Author: Dann Corbit
Date: 18:30:52 02/05/02
Go up one level in this thread
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.
[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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