Why does adding 'L' to one operand in a Java multiplication change the result?
Why Different Multiplication Results with and without "L"?
In Java, the multiplication operation produces different results when using the "L" suffix for one of the operands.
Adding "L" for Correct Long Value
Adding "L" ensures that the result is a long value. When multiplying integers, the result is also an integer. If the result exceeds the range of an integer (2147483647 to -2147483648), it overflows and becomes negative.
By adding "L" to 365 in the expression, you explicitly specify that it's a long value. This ensures that the result of multiplying it with the other integers is a long value and is not truncated to the integer range.
Without "L": Incorrect Integer Result
When multiplying integers without "L," the result is an integer. If the result exceeds the integer range, it "wraps around" and becomes a different value.
For example, the multiplication 1000606024365 without adding "L" gives the incorrect result 1471228928. This is because the result 31536000000 overflows the integer range and becomes -1702967296 (the 2's complement representation of 31536000000).
Binary Representation
The binary representations of the two results show the difference:
- 1000606024365L: 011101010111101100010010110000000000
- 1000606024365: 01111111111111111111111111111111
When you don't add "L," the four most significant bits are truncated, resulting in the incorrect representation 01010111101100010010110000000000, which corresponds to the incorrect result.
Other Considerations
- If the multiplication results in a value within the integer range, adding "L" won't make a difference.
- Overflows can occur during intermediate calculations. Even if the final result fits in the integer range, it's important to consider overflows when performing arithmetic operations with large values.
- Moving "L" to the start of the expression, such as 365L1000606024*30, ensures correctness for multiplications that might overflow an integer range.
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