Re: [PATCH bpf-next v1 3/5] selftests/bpf: Test referenced kptr arguments of struct_ops programs

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On 2/14/25 8:45 AM, Amery Hung wrote:
diff --git a/tools/testing/selftests/bpf/progs/struct_ops_refcounted_fail__global_subprog.c b/tools/testing/selftests/bpf/progs/struct_ops_refcounted_fail__global_subprog.c
new file mode 100644
index 000000000000..ae074aa62852
--- /dev/null
+++ b/tools/testing/selftests/bpf/progs/struct_ops_refcounted_fail__global_subprog.c
@@ -0,0 +1,39 @@
+#include <vmlinux.h>
+#include <bpf/bpf_tracing.h>
+#include "../test_kmods/bpf_testmod.h"
+#include "bpf_misc.h"
+
+char _license[] SEC("license") = "GPL";
+
+extern void bpf_task_release(struct task_struct *p) __ksym;
+
+__noinline int subprog_release(__u64 *ctx __arg_ctx)
+{
+	struct task_struct *task = (struct task_struct *)ctx[1];
+	int dummy = (int)ctx[0];
+
+	bpf_task_release(task);
+
+	return dummy + 1;
+}
+
+/* Test that the verifier rejects a program that contains a global
+ * subprogram with referenced kptr arguments
+ */
+SEC("struct_ops/test_refcounted")
+__failure __log_level(2)
+__msg("Validating subprog_release() func#1...")
+__msg("invalid bpf_context access off=8. Reference may already be released")
+int refcounted_fail__global_subprog(unsigned long long *ctx)
+{
+	struct task_struct *task = (struct task_struct *)ctx[1];
+
+	bpf_task_release(task);
+
+	return subprog_release(ctx);

One question, swap the subprog_release and bpf_task_release order will still be the same failure, right? Meaning:

	subprog_release(ctx);

	bpf_task_release(task);

	return 0;

which is fine based on the changes in the do_check_common() in patch 2. Just want to confirm my understanding.

Other than that,

Acked-by: Martin KaFai Lau <martin.lau@xxxxxxxxxx>




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