Multiple choice technology programming languages

In Java if an arithmetic operation is performed then , In case the operands are of different types the resulting type is always the widest of two types

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The statement is false. Java's arithmetic type promotion rules are more nuanced than simply taking the 'widest' type. For binary operations, Java uses 'widening primitive conversion' which promotes smaller types to int or long specifically, not just to the widest operand. For example, byte + byte produces int, not byte. char + byte produces int. Only if both operands are at least int does the result type become the wider of the two. The rule is: if either operand is double, result is double; else if float, result is float; else if long, result is long; otherwise, result is int.

AI explanation

Java's binary numeric promotion rules are more nuanced than simply 'the wider of the two operand types.' For example, byte + byte does not yield a byte or promote based on the operands' own width — both are promoted to int regardless, since byte/short/char always widen to at least int in arithmetic. So the claim that the result is 'always' the widest of the two original types is false; the actual rule involves specific promotion steps (to int as a floor, then further widening if one operand is long/float/double).

Keep practicing — related questions