[Solved] Why does Integer.parseInt("08", 8) throw a NumberFormatException in Java? — Ans: Because "08" is interpreted as octal and 8 is invalid digit in octal | Java

Java - Command Line Arguments

Why does Integer.parseInt("08", 8) throw a NumberFormatException in Java?

ABecause leading zeros are not allowed in parseInt

BBecause parseInt cannot parse strings with more than one digit

CBecause parseInt only accepts positive numbers without zeros

DBecause "08" is interpreted as octal and 8 is invalid digit in octal

Step-by-Step Solution

Solution:

Step 1: Understand parseInt with radix and leading zeros
When using parseInt with radix 8 (octal), digits must be 0-7; "08" is invalid octal.
Step 2: Check default parseInt behavior
parseInt("08") with default radix 10 works fine; error occurs if radix 8 is specified.
Final Answer:
Because "08" is interpreted as octal and 8 is invalid digit in octal -> Option D
Quick Check:
Leading zero triggers octal radix, invalid digit causes exception [OK]

Quick Trick: Leading zero may cause octal parsing errors [OK]

Common Mistakes:

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Why does Integer.parseInt("08", 8) throw a NumberFormatException in Java?