Left shift operation with Ones' complement binary

Question

I'm in researching about bitwise operation and signed number representations​. I've recognized that if we do left shift on ones' complement pattern. It does not correctly multiply the original number.

For example (Ones' complement):

11100101 (-26) << 1 = 11001010 (-53)

11110100 (-11) << 2 = 11010000 (-47)

-26 left shift 1 bit give -53 (not -52), -11 left shift 2 bits give -47 (not -44). So is this a reason why people choose two's complement for precision number operation. I've searched on google but no post mention about the left shift and ones' complement. Almost mention left shift with two's complement

Answer 1

In two's complement, a negative number A is coded by the positive number 2ⁿ-|A| and it is possible to find its value by -2^n-1 × a_n-1 + ∑₀^n-2 2ⁱ × a_i It is easy to show that shifting this value left by k bits will give the code of A×2^k, provided there is no overflow (ie only zeroes or only ones are shifted out).

In ones complement, a negative number A is coded by (2's complement of A)-1. Its value is -2^n-1 × a_n-1 + ∑₀^n-2 2ⁱ × a_i-1. If we left shift it by k, the numeric value of the result is (2's complement of 2^k*A)-2^k*1 (provided there is no overflow). It is different by 2^k-1 from the expected result which would be (2's complement of 2^k*A)-1

We can verify it on your examples:

C1(-26)<<1=-53(=-52-(2¹-1))
C1(-11)<<2=-47(=-44-(2²-1))

So to multiply a negative number coded in 1's complement by 2^k, you need to left shift it by k and to add to your result 2^k-1

In general, only two's complement gives simple arithmetic operations. Other codes (excess k, one's complement, sign absolute value) always require a correction (and it is the reason why they are very rarely used).