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Proper Subnormal Rounding When Interpreting 

During 25-bit rounding, subnormals are "normalized"
This would normally mean that the exponent needs to be able to be <-1023
Instead, you can modify at what bit you round and get the same results!
This is done by finding the highest bit and shifting right the round bit

Co-Authored-By: JosJuice <josjuice@gmail.com>
This commit is contained in:
Geotale 2024-05-27 17:38:35 -05:00
parent 8582644058
commit ffa680b15c
1 changed files with 27 additions and 1 deletions
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28
Source/Core/Core/PowerPC/Interpreter/Interpreter_FPUtils.h
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@ -92,7 +92,33 @@ inline double Force25Bit(double d)
{
u64 integral = std::bit_cast<u64>(d);
integral = (integral & 0xFFFFFFFFF8000000ULL) + (integral & 0x8000000);
u64 exponent = integral & Common::DOUBLE_EXP;
u64 fraction = integral & Common::DOUBLE_FRAC;
if (exponent == 0 && fraction != 0)
{
// Subnormals get "normalized" before they're rounded
// In the end, this practically just means that the rounding is
// at a different bit
s64 keep_mask = 0xFFFFFFFFF8000000LL;
u64 round = 0x8000000;
// Shift the mask and rounding bit to the right until
// the fraction is "normal"
// That is to say shifting it until the MSB of the fraction
// would escape into the exponent
u32 shift = std::countl_zero(fraction) - (63 - Common::DOUBLE_FRAC_WIDTH);
keep_mask >>= shift;
round >>= shift;
// Round using these shifted values
integral = (integral & keep_mask) + (integral & round);
}
else
{
integral = (integral & 0xFFFFFFFFF8000000ULL) + (integral & 0x8000000);
}
return std::bit_cast<double>(integral);
}