((int*)tiledp)[(j * tile_w) + i] = tmp / 2;
while v2 had:
((OPJ_INT32*)tiledp)[(j * tile_w) + i] = tmp >> 1;
Divide by two and a right shift operation are only equivalent when the data
is unsigned. In this case the data is signed, so the right shift operation
is incorrectly clearing the sign bit.
Patch from: Sheet Spotter
for (j = 0; j < cblk_h; ++j) {
for (i = 0; i < cblk_w; ++i) {
OPJ_INT32 tmp = datap[(j * cblk_w) + i];
- ((OPJ_INT32*)tiledp)[(j * tile_w) + i] = tmp >> 1;
+ ((OPJ_INT32*)tiledp)[(j * tile_w) + i] = tmp / 2;
}
}
} else { /* if (tccp->qmfbid == 0) */