linux/samples/seccomp/bpf-helper.c

90 lines
2.2 KiB
C

/*
* Seccomp BPF helper functions
*
* Copyright (c) 2012 The Chromium OS Authors <chromium-os-dev@chromium.org>
* Author: Will Drewry <wad@chromium.org>
*
* The code may be used by anyone for any purpose,
* and can serve as a starting point for developing
* applications using prctl(PR_ATTACH_SECCOMP_FILTER).
*/
#include <stdio.h>
#include <string.h>
#include "bpf-helper.h"
int bpf_resolve_jumps(struct bpf_labels *labels,
struct sock_filter *filter, size_t count)
{
struct sock_filter *begin = filter;
__u8 insn = count - 1;
if (count < 1)
return -1;
/*
* Walk it once, backwards, to build the label table and do fixups.
* Since backward jumps are disallowed by BPF, this is easy.
*/
filter += insn;
for (; filter >= begin; --insn, --filter) {
if (filter->code != (BPF_JMP+BPF_JA))
continue;
switch ((filter->jt<<8)|filter->jf) {
case (JUMP_JT<<8)|JUMP_JF:
if (labels->labels[filter->k].location == 0xffffffff) {
fprintf(stderr, "Unresolved label: '%s'\n",
labels->labels[filter->k].label);
return 1;
}
filter->k = labels->labels[filter->k].location -
(insn + 1);
filter->jt = 0;
filter->jf = 0;
continue;
case (LABEL_JT<<8)|LABEL_JF:
if (labels->labels[filter->k].location != 0xffffffff) {
fprintf(stderr, "Duplicate label use: '%s'\n",
labels->labels[filter->k].label);
return 1;
}
labels->labels[filter->k].location = insn;
filter->k = 0; /* fall through */
filter->jt = 0;
filter->jf = 0;
continue;
}
}
return 0;
}
/* Simple lookup table for labels. */
__u32 seccomp_bpf_label(struct bpf_labels *labels, const char *label)
{
struct __bpf_label *begin = labels->labels, *end;
int id;
if (labels->count == 0) {
begin->label = label;
begin->location = 0xffffffff;
labels->count++;
return 0;
}
end = begin + labels->count;
for (id = 0; begin < end; ++begin, ++id) {
if (!strcmp(label, begin->label))
return id;
}
begin->label = label;
begin->location = 0xffffffff;
labels->count++;
return id;
}
void seccomp_bpf_print(struct sock_filter *filter, size_t count)
{
struct sock_filter *end = filter + count;
for ( ; filter < end; ++filter)
printf("{ code=%u,jt=%u,jf=%u,k=%u },\n",
filter->code, filter->jt, filter->jf, filter->k);
}