Re: How can I link the kernel with libgcc ?

[linux-os (Dick Johnson)]

On Fri, 10 Mar 2006, Carlos Munoz wrote:

> Denis Vlasenko wrote:
>
>> On Friday 10 March 2006 05:47, Carlos Munoz wrote:
>>
>>
>>> Lee Revell wrote:
>>>
>>>
>>>
>>>> On Thu, 2006-03-09 at 19:25 -0800, Carlos Munoz wrote:
>>>>
>>>>
>>>>
>>>>
>>>>> I figured out how to get the driver to use floating point operations.
>>>>> I included source code (from an open source math library) for the
>>>>> log10 function in the driver. Then I added the following lines to the
>>>>> file arch/sh/kernel/sh_ksyms.c:
>>>>>
>>>>>
>>>>>
>>>>>
>>>> Where is the source code to your driver?
>>>>
>>>> Lee
>>>>
>>>>
>>>>
>>>>
>>>>
>>> Hi Lee,
>>>
>>> Be warned. This driver is in the early stages of development. There is
>>> still a lot of work that needs to be done (interrupt, dma, etc, etc).
>>>
>>>
>>
>> What? You are using log10 only twice!
>>
>>        if (!(siu_obj_status & ST_OPEN)) {
>> 		...
>>                /* = log2(over) */
>>                ydef[22] = (u_int32_t)(log10((double)(over & 0x0000003f)) /
>>                                       log10(2));
>> 		...
>>        }
>>        else {
>> 		...
>>                if (coef) {
>>                        ydef[16] = 0x03045000 | (over << 26) | (tap - 4);
>>                        ydef[17] = (tap * 2 + 1);
>>                        /* = log2(over) */
>>                        ydef[22] = (u_int32_t)
>>                                (log10((double)(over & 0x0000003f)) / log10(2));
>>                }
>>
>> Don't you think that log10((double)(over & 0x0000003f)) / log10(2)
>> can have only 64 different values depending on the result of (over & 0x3f)?
>>
>> Obtain them from precomputed uint32_t log10table[64].
>> --
>> vda
>>
>>
> Hi Denis,
>
> Yes, the driver code so far only uses log10 twice, but there will be
> more uses for it as I populate the rest of the tables. However, I think
> its use will be some what limited. I wasn't aware that the floating
> point registers are not saved. I'll investigate a way to create a table
> with pre-calculated log10 values.
>
> Thanks,
>
>
> Carlos

Since the log in base n is the log in any base times a constant,
you can probably use log base 2 (binary bit position) and multiply
the result by a constant, which may simply be shifts and adds.

I assume you are using 16-bit audio. If so, the dynamic range
is only 20 * log10(2^16) = 96.3 dB. That means that attenuation
from mininum to maximum, in 1 dB steps, requires only 94 values.

Your code shows something whacked off at 0x3f = 0->0x40 = 64
20 * log10(64) = 36 dB for only 36 values. Clearly, you don't
need floating point, just some thought ahead of time.

Cheers,
Dick Johnson
Penguin : Linux version 2.6.15.4 on an i686 machine (5589.54 BogoMips).
Warning : 98.36% of all statistics are fiction, book release in April.
_


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