I'm reading about signed number representation and the book says
-6'd3 // 8-bit negative number stored as 2's complement of 3
Is there a reason as to why a number declared as 6-bit is stored as an 8-bit?
I don't think it is the "sign bits" taking extra bits, as it clearly says 2's complement is used. Even if it is sign bits, then only 1 bit would be required.