Nicolas Pitre <[email protected]> writes:
>> The basic structure of the code (without the "SQUASH" we discussed)
>> looks like this:
>>
>> strbuf_addf(&outbuf, "%s", PREFIX);
>> while (retval == 0) {
>> len = packet_read(in_stream, NULL, NULL, buf, LARGE_PACKET_MAX,
>> 0);
>> ...
>> band = buf[0] & 0xff;
>> switch (band) {
>> case 3:
>> ... /* emergency exit */
>> case 2:
>> while ((brk = strpbrk(b, "\n\r"))) {
>> int linelen = brk - b;
>>
>> if (linelen > 0) {
>> strbuf_addf(&outbuf, "%.*s%s%c",
>> linelen, b, suffix, *brk);
>> } else {
>> strbuf_addf(&outbuf, "%c", *brk);
>> }
>> fprintf(stderr, "%.*s", (int)outbuf.len,
>> outbuf.buf);
>> strbuf_reset(&outbuf);
>> strbuf_addf(&outbuf, "%s", PREFIX);
>> b = brk + 1;
>> }
>> if (*b)
>> strbuf_addf(&outbuf, "%s", b);
>> break;
>> ...
>> }
>> }
>>
>> if (outbuf.len > 0)
>> fprintf(stderr, "%.*s", (int)outbuf.len, outbuf.buf);
>> ...
> That won't work. If at this point there is the beginning of a partial
> line queued in the buffer from the previous round waiting to see its
> line break, you just deleted the beginning of that line.
Ahh, OK, a single logical line split into two and sent in two
packets--the first one would not result in any output (just does "if
(*b) strbuf_addf(...)" to buffer it) and then the second one finally
finds a LF. Yeah, that won't work if we cleared.
But then my observation still holds, no?
while () {
read();
switch () {
case 2:
while ((brk = strpbfk())) {
/* we have LF! we'll say something! */
strbuf_splice(&outbuf, 0, 0,
PREFIX, strlen(PREFIX));
strbuf_addf(&outbuf, ...);
fwrite(...);
b = brk + 1;
}
if (*b)
strbuf_addstr(&outbuf, b);
break;
}
}
if (outbuf.len) {
/* we still have something to say */
strbuf_splice(&outbuf, 0, 0, PREFIX,
strlen(PREFIX));
fwrite(...);
}
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