[Exolent[jNr]]

Originally Posted by ehha Both signed and unsigned should hold the same amount of numbers, in you example, signed holds 2x.

Wow. You are so wrong with your math, but the logic is right.

For n-bit signed integers, it is -(2^(n-1)) to (2^(n-1)-1). For n-bit unsigned integers, it is 0 to (2^n-1).

For example, if n = 8 for 8bit integers.

signed = -(2^(8-1)) to (2^(8-1)-1) = -(2^7) to (2^7-1) = -128 to 127
unsigned = 0 to (2^8-1) = 0 to 255

127 - (-128) = 127 + 128 = 255
255 - 0 = 255