On Tue, Jun 07, 2011 at 12:17:45PM -0600, Weddington, Eric wrote:
>
> > uint32_t t1;
> > int main(){
> > t1=0xf << (12); // t1 gets fffff000 instead of 0000f000
> > }
>
> uint32_t t1;
> int main(){
> t1=0xfUL << (12); // t1 gets fffff000 instead of 0000f000
> }
Celio, 0x0f is getting shifted left 12 bits into an int16_t (the AVR
default variable type) before being promoted to an uint32_t which is
when the sign bit gets stretched out.
Eric's version expands 0x0f into uint32_t first, before the shift.
--
David Kelly N4HHE, dke...@hiwaay.net
========================================================================
Whom computers would destroy, they must first drive mad.
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