Re: about udelay in mips

[loody <miloody@gmail.com>]

hi all:
thanks for your kind help :)

2011/1/17 Winfred Lu <winfred.lu@gmail.com>:
> 2011/1/16 loody <miloody@gmail.com>
>>
>> below is the formula about calculating the delay
>> (us  *   0x000010c7   *   HZ   *   lpj   )) >> 32)
>> I cannot figure out why we need to multiply 0x10c7, and what lpj mean?
>
> 0x10c7 is the rounded up value of (2 ^ 32) / 1,000,000.
>
> The 1,000,000 comes from that 1 second is equal to 1,000,000 micro seconds.
> That the value multiplied by (2 ^ 32) and shifted 32 bit right after is
> equal to
> that the value multiplied by 1.
I found my kernel will compile udelay(xx) as zero no matter what xx I filled in.
below are the dis-assemblies:
(as you can see the v0 = v1 = zero.)
My version is 2.6.30.9:
void __udelay(unsigned long us)
{
        unsigned int lpj = current_cpu_data.udelay_val;

        __delay(((unsigned long long)(us * 0x000010c7 * HZ * lpj)) >> 32);
80306ed0:       00001821        move    v1,zero
80306ed4:       00601021        move    v0,v1
#include <asm/compiler.h>
#include <asm/war.h>

inline void __delay(unsigned int loops)
{
        __asm__ __volatile__ (
80306ed8:       1440ffff        bnez    v0,80306ed8 <__udelay+0x8>
80306edc:       2442ffff        addiu   v0,v0,-1
void __udelay(unsigned long us)

I have double checked the __delay source code of 2.6.33.4
and the dis-assemblies:

void __udelay(unsigned long us)
{
        unsigned int lpj = current_cpu_data.udelay_val;

        __delay((us * 0x000010c7ull * HZ * lpj) >> 32);
802f7310:       3c02804f        lui     v0,0x804f
802f7314:       8c429360        lw      v0,-27808(v0)
802f7318:       3c050010        lui     a1,0x10
802f731c:       34a56256        ori     a1,a1,0x6256
802f7320:       00450019        multu   v0,a1
802f7324:       00002821        move    a1,zero
802f7328:       00001012        mflo    v0
802f732c:       00001810        mfhi    v1
802f7330:       00a20018        mult    a1,v0
802f7334:       70640000        madd    v1,a0
802f7338:       00003012        mflo    a2
802f733c:       00440019        multu   v0,a0
802f7340:       00001810        mfhi    v1
802f7344:       00c31021        addu    v0,a2,v1
#include <asm/compiler.h>
#include <asm/war.h>

inline void __delay(unsigned int loops)
{
        __asm__ __volatile__ (
802f7348:       1440ffff        bnez    v0,802f7348 <__udelay+0x38>
802f734c:       2442ffff        addiu   v0,v0,-1
void __udelay(unsigned long us)
{
        unsigned int lpj = current_cpu_data.udelay_val;

        __delay((us * 0x000010c7ull * HZ * lpj) >> 32);
}
802f7350:       03e00008        jr      ra

Does that mean "unsigned long long" not workable?
appreciate your help,
miloody