On Wed, Mar 12, 2014 at 09:27:02AM +0800, arthur wrote:
> Description
> in dump_dmesg_structured() the out_buf size is 4096, and if the
> length is less than 4080( 4096-16 ) it won't really write out.
> Normally, after writing one or four chars to the out_buf, it will
> check the length of out_buf. But in extreme cases, 19 chars was
> written to the out_buf before checking the length. This may cause
> the stack corruption. If the length was 4079 (won't realy write out),
> and then write 19 chars to it. the out_buf will overflow.
>
> Solution
> Change 16 to 64 thus can make sure that always have 64bytes before
> moving to next records. why using 64 is that a long long int can take
> 20 bytes. so the length of timestamp can be 44 ('[','.',']',' ') in
> extreme case.
>
> Signed-off-by: arthur <zzou at redhat.com>
> ---
Looks good to me.
Acked-by: Vivek Goyal <vgoyal at redhat.com>
Thanks
Vivek
> vmcore-dmesg/vmcore-dmesg.c | 2 +-
> 1 file changed, 1 insertion(+), 1 deletion(-)
>
> diff --git a/vmcore-dmesg/vmcore-dmesg.c b/vmcore-dmesg/vmcore-dmesg.c
> index 0345660..e15cd91 100644
> --- a/vmcore-dmesg/vmcore-dmesg.c
> +++ b/vmcore-dmesg/vmcore-dmesg.c
> @@ -674,7 +674,7 @@ static void dump_dmesg_structured(int fd)
> else
> out_buf[len++] = c;
>
> - if (len >= OUT_BUF_SIZE - 16) {
> + if (len >= OUT_BUF_SIZE - 64) {
> write_to_stdout(out_buf, len);
> len = 0;
> }
> --
> 1.8.4.2
>
>
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