htx_reserve_max_data() computed the available room up front but tested
the empty-message case before testing that room:
int32_t len = htx_free_data_space(htx);
if (htx->head == -1)
goto rsv_new_block;
if (!len)
return (struct htx_ret){.ret = 0, .blk = NULL};
So a message which is both empty and has no room at all went to
rsv_new_block with len == 0. That matters because htx_reserve_nxblk()
only rejects a reservation when "blksz > htx_free_data_space(htx)", so
a zero-sized one passes even when the free space is zero, and the
"Empty message" path then evaluates htx_get_blk(htx, 0), that is
htx->blocks + (htx->size - sizeof(struct htx_blk)). For any size below
sizeof(struct htx_blk) this subtraction underflows the uint32_t and
yields a pointer roughly 4 GB away, which is immediately written to.
The case that is trivial to reach in theory is the shared empty HTX
(htx_empty, size 0) that htxbuf() and htx_from_buf() return for an
unallocated buffer. No caller can do it today: the only caller,
h1_parse_full_contig_chunks(), returns earlier when the maximum is not
larger than sizeof(struct htx_blk), and the H1 mux allocates the
destination buffer before parsing into it. So this is not a fix for a
reachable bug, just one less trap for whoever adds the next caller.
Swapping the two tests is transparent for every other case: when the
message is empty but its buffer is allocated, the free space is
necessarily non-zero, so the early return cannot trigger and we still
reach rsv_new_block as before.
int32_t len = htx_free_data_space(htx);
uint32_t flags = 0;
- if (htx->head == -1)
- goto rsv_new_block;
-
if (!len)
return (struct htx_ret){.ret = 0, .blk = NULL};
+ if (htx->head == -1)
+ goto rsv_new_block;
+
/* get the tail and head block */
tailblk = htx_get_tail_blk(htx);
if (tailblk == NULL)