Question

I have some sliders in my application that allows the user to change ARGB colors, however I need to convert these values to a hex value like 0xff000000, which is solid black.

This is what I have so far:

protected int toHex(Color col) {
    String as = pad(Integer.toHexString(col.getAlpha()));
    String rs = pad(Integer.toHexString(col.getRed()));
    String gs = pad(Integer.toHexString(col.getGreen()));
    String bs = pad(Integer.toHexString(col.getBlue()));
    String hex = "0x" + as + rs + gs + bs;
    return Integer.parseInt(hex, 16);
}

private static final String pad(String s) {
    return (s.length() == 1) ? "0" + s : s;
}

However upon getting the Integer value like below, I get a NumberFormatException for input string: "0xccffffff":

int color = toHex(new Color(153f, 153f, 153f, 0.80f));

Any ideas on how to get this to an Integer? Thanks.

Was it helpful?

Solution

The Color parameters must be floats between 1f and 0f. So this is a valid color:

int color = toHex(new Color(1f, 1f, 1f, 1f));

Which is white.

OTHER TIPS

The problem is that you are including alpha values. So your maximum color code is #FFFFFFFF (8 digits).

The method Integer.parseInt will let you parse value from -0x80000000 to 0x7FFFFFFF. In order to get your value 0xCC999999 from it, you would have to negate the value and input -0x33666667 - which is of course not useful at all.

The clunky but stable workaround is using Long.

(int) Long.parseLong(text, 16)
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