Author: Pham Minh Tri
Date: 18:59:22 02/05/02
Go up one level in this thread
On February 05, 2002 at 21:30:52, Dann Corbit wrote:
>On February 05, 2002 at 21:13:20, Pham Minh Tri wrote:
>
>>Note that his question is "without using none kind of loops" so struct
>>assignment and memcpy are surely not answer because they use some assembly loop
>>;)
>>
>>I think even they are the fastest solutions for his problem but the answer for
>>that question is switch statement.
>
>I doubt it.
>
>Switch is a construct you should use only if you have to[1]. And if that
>particular area is a bottleneck, I doubt very much if it will help. A
>mispredicted branch is one of the worst things that can go wrong on a modern
>CPU, and switch statements are mis-predicted branch generators.
Dann, I agree that switch is slow and should not use to replace memcpy (struct
assignment is also memcpy) in that case. Just joke because he did not ask about
effective or faster solution, but not using loops.
>
>[1] I am talking about performance solutions here. If a switch is clearest,
>write using a switch. But if you have a slow spot, it is probably not going to
>help to change to a switch. Try this on your handy-dandy compiler and tell me
>how it turns out:
>
>#include <math.h>
>#include <stdio.h>
>
>typedef double (*f_t) (double);
>static f_t f[] = {log, log10, sqrt, cos, cosh, exp, sin, sinh, tan, tanh,
>0};
>
>static double accum0 = 0;
>static double accum1 = 0;
>static double accum2 = 0;
>
>
>void arr(void)
>{
> int i;
> double d = 0;
> for (i = 0; f[i]; i++) {
> d += f[i] (0.5);
> }
> accum0 += d;
>}
>
>void poi(void)
>{
> f_t *flist = f;
> double d = 0;
> while (*flist) {
> f_t ff = *flist;
> d += ff(0.5);
> flist++;
> }
> accum1 += d;
>}
>
>void swi(void)
>{
> int i;
> double d = 0;
> for (i = 0; f[i]; i++) {
> switch (i) {
> case 0:
> d += f[0] (0.5);
> break;
> case 1:
> d += f[1] (0.5);
> break;
> case 2:
> d += f[2] (0.5);
> break;
> case 3:
> d += f[3] (0.5);
> break;
> case 4:
> d += f[4] (0.5);
> break;
> case 5:
> d += f[5] (0.5);
> break;
> case 6:
> d += f[6] (0.5);
> break;
> case 7:
> d += f[7] (0.5);
> break;
> case 8:
> d += f[8] (0.5);
> break;
> case 9:
> d += f[9] (0.5);
> break;
> default:
> break;
> }
> }
> accum2 += d;
>}
>
>int main(void)
>{
> long i;
> for (i = 0; i < 1000000; i++) {
> arr();
> poi();
> swi();
> }
> printf("%.20g, %.20g, %.20g\n", accum0, accum1, accum2);
> return 0;
>}
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