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-rw-r--r--kernel/bpf/log.c198
1 files changed, 189 insertions, 9 deletions
diff --git a/kernel/bpf/log.c b/kernel/bpf/log.c
index 1974891fc324..92b1c8ad6601 100644
--- a/kernel/bpf/log.c
+++ b/kernel/bpf/log.c
@@ -8,6 +8,7 @@
#include <linux/types.h>
#include <linux/bpf.h>
#include <linux/bpf_verifier.h>
+#include <linux/math64.h>
bool bpf_verifier_log_attr_valid(const struct bpf_verifier_log *log)
{
@@ -32,23 +33,202 @@ void bpf_verifier_vlog(struct bpf_verifier_log *log, const char *fmt,
return;
}
- n = min(log->len_total - log->len_used - 1, n);
- log->kbuf[n] = '\0';
- if (!copy_to_user(log->ubuf + log->len_used, log->kbuf, n + 1))
- log->len_used += n;
- else
- log->ubuf = NULL;
+ if (log->level & BPF_LOG_FIXED) {
+ n = min(log->len_total - bpf_log_used(log) - 1, n);
+ log->kbuf[n] = '\0';
+ n += 1;
+
+ if (copy_to_user(log->ubuf + log->end_pos, log->kbuf, n))
+ goto fail;
+
+ log->end_pos += n - 1; /* don't count terminating '\0' */
+ } else {
+ u64 new_end, new_start, cur_pos;
+ u32 buf_start, buf_end, new_n;
+
+ n += 1;
+
+ new_end = log->end_pos + n;
+ if (new_end - log->start_pos >= log->len_total)
+ new_start = new_end - log->len_total;
+ else
+ new_start = log->start_pos;
+ new_n = min(n, log->len_total);
+ cur_pos = new_end - new_n;
+
+ div_u64_rem(cur_pos, log->len_total, &buf_start);
+ div_u64_rem(new_end, log->len_total, &buf_end);
+ /* new_end and buf_end are exclusive indices, so if buf_end is
+ * exactly zero, then it actually points right to the end of
+ * ubuf and there is no wrap around
+ */
+ if (buf_end == 0)
+ buf_end = log->len_total;
+
+ /* if buf_start > buf_end, we wrapped around;
+ * if buf_start == buf_end, then we fill ubuf completely; we
+ * can't have buf_start == buf_end to mean that there is
+ * nothing to write, because we always write at least
+ * something, even if terminal '\0'
+ */
+ if (buf_start < buf_end) {
+ /* message fits within contiguous chunk of ubuf */
+ if (copy_to_user(log->ubuf + buf_start,
+ log->kbuf + n - new_n,
+ buf_end - buf_start))
+ goto fail;
+ } else {
+ /* message wraps around the end of ubuf, copy in two chunks */
+ if (copy_to_user(log->ubuf + buf_start,
+ log->kbuf + n - new_n,
+ log->len_total - buf_start))
+ goto fail;
+ if (copy_to_user(log->ubuf,
+ log->kbuf + n - buf_end,
+ buf_end))
+ goto fail;
+ }
+
+ log->start_pos = new_start;
+ log->end_pos = new_end - 1; /* don't count terminating '\0' */
+ }
+
+ return;
+fail:
+ log->ubuf = NULL;
}
-void bpf_vlog_reset(struct bpf_verifier_log *log, u32 new_pos)
+void bpf_vlog_reset(struct bpf_verifier_log *log, u64 new_pos)
{
char zero = 0;
+ u32 pos;
+
+ if (WARN_ON_ONCE(new_pos > log->end_pos))
+ return;
if (!bpf_verifier_log_needed(log))
return;
- log->len_used = new_pos;
- if (put_user(zero, log->ubuf + new_pos))
+ /* if position to which we reset is beyond current log window,
+ * then we didn't preserve any useful content and should adjust
+ * start_pos to end up with an empty log (start_pos == end_pos)
+ */
+ log->end_pos = new_pos;
+ if (log->end_pos < log->start_pos)
+ log->start_pos = log->end_pos;
+ div_u64_rem(new_pos, log->len_total, &pos);
+ if (put_user(zero, log->ubuf + pos))
+ log->ubuf = NULL;
+}
+
+static void bpf_vlog_reverse_kbuf(char *buf, int len)
+{
+ int i, j;
+
+ for (i = 0, j = len - 1; i < j; i++, j--)
+ swap(buf[i], buf[j]);
+}
+
+static int bpf_vlog_reverse_ubuf(struct bpf_verifier_log *log, int start, int end)
+{
+ /* we split log->kbuf into two equal parts for both ends of array */
+ int n = sizeof(log->kbuf) / 2, nn;
+ char *lbuf = log->kbuf, *rbuf = log->kbuf + n;
+
+ /* Read ubuf's section [start, end) two chunks at a time, from left
+ * and right side; within each chunk, swap all the bytes; after that
+ * reverse the order of lbuf and rbuf and write result back to ubuf.
+ * This way we'll end up with swapped contents of specified
+ * [start, end) ubuf segment.
+ */
+ while (end - start > 1) {
+ nn = min(n, (end - start ) / 2);
+
+ if (copy_from_user(lbuf, log->ubuf + start, nn))
+ return -EFAULT;
+ if (copy_from_user(rbuf, log->ubuf + end - nn, nn))
+ return -EFAULT;
+
+ bpf_vlog_reverse_kbuf(lbuf, nn);
+ bpf_vlog_reverse_kbuf(rbuf, nn);
+
+ /* we write lbuf to the right end of ubuf, while rbuf to the
+ * left one to end up with properly reversed overall ubuf
+ */
+ if (copy_to_user(log->ubuf + start, rbuf, nn))
+ return -EFAULT;
+ if (copy_to_user(log->ubuf + end - nn, lbuf, nn))
+ return -EFAULT;
+
+ start += nn;
+ end -= nn;
+ }
+
+ return 0;
+}
+
+bool bpf_vlog_truncated(const struct bpf_verifier_log *log)
+{
+ if (log->level & BPF_LOG_FIXED)
+ return bpf_log_used(log) >= log->len_total - 1;
+ else
+ return log->start_pos > 0;
+}
+
+void bpf_vlog_finalize(struct bpf_verifier_log *log)
+{
+ u32 sublen;
+ int err;
+
+ if (!log || !log->level || !log->ubuf)
+ return;
+ if ((log->level & BPF_LOG_FIXED) || log->level == BPF_LOG_KERNEL)
+ return;
+
+ /* If we never truncated log, there is nothing to move around. */
+ if (log->start_pos == 0)
+ return;
+
+ /* Otherwise we need to rotate log contents to make it start from the
+ * buffer beginning and be a continuous zero-terminated string. Note
+ * that if log->start_pos != 0 then we definitely filled up entire log
+ * buffer with no gaps, and we just need to shift buffer contents to
+ * the left by (log->start_pos % log->len_total) bytes.
+ *
+ * Unfortunately, user buffer could be huge and we don't want to
+ * allocate temporary kernel memory of the same size just to shift
+ * contents in a straightforward fashion. Instead, we'll be clever and
+ * do in-place array rotation. This is a leetcode-style problem, which
+ * could be solved by three rotations.
+ *
+ * Let's say we have log buffer that has to be shifted left by 7 bytes
+ * (spaces and vertical bar is just for demonstrative purposes):
+ * E F G H I J K | A B C D
+ *
+ * First, we reverse entire array:
+ * D C B A | K J I H G F E
+ *
+ * Then we rotate first 4 bytes (DCBA) and separately last 7 bytes
+ * (KJIHGFE), resulting in a properly rotated array:
+ * A B C D | E F G H I J K
+ *
+ * We'll utilize log->kbuf to read user memory chunk by chunk, swap
+ * bytes, and write them back. Doing it byte-by-byte would be
+ * unnecessarily inefficient. Altogether we are going to read and
+ * write each byte twice, for total 4 memory copies between kernel and
+ * user space.
+ */
+
+ /* length of the chopped off part that will be the beginning;
+ * len(ABCD) in the example above
+ */
+ div_u64_rem(log->start_pos, log->len_total, &sublen);
+ sublen = log->len_total - sublen;
+
+ err = bpf_vlog_reverse_ubuf(log, 0, log->len_total);
+ err = err ?: bpf_vlog_reverse_ubuf(log, 0, sublen);
+ err = err ?: bpf_vlog_reverse_ubuf(log, sublen, log->len_total);
+ if (err)
log->ubuf = NULL;
}