Left Shift Operator, Java : How does 0 / 2 doesn't throws error

Question

I was looking into this video, for a really famous question for bit manipulation

He wrote this statement

count -= (1 << ((length - 1) / 2));

I am not able to understand. Consider if I pass length as 1, why does this statement doesn't throws ArithmeticException for 0 / 2.

I am missing out the core fundamental over here. Please help me out here.

Here is the full code for reference:

public int solve(int A) {
        if(A == 0)
            return 0;
        
        int result = 1;    
        int length = 1;
        int count = 1;
        
        while(count < A) {
            length++;
            count += (1 << ((length - 1) / 2));
            
        }
        
        count -= (1 << ((length - 1) / 2));
        
        int offset = A - count - 1;
        
        result |= (1 << (length - 1));
        result |= offset << (length / 2);
        
        int halfShiftedNumber = (result >> (length / 2));
        int reversedNumber = getReversedNumber(halfShiftedNumber);
        
        result |= reversedNumber;
        
        return result;
    }
    
    int getReversedNumber(int A) {
        int result = 0;
        
        while(A > 0) {
            int lsb = (A & 1);
            
            result |= lsb;
            
            result = result << 1;
            A = A >> 1;
        }
        result = result >> 1;
        return result;
    }

What if I pass 1 as an input to the function.. it should throw ArithmeticException. But its not..

Can anyone help me, explaining this basic stuff.

Thanks!

Answer 1

0 / NaturalNumber is a valid mathematical operation that will result a 0. So programming languages support the operation. (this applies to negative and floating point numbers too)

In the related context, 0 as denominator is undefined.

So, for AnyNumber / 0 operation, programming languages can throw exception.

Issues to consider

Please be aware that when length is zero or negative (so that length-1 is negative), then it will result in unexpected result.

1. In the context of integer, somenumber << -value is similar to somenumber << ((32 - value) % 32)
2. In the context of long, somenumber << -value is similar to somenumber << ((64 - value) % 64)