Question

I need to parse string hex value to the Integer value. Like this:

String hex = "2A"; //The answer is 42  
int intValue = Integer.parseInt(hex, 16);

But when I insert an incorrect hex value (for example “LL”) then I get java.lang.NumberFormatException: For input string: "LL" How I can avoid it (for example return 0)?

Answer 1

Just catch the exception and set the default value. However, you need to declare the variable outside the try block.

int intValue;
try {
    intValue = Integer.parseInt(hex, 16);
} catch (NumberFormatException e) {
    intValue = 0;
}

If you need to set the value using an initializer expression (say, for a final variable), you’ll have to package up the logic in a method:

public int parseHex(String hex) {
    try {
        return Integer.parseInt(hex, 16);
    } catch (NumberFormatException e) {
        return 0;
    }
}

// elsewhere...
final int intValue = parseHex(hex);