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From: e.legerov at (Evgeny Legerov)
Subject: Advisory: Multiple Vulnerabilities in Monit


 > wrote:

>Multiple Vulnerabilities in Monit
>* UPDATE: Integer Overflow in POST Input Handler (Initially discovered by
>S-Quadra discovered that a large HTTP POST would cause an xmalloc() call
>within the WBA to fail.  This issue was fixed in 4.2.1 as a denial of
>service.  In fact, this code also contained an exploitable integer
>overflow.  By specifying a Content-Length header of -1, a zero-byte heap
>allocation is performed.  An attacker can then input an arbitrary amount of
>data, overwriting significant portions of the heap.  My research suggests
>that this issue could also be exploited.
It seems to me that the above statement is bit incorrect, lets follow 
the code path to see why:

$ cat monit-4.0/http/protocol.c
static int create_parameters(HttpRequest req) {

  int alloc= FALSE;
  char *query_string= NULL;

  if(IS(req->method, METHOD_POST)) {

    Socket_T S= req->S;

    int len; char *l= get_header(req, "Content-Length");

    if(l && (len= atoi(l))) {

[1]      query_string= xmalloc(sizeof(char) * (len + 1));

[2]     if(socket_read(S, query_string, len) <= 0) {

        return FALSE;


      alloc= TRUE;


  } else if(IS(req->method, METHOD_GET)) {

On line #1 we see that indeed, by specifying Content-Length header of 
-1, we can perform 'zero-byte heap allocation',
on #2 (it translates to socket_read(S, query_string, -1) ) we supposedly 
should be able to 'input an arbitrary amount of data, overwriting 
significant portions of the heap', but lets look at socket_read() function:

$ cat monit-4.0/socket.c
int socket_read(Socket_T S, void *b, int size) {

  int n= 0;
  void *p= b;
  int timeout= 0;


  timeout= S->timeout;

  while(size > 0) {
    if(S->ssl) {
      n= recv_ssl_socket(S->ssl, p, size, timeout);
    } else {
      n= sock_read(S->socket, p, size, timeout);
    if(n <= 0) break;
    p+= n;
    size-= n;
    timeout= 0;

  if(n < 0 && p==b) {
    return -1;

  return (int) p - (int) b;


We can clearly see that while loop will not be executed (as long as size 
< 0) ...

Best regards,
Evgeny Legerov
Director of simple source code review department, S-Quadra

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