CODEWITHSUNDEEP
javanumberformatexceptionparseint
Germán
Germán

Reputation: 23

Java: How to turn Hex into int?

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

Upvotes: 0

Views: 187

Answers (3)

David Conrad
David Conrad

Reputation: 16359

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).

Upvotes: 0

Aify
Aify

Reputation: 3537

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)

Upvotes: 0

Reimeus
Reimeus

Reputation: 159774

You could do

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

Upvotes: 3

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