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Render thread: never skip two presents in a row

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R4 races present new content every other vblank, so at a saturated 4K
widescreen frame two vblanks behind was common and about half the
presents were dropped. Skipping is a catch-up step, not a frame-rate
policy: one skipped present is always followed by a shown one.

Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com>
This commit is contained in:
Shokunin
2026-10-06 00:48:50 -07:00
parent 479cbb789c
commit 4dc26a2bb2
1 changed files with 7 additions and 2 deletions
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runtime/src/gpu_gl_renderer.c
+7 -2
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@@ -9283,15 +9283,20 @@ static int rth_record_present(uint16_t op, int n, const int32_t *v) {
static void rth_replay_present(const RtCmd *c, const uint8_t *p) {
RthOvHdr hd;
memcpy(&hd, p, sizeof hd);
if (rt_frames_ahead() >= 2) {
static int skipped_last = 0;
if (rt_frames_ahead() >= 2 && !skipped_last) {
/* Two later frames are already recorded, so the emulation thread is
* at (or near) the in-flight bound waiting on us. Showing this frame
* would only delay those; skip its present (its drawing has been
* replayed and stays). One frame behind is the normal pipelined
* state and still presents every frame. */
* state and still presents every frame, and never two presents in a
* row are skipped, so a saturated render thread keeps showing at
* least every other frame. */
s_rth_presents_stale++;
skipped_last = 1;
return;
}
skipped_last = 0;
const uint8_t *q = p + sizeof hd;
memset(&s_rth_ov, 0, sizeof s_rth_ov);
s_rth_ov.valid = 1;