How to shift bits in a color (hex value)

I have the following hex value store 开发者_如何学Goin a variable:

0x04a8f5

I want to convert the value to:

0xff04a8f5

How can I accomplish this? I've tried to do this by the following operation:

int result = 0x04a8f5 >> 8;

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Use the following example as a guideline.

val = 0x04a8f5; //Your value
val |= 0xFF000000; //OR 0xFF000000 with your value, and assign the new value to val

Note, this isn't bit shifting because if your original value is a 32 bit (or larger) integer, then there is already a higher order byte available that can store the FF value. In other words, your original variable is actually 0x0004a8f5. Using an |= assignment will OR FF with the byte that you are wanting to change. No shifting necessary.

Also, shifting 0x0004a8f5 by 8 bits would result in 0x000004a8.

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Because you want to prepend FF (1111 1111) to the front of your number, this isn't really a bit shift at all. You are just adding a constant to your color value.

As long as your color value is never going to take more than 6 hex digits to represent, you can just do:

color |= 0xFF000000