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

Are these strings coming from an external source? Because if they're not, just use the hex literals 0xffff00fe and 0xffaaaaaa — Ian McLaird
– Ian McLaird, Commented Feb 4, 2015 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
– David Conrad, Commented Feb 4, 2015 at 23:14
0xffff00fe is a negative number when interpreted as a 32-bit int. What is the problem? — Jim Garrison
– Jim Garrison, Commented Feb 5, 2015 at 3:17

Reimeus · Accepted Answer · 2015-02-04 22:59:08Z

3

You could do

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

answered Feb 4, 2015 at 22:59

Reimeus

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1 Comment

Germán Over a year ago

this is awkward, last time I tried it didn't work, but now it did, thanks

David Conrad · Accepted Answer · 2015-02-04 23:05:58Z

0

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).

answered Feb 4, 2015 at 23:05

David Conrad

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Comments

Aify · Accepted Answer · 2015-02-04 23:00:27Z

0

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)

answered Feb 4, 2015 at 23:00

Aify

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