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I have a 16 bit register (0-15). I want to store a 10 bit number in bits 5-14 as bits 0-4 are reserved. When I use the left shift operator the number multiplies by 32, therefore when I read the register I have stored a different value. Is there a way around this?
is this a hardware register or a defined by you register. If the latter you can define it as a bit field then you will be able to do what you want with the bits you want to play with and leave the reserved bits alone.
If you want to preserve the bits already there, then you need to do a read/modify/write type operation.
> If the latter you can define it as a bit fieldCode:int temp = reg; // read your register
temp = temp & 0x1F; // mask off all the bits you want to change
temp = temp | (new_value << 5); // add your new bits, shifted into the correct place
reg = temp; // assign back to your reg.
If its an internal data structure, then this is OK, but it's far from portable if its an external register/data structure.
The layout of bit fields is compiler specific.