On Mon, 2005-11-28 at 15:10 +0100, Jules Colding wrote:
> Hi,
>
> Consider the camel_header_unfold() function:
>
> char *
> camel_header_unfold(const char *in)
> {
> char *out = g_malloc(strlen(in)+1);
> const char *inptr = in;
> char c, *o = out;
>
> o = out;
> while ((c = *inptr++)) {
> if (c == '\n') {
> if (camel_mime_is_lwsp(*inptr)) {
> do {
> inptr++;
> } while (camel_mime_is_lwsp(*inptr));
> *o++ = ' ';
> } else {
> *o++ = c;
> }
> } else {
> *o++ = c;
> }
> }
> *o = 0;
>
> return out;
> }
>
> This function will skip any sequence of space (' '), tabs ('\t') and
> newlines ('\n') as long as it starts with '\n'. How can that be right?
>
> RFC 822, and RFC 2822 in particular, explicitly states that unfolding is
> done by removing any CRLF ('\r' followed by '\n') that is immediately
> followed by ' ' or '\t'.
>
> I really can't see how camel_header_unfold() can unfold a message header
> correctly.
>
I see what you are thinking, but by the time this code is run on any
input, the \r has already been stripped and you cannot, by definition,
have \n\n in a header (it terminates the header block so we don't have
to worry about that).
I suppose the following modification could be made to be more clear:
char *
camel_header_unfold(const char *in)
{
char *out = g_malloc(strlen(in)+1);
const char *inptr = in;
char c, *o = out;
o = out;
while ((c = *inptr++)) {
if (c == '\n') {
if (*inptr == ' ' || *inptr == '\t') {
do {
inptr++;
} while (*inptr == ' ' || *inptr == '\t');
*o++ = ' ';
} else {
*o++ = c;
}
} else {
*o++ = c;
}
}
*o = 0;
return out;
}
--
Jeffrey Stedfast
Evolution Hacker - Novell, Inc.
[EMAIL PROTECTED] - www.novell.com
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