V579. The 'Foo' function receives the pointer and its size as arguments. It is possibly a mistake. Inspect the N argument.


The analyzer detected an odd function call in code. A pointer and the size of the pointer are passed into a function as its arguments. Actually it is a common case when developers want to pass a buffer size instead of a pointer size into a function.

Let's see how an error like that can appear in code. Assume we had the following code in the beginning:

char buf[100];
...
memset(buf, 0, sizeof(buf));

The code is correct. The memset() function clears an array of 100 bytes. Then the code was changed and the buffer became variable-sized. The programmer forgot to change the code of buffer clearing:

char *buf = new char[N];
...
memset(buf, 0, sizeof(buf));

Now the code is incorrect. The sizeof() operator returns the pointer size instead of the size of the buffer with data. As a result, the memset() function clears only part of the array.

Let's consider another sample taken from a real application:

apr_size_t ap_regerror(int errcode,
  const ap_regex_t *preg, char *errbuf,
  apr_size_t errbuf_size)
{
  ...
  apr_snprintf(errbuf, sizeof errbuf,
    "%s%s%-6d", message, addmessage,
    (int)preg->re_erroffset);
  ...
}

It is not easy to notice the error in this code. The apr_snprintf() function accepts the 'errbuf' pointer and the size of this pointer 'sizeof errbuf' as arguments. The analyzer considers this code odd and is absolutely right. The buffer size is stored in the 'errbuf_size' variable and it is this variable that should be used. This is the correct code:

apr_snprintf(errbuf, errbuf_size,
  "%s%s%-6d", message, addmessage,
  (int)preg->re_erroffset);


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