In Bash arithmetic, what does echo $((++a)) print?

• A. The value before the increment.
• B. The value after the increment.
• C. It prints a's old value.
• D. It prints 0 if a is undefined.

Answer: (B)

Pre-increment in Bash arithmetic expands increments the variable before producing its value. So echo $((++a)) increments a by 1 and then prints that new value. If a is undefined, Bash treats it as 0 in arithmetic, so ++a makes it 1 and prints 1 (and a becomes 1). This is why the result is the incremented value, not the original one. The other options either describe post-increment behavior or rely on a different undefined-variable rule that doesn’t apply here.