Getting Overflow in constant value computation

Question

The following code:

private const uint FIRMWARE_DOWNLOAD_ADDRESS = 0x00001800;

public void someFunc(){
    byte[] command = new byte[16];


    command[11] = (byte)(FIRMWARE_DOWNLOAD_ADDRESS >> 24);
    command[10] = (byte)(FIRMWARE_DOWNLOAD_ADDRESS >> 16);
    command[9] = (byte)(FIRMWARE_DOWNLOAD_ADDRESS >> 8);
    command[8] = (byte)(FIRMWARE_DOWNLOAD_ADDRESS); //error: Overflow in constant value computation
}

throws an error Overflow in constant value computation.

Why? From what I understand 0x00001800 <= 0xffffffff so there should be no overflow happening.

And why don't the other 3 lines throw an error? I tried to do:

command[8] = (byte)(FIRMWARE_DOWNLOAD_ADDRESS >>0);

thinking that the right shift operator was somehow checking for the overflow condition but this still gives the same error.

MAV · Accepted Answer · 2016-03-07 17:17:47Z

1

You get the error because the value you are trying to cast to a byte cannot be represented by a byte.

A byte's max value is 0x000000FF (or 255). But you are trying to cast 0x00001800 (or 6144). A byte simply cannot contain that value.

The remaining works fine since, after the bit shift, the value is small enough to be represented by a byte

FIRMWARE_DOWNLOAD_ADDRESS >> 24 = 0
FIRMWARE_DOWNLOAD_ADDRESS >> 16 = 0
FIRMWARE_DOWNLOAD_ADDRESS >> 8  = 24

It seems like you are thinking about an unsigned integer's max value, which is 0xFFFFFFFF (or 4294967295)

answered Mar 7, 2016 at 17:17

MAV

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3 Comments

stackErr Over a year ago

Thanks! So how would I go about programatically getting the last 8 bits of my constant value?

stackErr Over a year ago

Doh! Use the binary AND operator with 0xFF!! Thanks again!

MAV Over a year ago

You found the answer before I could write my comment. :) You're welcome.

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