jas_stream_t *jas_stream_memopen2(char *buf, size_t bufsize)
{
	jas_stream_t *stream;
	jas_stream_memobj_t *obj;

	JAS_DBGLOG(100, ("jas_stream_memopen2(%p, %zu)\n", buf, bufsize));

	assert((buf && bufsize > 0) || (!buf));

	if (!(stream = jas_stream_create())) {
		return 0;
	}

	/* A stream associated with a memory buffer is always opened
	for both reading and writing in binary mode. */
	stream->openmode_ = JAS_STREAM_READ | JAS_STREAM_WRITE | JAS_STREAM_BINARY;

	/* Since the stream data is already resident in memory, buffering
	is not necessary. */
	/* But... It still may be faster to use buffering anyways. */
	jas_stream_initbuf(stream, JAS_STREAM_FULLBUF, 0, 0);

	/* Select the operations for a memory stream. */
	stream->ops_ = &jas_stream_memops;

	/* Allocate memory for the underlying memory stream object. */
	if (!(obj = jas_malloc(sizeof(jas_stream_memobj_t)))) {
		jas_stream_destroy(stream);
		return 0;
	}
	stream->obj_ = (void *) obj;

	/* Initialize a few important members of the memory stream object. */
	obj->myalloc_ = 0;
	obj->buf_ = 0;

	/* If the buffer size specified is nonpositive, then the buffer
	is allocated internally and automatically grown as needed. */
	if (!bufsize) {
		obj->bufsize_ = 1024;
		obj->growable_ = 1;
	} else {
		obj->bufsize_ = bufsize;
		obj->growable_ = 0;
	}
	if (buf) {
		obj->buf_ = JAS_CAST(unsigned char *, buf);
	} else {
		obj->buf_ = jas_malloc(obj->bufsize_);
		obj->myalloc_ = 1;
	}
	if (!obj->buf_) {
		jas_stream_close(stream);
		return 0;
	}
	JAS_DBGLOG(100, ("jas_stream_memopen2 buffer buf=%p myalloc=%d\n",
	  obj->buf_, obj->myalloc_));

	if (bufsize > 0 && buf) {
		/* If a buffer was supplied by the caller and its length is positive,
		  make the associated buffer data appear in the stream initially. */
		obj->len_ = bufsize;
	} else {
		/* The stream is initially empty. */
		obj->len_ = 0;
	}
	obj->pos_ = 0;
	
	return stream;
}
The software performs a calculation that can produce an integer overflow or wraparound, when the logic assumes that the resulting value will always be larger than the original value. This can introduce other weaknesses when the calculation is used for resource management or execution control.