Java: How to turn Hex into int?

Question

I'm working with BufferedImage (in PNG) and want to replace a colour with another. I have all colours stored as strings for easy handling but...

for(int x=0;x<output.getWidth();x++)
    for(int y=0;y<output.getHeight();y++)
        if(output.getRGB(x,y)==Integer.parseInt("ffff00fe",16))
            output.setRGB(x,y,Integer.parseInt("ffaaaaaa",16));

the resultant integers should be negative numbers, but it throws NumberFormatException

when I do output.getRGB(x,y) it returns negative numbers on non-transparent pixels

possible duplicate of [Convert hex string to int](http://stackoverflow.com/questions/11194513/convert-hex-string-to-int) — karvoynistas, Feb 04 '15 at 23:01
Are these strings coming from an external source? Because if they're not, just use the hex literals `0xffff00fe` and `0xffaaaaaa` — Ian McLaird, Feb 04 '15 at 23:05
Even if not, unless the strings are changing each time around the loop it would be better to hoist the conversion out of the loop. (The JIT *may* be smart enough to do that, but I wouldn't count on that.) If you want to see the positive (unsigned) value for those pixels, you can also use `Integer.toUnsignedString(output.getRGB(x, y))` or `Integer.toUnsignedString(output.getRGB(x, y), 16)`. — David Conrad, Feb 04 '15 at 23:14
`0xffff00fe` _is_ a negative number when interpreted as a 32-bit int. What is the problem? — Jim Garrison, Feb 05 '15 at 03:17

score 4 · Answer 1 · answered Feb 04 '15 at 22:59

4

You could do

int number = (int)Long.parseLong("ffff00fe", 16);

answered Feb 04 '15 at 22:59

Reimeus

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this is awkward, last time I tried it didn't work, but now it did, thanks – Germán Feb 04 '15 at 23:03

score 0 · Answer 2 · answered Feb 04 '15 at 23:00

The number 2,147,483,647 (or hexadecimal 7FFFFFFF) is the maximum positive value for a 32-bit signed binary integer. What you're trying to convert is something almost double that, which means the first bit of the binary number is a 1; in a signed binary integer the first bit being a 1 means it's a negative number.

Basically, you need something bigger to parse it. Try (int) Long.parseLong("ffff00fe", 16) instead of Integer.parseInt("ffff00fe",16)

score 0 · Accepted Answer · answered Feb 04 '15 at 23:05

Values greater that 0x7fff_ffff are too large to be handled as signed ints. Java 8 has added methods for dealing with ints as if they contained unsigned values. Simply replace parseInt with parseUnsignedInt:

Integer.parseUnsignedInt("ffaaaaaa", 16)

If you need to work with Java 7 and earlier, you can parse it as a long and then cast it to int. Or, if the values are constants, you can write them as numeric constants such as 0xffaaaaaa or even 0xffaa_aaaa and avoid dealing with string conversions (the underscores in numbers are allowed since Java 7 and can make them much easier to read).

Java: How to turn Hex into int?

3 Answers3