4 ms·
float32 (1+m)×2^e as int32 = e<<23 | m (1+m1)×2^e1 × (1+m2)×2^e2 = (1+m1+m2+m1×m2)×2^(e1+e2) If m1×m2 is small, that's approximately float32(m1+m2, e1+e2).
by akomtu 2y ago
float32 (1+m)×2^e as int32 = e<<23 | m
(1+m1)×2^e1 × (1+m2)×2^e2 = (1+m1+m2+m1×m2)×2^(e1+e2)
If m1×m2 is small, that's approximately float32(m1+m2, e1+e2).