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