Let's get to the core of adding two integers without using the ' ' operator. This requires binary manipulation.
We'll approach this like regular addition, but using binary.
The XOR (^) bitwise operator handles this initial sum perfectly:
This aligns with our needs: 1 1 → 0 (with a carry), 0 1 or 1 0 → 1, and 0 0 → 0.
Now, let's address the carries. The AND (&) operator helps us find them:
To shift the carry to the left, we'll use a left bit shift.
sum = a ^ b
(XOR for sum without carry)carry = (a & b)
(AND for carry)carry == 0
:a = sum
b = carry << 1
(left bit shift for carry)sum = 0101 ^ 0011 = 0110
carry = 0101 & 0011 = 0001
sum = 0110 ^ 0010 = 0100
carry = 0110 & 0010 = 0010
sum = 0100 ^ 00100 = 0000
carry = 0100 & 0100 = 0100
sum = 0000 ^ 1000 = 1000
carry = 0000 & 1000 = 0000
The carry is 0, so the final sum is 1000 (8).
Python's unbounded integers cause issues with negative numbers. The left bit shift can lead to infinite growth. To fix this, we need to simulate fixed-size integers (e.g., 32-bit).
We'll use a 32-bit mask (0xFFFFFFFF) to limit the number of bits:
This ensures that only the last 32 bits are considered, preventing infinite growth. We also handle potential negative results by converting them to their 32-bit two's complement representation if necessary.
This approach effectively simulates 32-bit integer arithmetic within Python, resolving the issue with unbounded integers and negative numbers. The if a > MAX_INT
condition ensures that the result remains within the 32-bit signed integer range. The example with -12 and -8 demonstrates how this correction works to produce the expected result of -20.
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