On 29 Oct 2011 at 0:40, Oli Glaser wrote:

> On 28/10/2011 20:48, Yigit Turgut wrote:
> > readEepromByte(0x00) + (readEepromByte(0x01)<<8);
> >
> > readEepromByte(0x00 + (readEepromByte(0x01)<<8));
> >
> > They are equivalent aren't they ? They should be because it still
> > doesn't respond logically (:
> >
> >
>=20
> The first will add the two function return values together (one left=20
> shifted by 8), whilst the second will use the (left shifted) return=20
> value from the second (right hand) function (+ 0x00) as the argument for=
=20
> the first function.

Not to mention that shifting left by 8 places is likely to zero an 8 bit in=
teger (if that is=20
the variable size returned from your function, and if your compiler has no =
good=20
reason to think otherwise). To ensure 16 bit width in the shifting and addi=
tion you=20
probably need to use some typecasting, eg:

(unsigned int) readEepromByte(0x01) << 8;

--=20
Brent Brown, Electronic Design Solutions
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eMail:  brent.brown@clear.net.nz



--=20
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