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Question

# Which of the following is not a valid Java integer ?25True0×2ANone of the above

A
25
B
True
C
0×2A
D
None of the above
Solution
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#### True is not a valid Java integer.However, the catch is that Integer.parseInt(str) will still fail if str represents a number that is outside range of legal int values. You can use Integer.parseInt() or Integer.valueOf() to get the integer from the string, and catch the exception if it is not a parsable int.Integer class provides a static method parseInt() which will throw NumberFormatException if the String does not contain a parsable int. We will catch this exception using catch block and thus confirm that given string is not a valid integer number.Below is the java program to demonstrate the same.To check if a String contains digit character which represent an integer, you can use Integer.parseInt() . To check if a double contains a value which can be an integer, you can use Math.floor() or Math.ceil() . You need to first check if it's a number. If so you can use the Math.Round method.int is a primitive type. Variables of type int store the actual binary value for the integer you want to represent. int.parseInt("1") doesn't make sense because int is not a class and therefore doesn't have any methods. Integer is a class, no different from any other in the Java language.

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