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When the end of a stream is reached (ie mix_eof() called) other 

possibly blocked streams are processed. If during this phase the end
of another stream is reached then stop the processing because the job
will be already finished by the second stream. Otherwise we may end up
running a destroyed stream.

help from Edward Wandasiewicz <w13ntd at googlemail.com>, thanks
This commit is contained in:
Alexandre Ratchov 2010-10-22 07:46:17 +02:00
parent 6cd91baafa
commit 1c465cd239
1 changed files with 14 additions and 8 deletions
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22
aucat/aproc.c
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@ -911,7 +911,7 @@ mix_out(struct aproc *p, struct abuf *obuf)
void
mix_eof(struct aproc *p, struct abuf *ibuf)
{
struct abuf *i, *inext, *obuf = LIST_FIRST(&p->outs);
struct abuf *i, *obuf = LIST_FIRST(&p->outs);
unsigned odone;
mix_setmaster(p);
@ -927,10 +927,13 @@ mix_eof(struct aproc *p, struct abuf *ibuf)
* Find a blocked input.
*/
odone = obuf->len;
for (i = LIST_FIRST(&p->ins); i != NULL; i = inext) {
inext = LIST_NEXT(i, ient);
LIST_FOREACH(i, &p->ins, ient) {
/*
* abuf_fill() may trigger mix_eof(), do the job
* and possibly reorder the list
*/
if (!abuf_fill(i))
continue;
return;
if (MIX_ROK(i) && i->r.mix.done < obuf->w.mix.todo) {
abuf_run(i);
return;
@ -1352,7 +1355,7 @@ sub_eof(struct aproc *p, struct abuf *ibuf)
void
sub_hup(struct aproc *p, struct abuf *obuf)
{
struct abuf *i, *inext, *ibuf = LIST_FIRST(&p->ins);
struct abuf *i, *ibuf = LIST_FIRST(&p->ins);
unsigned idone;
if (!aproc_inuse(p)) {
@ -1366,10 +1369,13 @@ sub_hup(struct aproc *p, struct abuf *obuf)
* Find a blocked output.
*/
idone = ibuf->len;
for (i = LIST_FIRST(&p->outs); i != NULL; i = inext) {
inext = LIST_NEXT(i, oent);
LIST_FOREACH(i, &p->outs, oent) {
/*
* abuf_flush() may trigger sub_hup(), do the job
* and possibly reorder the list
*/
if (!abuf_flush(i))
continue;
return;
if (SUB_WOK(i) && i->w.sub.done < ibuf->used) {
abuf_run(i);
return;