Given the attached sample code.
It tries to profile the difference between accessing a native mapping and
accessing it through a class which is declared inline as much as possible.
When I run it, I find that the native mapping is about twice as fast and
the inlined amapping class.
What would it take to make an amapping class less than a factor 1.2 slower than
the native mapping?
Would being able to inherit from some kind of "_Builtin._mapping" class be of
any use?
I ran the tests on Pike 8.1.
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#!/usr/local/bin/pike
#define REPEAT 10000
class timeit {
protected void create(string name, function(:void) fn) {
int i,t1,t2; mixed e;
t1 = gethrtime();
e = catch {
for (i=REPEAT; i>0; i--)
fn();
};
if (e)
write("Skipped %s\n", name);
else {
t2 = gethrtime();
write("%f %s\n", (t2-t1)/(float)REPEAT, name);
}
}
}
class amapping {
private mapping values;
inline protected void create(mapping v) {
values = v;
}
inline private mixed `[](string key) {
return values[key];
}
inline private mixed `->(string key) {
return values[key];
}
inline private mixed `[]=(string key, mixed v) {
return values[key] = v;
}
inline private mixed `->=(string key, mixed v) {
return values[key] = v;
}
inline private Iterator _get_iterator() {
return get_iterator(values);
}
inline private array(string) _indices() {
return indices(values);
}
private string _sprintf(int type, void|mapping flags) {
string res = UNDEFINED;
switch (type) {
case 'O':
res = sprintf("%O", values);
int indent;
if (flags && (indent = flags->indent))
res = replace(res, "\n", "\n" + " " * indent);
break;
}
return res;
}
}
int main(int argc, array(string) argv) {
mapping m = ([
"abc":8,
"acc":1,
"adc":2,
"aec":3,
"afc":4,
"agc":5,
"ahc":6,
"adc":2,
"1dc":3,
"3dc":2,
"4dc":2,
"5dc":8,
"6dc":7,
"7dc":2,
"8dc":3,
"rdc":2,
]);
amapping am = amapping(m);
timeit("mapping", lambda(){
int total = 0;
foreach (indices(m);; string id)
total += m[id];
return total;
});
timeit("amapping", lambda(){
int total = 0;
foreach (indices(m);; string id)
total += am[id];
return total;
});
return 0;
}
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--
Stephen.
