bpf: Mark bpf prog stack with kmsan_unposion_memory in interpreter mode

In the Linux kernel, the following vulnerability has been resolved: bpf: Mark bpf prog stack with kmsan_unposion_memory in interpreter mode syzbot reported uninit memory usages during map_{lookup,delete}_elem. ========== BUG: KMSAN: uninit-value in __dev_map_lookup_elem kernel/bpf/devmap.c:441 [inline] BUG: KMSAN: uninit-value in dev_map_lookup_elem+0xf3/0x170 kernel/bpf/devmap.c:796 __dev_map_lookup_elem kernel/bpf/devmap.c:441 [inline] dev_map_lookup_elem+0xf3/0x170 kernel/bpf/devmap.c:796 ____bpf_map_lookup_elem kernel/bpf/helpers.c:42 [inline] bpf_map_lookup_elem+0x5c/0x80 kernel/bpf/helpers.c:38 ___bpf_prog_run+0x13fe/0xe0f0 kernel/bpf/core.c:1997 __bpf_prog_run256+0xb5/0xe0 kernel/bpf/core.c:2237 ========== The reproducer should be in the interpreter mode. The C reproducer is trying to run the following bpf prog: 0: (18) r0 = 0x0 2: (18) r1 = map[id:49] 4: (b7) r8 = 16777216 5: (7b) *(u64 *)(r10 -8) = r8 6: (bf) r2 = r10 7: (07) r2 += -229 ^^^^^^^^^^ 8: (b7) r3 = 8 9: (b7) r4 = 0 10: (85) call dev_map_lookup_elem#1543472 11: (95) exit It is due to the "void *key" (r2) passed to the helper. bpf allows uninit stack memory access for bpf prog with the right privileges. This patch uses kmsan_unpoison_memory() to mark the stack as initialized. This should address different syzbot reports on the uninit "void *key" argument during map_{lookup,delete}_elem.

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Linux kernel 安全漏洞

Linux kernel是美国Linux基金会的开源操作系统Linux所使用的内核。 Linux kernel存在安全漏洞，该漏洞源于在解释器模式下，bpf程序堆栈使用未初始化的内存。

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