For this problem: 1111 1111 + 0000 0001, Cn does equal Cn-1, but if only 8 bits are available, doesn't the last carry bit that has to end up in the extra bit mean that there is an overflow? But according to your logic, it should not have an overflow. Please clarify!
No, the overflow bit alone does not detect an overflow in 2's Complement addition. There is no overflow on your operation (-1 + 1 = 0). Most CPUs have an overflow flag (1 bit), this is where the bit will end up after the operation.
@@tunez4moodz34 When you work with the overflow flag, you need to ckeck always the MSB of your binary numbers. If MSB=1 means it's a negative number; if MSB=0 means a positive number. So based on this logic, 1111 is a negative number but How is 1111= -1? that can be proved using the two's complement method. Write 1 in binary(+1=0001) then change bit0 to 0 and bits1,2&3 to 1 so it will be equal to 1110 after that you add 1 to 1110 so the result is 1111.This how you know that (-1=1111).
@@tunez4moodz34 You can retry the method with the number 8. 8 in binary equals 1000 but we are checking the overflow flag so here the MSB is 1 then 1000 is a negative number. Again to calculate -8 we need two's complement, you write +8=1000 then you change bit3 to 0 and bits2,1&0 to 1, so now you have 0111.Then add 1 to 0111 (0111+1=1000). That means -8=1000
