generator: skip fsck if fsck command is missing

This is useful for systems which don't have any fsck.

We already skip emitting the fsck dependency when the fsck.$fstype helper
is missing, but fstab-generator doesn't necessarily know the fstype when
handling the root= parameter.

Previously, systemd-fsck was started for these mounts and then exited
immediately because it couldn't find the fsck.$fstype helper.

src/basic/path-util.c

@@ -782,14 +782,23 @@ static int executable_is_good(const char *executable) {
                                "/dev/null");
 }
 
-int fsck_exists(const char *fstype) {
+int fsck_exists(void) {
+        return executable_is_good("fsck");
+}
+
+int fsck_exists_for_fstype(const char *fstype) {
         const char *checker;
+        int r;
 
         assert(fstype);
 
         if (streq(fstype, "auto"))
                 return -EINVAL;
 
+        r = fsck_exists();
+        if (r <= 0)
+                return r;
+
         checker = strjoina("fsck.", fstype);
         return executable_is_good(checker);
 }

src/basic/path-util.h

@@ -103,7 +103,8 @@ static inline int find_executable(const char *name, char **ret_filename) {
 
 bool paths_check_timestamp(const char* const* paths, usec_t *paths_ts_usec, bool update);
 
-int fsck_exists(const char *fstype);
+int fsck_exists(void);
+int fsck_exists_for_fstype(const char *fstype);
 
 /* Iterates through the path prefixes of the specified path, going up
  * the tree, to root. Also returns "" (and not "/"!) for the root