Converting ints to hex and bit shifting

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Hi,

I did not study software at uni, I did physics, so i've never really got to grips with bits, bytes and bit shifting, and it has fortunately not cropped up very often. However, I now need to do something that I am stuck on how to start.

I have four decimal integers that describe a colour for display on a GUI. I need to convert these into a hex number in the format 0xTTRRGGBB. TT is transparency, RR is red, GG is green, BB is blue. Each of these values can be 0-FF in hex, so my decimal values are 0-255.

Could somebody help me with this please? I know that a number is not really hex or decimal, as it is ultimately stored in bits on the PC and whether or not it is diplayed to me in hex or dec is irrelevant as far as the code is concerned. So I guess I can just bit-shift the decimal values then add(?) them - is that right?

So far, I have tried this:

int redVal = 255; // in hex is 0x000000FF
redVal << 2; // I thought this would change redVal to 0x0000FF00, but in debug //mode, this line is skipped over and redVal remains as 0x000000FF - why is this?

Any help, very gratefully received.

Posted 28-Oct-12 23:43pm

Jackie Lloyd

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Mohibur Rashid 29-Oct-12 5:55am

The reason debugger skipped that part because you did not store that value any where. As OrginalGriff shows, store it somewhere.

Jackie Lloyd 29-Oct-12 5:58am

yeah, thanks, I just spotted that and was about to update my answer but you beat me to it! Thankyou :)

2 solutions

Solution 2

Quote:
int redVal = 255; // in hex is 0x000000FF
redVal << 2; // I thought this would change redVal to 0x0000FF00, but in debug //mode, this line is skipped over and redVal remains as 0x000000FF - why is this?

The result of redVal << 2 is 0x000003FC (as noted by our Original two bits are shifted, not two hex digits), but you never see that because it's a temporary (you didn't assign it).

Posted 29-Oct-12 0:07am

CPallini

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OriginalGriff · Accepted Answer · 2012-10-28T23:46:00

Solution 1

No, not quite! :laugh:
x << 2 only shifts the value by two bits - the equivalent of multiplying it by 4 ( by 2 for each bit shifted, since each bit can have two values).

If you want to change 0x00FF to 0xFF00, then you need to shift it by 8 bits:

C++

int x = 0x00FF << 8;

- the equivalent of multiplying it by 256!

Posted 28-Oct-12 23:46pm

OriginalGriff

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Jackie Lloyd 29-Oct-12 6:00am

ahhh - I see now, thankyou so much, it seems so obvious now but I'd never had solved it by myself.

OriginalGriff 29-Oct-12 6:04am

You're welcome!