3 ms·
internally the hardware that's doing this for you has to carry that extra precision (in a carry bit usually) in order to do this - it may not be architecturally
by Taniwha 8y ago
internally the hardware that's doing this for you has to carry that extra precision (in a carry bit usually) in order to do this - it may not be architecturally visible but it has to be there
- Const-me 8y ago> it has to be there No it doesn't. For example, for single bit of precision, saturated addition is bitwise or, saturated subtraction is bitwise andnot. Obviously, the electronic doing these bitwise operations do not use any extra bits. Similarly, for higher count of bits, the electronic doing the math may or may not have that extra precision.