Re: Habitual optimization

From:
"dhruv" <dhruvbird@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
21 Dec 2006 12:50:12 -0500
Message-ID:
<1166711593.692565.237120@79g2000cws.googlegroups.com>
Steven T. Hatton wrote:

I have long wondered about certain programming styles/strategies and how
they can impact performance.

[snip]....

const int dTick((rect.width() - 1)/settings.numXTicks);

and then writing:

int x = rect.left() + i * dTick;?

Or even `int x(rect.left());' outside the loop, and `x+=dTick'; inside the
loop?


Yes, definitely there is a gain to use the additive operation inside of
the loop.

Two in fact:

[1] You are executing an add operation instead of a multiply every
time, which is less time consuming.

[2] You are touching lesser number of variables.
Typically, on an x86 machine, the former would be:

(a) Load dTick into a temporary.
(b) Multiply i and dTick.
(c) Store the result of rect.left() in a temporary.
(d) Add the result of the above 2 temporaries.

However, the latter would be:
(a) Store the result of rect.left() in a temporary.
(b) Add the above temporary and the value of dTick.

[snip]....

One additional question I have deals with for-loops.
std::vector<Object> v;
// fill the vector
for(unsigned i(0); i<v.size(); ++i){
 //do something.
}

Is it likely that an optimizer could improve performance if it could
determine that v.size() is unchanged during the loop, and thus eliminate
calls to v.size()?


Yes, definitely it would result in a performance gain.
The C++ standard specifies that the size() member function should be
constant time. It may be implemented as a difference of 2 pointers,
which boils down to difference followed by a divide with the type's
size. If these 2 operations can be eliminated(which btw involve an AGI)
then there's nothing better for speed!!!!

Regards,
-Dhruv.

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