Why does adding an 'L' to a numeric value in Java drastically change the result of a multiplication operation?

Understanding the Numerical Value Discrepancy

In the code snippet provided, the apparent discrepancy between the two multiplication operations stems from the absence or presence of an "L" character appended to the numerical value:

long oneYearWithL = 1000*60*60*24*365L;
long oneYearWithoutL = 1000*60*60*24*365;

Appending "L" to Ensure Long Value

In the first expression, the "L" suffix denotes a long integer literal, indicating that the value should be treated as a 64-bit integer. The result of this multiplication is a long value of 31536000000 milliseconds per year.

Absence of "L" Results in Integer Value

In the second expression, without the "L" suffix, the multiplication operation produces an integer value. In this case, the result is 1471228928 when the expected value should be over 3 billion milliseconds.

Integer Overflow and Truncation

The discrepancy occurs because the result of multiplying four 32-bit integers (1000, 60, 60, and 24) by a fifth 32-bit integer (365) exceeds the range of an integer, which is [-2^31, 2^31-1]. The result wraps around, producing the incorrect value of 1471228928 in the second expression.

Conversion to 2's Complement Representation

In the absence of the "L" suffix, if the result of multiplying the integers is negative and exceeds the integer range, it is converted to its 2's complement representation before being assigned to the integer variable. This conversion ensures that the negative value can be represented within the integer range.

Ensuring Correct Results

To obtain the correct result, it is essential to declare the numerical value that exceeds the integer range as a long value by appending "L" to ensure proper representation and avoid truncation or conversion to 2's complement representation.

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