selftests/ftrace: Fix unique symbol check in kprobe_non_uniq_symbol.tcgrafted

The current regex also matches symbols in modules, which makes the
test fail on s390 where name_show is present only once in the kernel,
but also multiple times in modules:

000001b1401cdc20 t name_show
000001b0c05e6c40 t name_show    [mdev]
000001b0c0495f30 t name_show    [i2c_core]

Fix this by changing the regular expression to only match the function
name.

Link: https://lore.kernel.org/all/20260909092954.2200558-1-svens@linux.ibm.com/

Fixes: 03b80ff8023a ("selftests/ftrace: Add new test case which checks non unique symbol")
Signed-off-by: Sven Schnelle <svens@linux.ibm.com>
Reviewed-by: Steven Rostedt <rostedt@goodmis.org>
Signed-off-by: Masami Hiramatsu (Google) <mhiramat@kernel.org>

1 files changed, 49 insertions, 0 deletions

diff --git a/lib/union_find.c b/lib/union_find.c
new file mode 100644
index 000000000..413b0f8ad
--- /dev/null
+++ b/lib/union_find.c
@@ -0,0 +1,49 @@
+// SPDX-License-Identifier: GPL-2.0
+#include <linux/union_find.h>
+
+/**
+ * uf_find - Find the root of a node and perform path compression
+ * @node: the node to find the root of
+ *
+ * This function returns the root of the node by following the parent
+ * pointers. It also performs path compression, making the tree shallower.
+ *
+ * Returns the root node of the set containing node.
+ */
+struct uf_node *uf_find(struct uf_node *node)
+{
+	struct uf_node *parent;
+
+	while (node->parent != node) {
+		parent = node->parent;
+		node->parent = parent->parent;
+		node = parent;
+	}
+	return node;
+}
+
+/**
+ * uf_union - Merge two sets, using union by rank
+ * @node1: the first node
+ * @node2: the second node
+ *
+ * This function merges the sets containing node1 and node2, by comparing
+ * the ranks to keep the tree balanced.
+ */
+void uf_union(struct uf_node *node1, struct uf_node *node2)
+{
+	struct uf_node *root1 = uf_find(node1);
+	struct uf_node *root2 = uf_find(node2);
+
+	if (root1 == root2)
+		return;
+
+	if (root1->rank < root2->rank) {
+		root1->parent = root2;
+	} else if (root1->rank > root2->rank) {
+		root2->parent = root1;
+	} else {
+		root2->parent = root1;
+		root1->rank++;
+	}
+}