Parsing a Hexadecimal String to an Integer throws a NumberFormatException?

java integer parseint hex
mattbdean picture mattbdean · Jul 7, 2012 · Viewed 72k times · Source

So, In Java, you know how you can declare integers like this:

int hex = 0x00ff00;

I thought that you should be able to reverse that process. I have this code:

Integer.valueOf(primary.getFullHex());

where primary is an object of a custom Color class. It's constructor takes an Integer for opacity (0-99) and a hex String (e.g. 00ff00).

This is the getFullHex method:

public String getFullHex() {
    return ("0x" + hex);
}

When I call this method it gives my this NumberFormatException:

java.lang.NumberFormatException: For input string: "0xff0000"

I can't understand what's going on. Can someone please explain?

Answer

Tomasz Nurkiewicz picture Tomasz Nurkiewicz · Jul 7, 2012

Will this help?

Integer.parseInt("00ff00", 16)

16 means that you should interpret the string as 16-based (hexadecimal). By using 2 you can parse binary number, 8 stands for octal. 10 is default and parses decimal numbers.

In your case Integer.parseInt(primary.getFullHex(), 16) won't work due to 0x prefix prepended by getFullHex() - get rid of and you'll be fine.

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