On 1/29/19 9:10 PM, Mark Cave-Ayland wrote:
>> So... the point of "half" was to not replicate knowledge out to VMRG.
>> Just use
>>
>> int i, half = ARRAY_SIZE(r->element) / 2;
>> for (i = 0; i < half; i++) {
>>
>> within VMRG_DO.
>
> Okay - I was a bit confused because in your example macro signature you
> added ofs which made me think you wanted its value to be determined outside,
> but nevermind.
>
> What about the following instead? With high set as part of the macro then
> the initial assignment to ofs should be inlined accordingly at compile time:
>
> #define VMRG_DO(name, element, access, high) \
> void helper_v##name(ppc_avr_t *r, ppc_avr_t *a, ppc_avr_t *b) \
> { \
> ppc_avr_t result; \
> int ofs, half = ARRAY_SIZE(r->element) / 2; \
> int i; \
> \
> if (high) { \
> ofs = 0; \
> } else { \
> ofs = half; \
> } \
Why are you over-complicating things? I thought I'd explicitly said this
twice, but perhaps not:
Pass the symbol "half" directly to VMRG_DO:
#define VMRG(suffix, element, access) \
VMRG_DO(mrgl##suffix, element, access, 0) \
VMRG_DO(mrgh##suffix, element, access, half)
You do not need another conditional within VMRG_DO.
Also, I'm pretty sure it's the "low" merge that wants the "ofs" to be non-zero,
since all of the Vsr* symbols implement big-endian indexing.
r~
