == fstype { exit 1 }' /proc/filesystems; then
+  if fstype_is_pseudofs "$fstype"; then
     optstring_remove_option "$varname" relatime
   fi