1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
|
// SPDX-License-Identifier: GPL-2.0
/*
* Three translation units linked with ThinLTO, two of which have a file-local
* helper of the same name.
*
* TU_C calls into both of the others, so ThinLTO imports entry_a and entry_b
* and with them the static helper each one calls. A file-local symbol which
* has to become visible is renamed helper.llvm.<hash>, and the hash is content
* derived -- so the two helpers get different hashes from each other, and
* TU_A's gets a different one again after the patch changes it. Only TU_A's
* changes: were both bodies to change, both would be cloned whichever way
* they were paired, and the pairing would not be observable.
*
* That leaves klp diff with two symbols in the original and two in the patched
* object, all four named differently, which have to be paired up correctly.
* Demangling alone gives "helper" for all of them; something else has to
* decide which is which.
*
* Only TU_A's helper changes. That is what makes a wrong pairing observable:
* paired correctly, one helper is changed and the other is not, so exactly one
* is cloned. Paired the wrong way round, both look changed -- or the wrong
* one does. If both bodies changed the outcome would be the same either way
* and the test would prove nothing.
*
* BASE differs between the two so their bodies are not identical to begin
* with.
*/
#if defined(TU_C)
extern int entry_a(int x);
extern int entry_b(int x);
int glue(int x) { return entry_a(x) + entry_b(x + 1); }
#else
#ifdef TU_B
#define ENTRY entry_b
#define BASE 5
#else
#define ENTRY entry_a
#define BASE 10
static const char __modinfo[]
__attribute__((section(".modinfo"), used, aligned(1))) = "\0name=vmlinux";
#endif
static __attribute__((noinline)) int helper(int x, int len)
{
int sum = 0, i;
for (i = 0; i < len; i++)
#if defined(PATCHED) && !defined(TU_B)
sum += i * 2 + BASE; /* only TU_A's helper changes */
#else
sum += i + BASE;
#endif
return sum + x;
}
int ENTRY(int x) { return helper(x, 4); }
#endif
|
