In V3 RTLinux we have "udelay" native.
What is happens when you use the Linux udelay? Note it is a busy waiting
loop, so if you call the Linux udelay from a RT context, Linux will
not advance.
On Sat, Jul 08, 2000 at 03:26:18PM +0800, dyd wrote:
> Hi all,
>
> I found udelay may cause dead in rtl , And Yodaiken said that udelay cannot
> be used in rtl context in a thread of Jun 18.(I paste the thread below.)
>
> But after examined the code of udelay I did not found any special operation.
> As far as I see, the code is nothing more than a busy loop, and it only use
> a piece of data in Linux ,the current_cpu_data.loops_per_sec.
> Who can ponit out where the mistake is? thanks a lot.
>
> #define udelay(n) (__builtin_constant_p(n) ? \
> __const_udelay((n) * 0x10c6ul) : \
> __udelay(n))
>
> inline void __const_udelay(unsigned long xloops)
> {
> int d0;
> __asm__("mull %0"
> :"=d" (xloops), "=&a" (d0)
> :"1" (xloops),"0" (current_cpu_data.loops_per_sec));
> __delay(xloops);
> }
>
> void __udelay(unsigned long usecs)
> {
> __const_udelay(usecs * 0x000010c6); /* 2**32 / 1000000 */
> }
>
> This is the thread I mentioned.
> On Sat, Jun 17, 2000 at 11:52:57PM +0800, Zaimin Zhong wrote:
> > Hi!
> >
> > Sorry to waste the bandwidth, but I could not link following functions
> >
> > check_region()
> > request_region()
> > udelay()
>
> These are exported kernel functions and are linked in dynamically by insmod
> but
> live in kernel space and cannot be statically linked.
> (BTW: they cannot be used in RT threads or interrupt handlers).
>
> >
> > statically into my RTL module. Which libary must be linked when above
> > functions are called.
> >
>
> dyd
>
>
>
>
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