public class Main { public static void main(String args[]) { blank }}Drag and drop the correct code which will cause NumberFormatException when the above code is compiled.int x = Integer.parseInt("5.6"); double pi=Double.parseDouble("3.14"); boolean status=Boolean.parseBoolean("TRUE");
Question
public class Main { public static void main(String args[]) { blank }}Drag and drop the correct code which will cause NumberFormatException when the above code is compiled.int x = Integer.parseInt("5.6"); double pi=Double.parseDouble("3.14"); boolean status=Boolean.parseBoolean("TRUE");
Solution
The correct code that will cause NumberFormatException when the above code is compiled is:
int x = Integer.parseInt("5.6");
Here's the step by step explanation:
-
The
Integer.parseInt()method is used to parse a string argument as a signed decimal integer in Java. -
In the given code, the string "5.6" is not a valid integer as it contains a decimal point.
-
Therefore, when the
Integer.parseInt("5.6")method is called, it will throw aNumberFormatExceptionbecause "5.6" cannot be parsed to an integer.
The other two lines of code will not cause a NumberFormatException:
-
double pi=Double.parseDouble("3.14");This line of code is correct because "3.14" is a valid double and can be parsed to a double value. -
boolean status=Boolean.parseBoolean("TRUE");This line of code is also correct because "TRUE" is a valid boolean and can be parsed to a boolean value.
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