Hello,
I have an exam this friday, not an C exam.
But one of the questions on the exam example is:
optimalise this piece of code. What do you guy think that
the BEST optimization is?
Code:void Increase(int ∗ list , unsigned int size, int extra) { int i; for( i =0; i <size ; i++) size[i] ∗= 36; for ( i =0; i < size ;i++) size[i] = size[i] ∗ extra + size[i] ∗ 5; }
I would start with code that actually compiled.
It isn't a programming exercise but trying to find the best optimization for this code and the reason wy.
So I need to add the 2 loops into one loop, but I am pretty sure i need to do something more, but I can't find what. Can somebody please help?
Thank you
I am sorry, I forgot to adapt the code, On the example there was a error.
Code:void Increase(int ∗ list , unsigned int size, int extra) { int i; for( i =0; i <size ; i++) list[i] ∗= 36; for ( i =0; i < size ;i++) list[i] = list[i] ∗ extra + list[i] ∗ 5; }
If you know you have to take one loop out, why don't you like, try it
I am sorry but i think i did not explained my question that good.
The given code becomes :
BUT I think this is to easy to be an exam question(university). So I believe there must be something more that can be optimised.Code:void Increase(int ∗ list , unsigned int size, int extra) { int i; for ( i =0; i < size ;i++){ list[i] ∗= 36; list[i] = list[i] ∗ extra + list[i] ∗ 5; } }
For example: ++i is faster then i++,
the use of bitinstructions,
lazy evaluation,
loop unrolling, loop fusion,
add switch statements that are most used to the front,
etc,...
But I cant figure out if I am missing something in the code above
Sorry if my question is ambiguous.
No, at least in C...For example: ++i is faster then i++,
Well, my book says: i++ copys the existing value to a temporal object, raises the internal value with one and returns that temporal value. But ++i raises the value and returns a reference.
Conceptually, yes. But when used as a standalone expression, a C compiler can be reasonably expected to generate the same code either way. In C++ it may be the case that a compiler will not apply such an optimisation for class types because operator++ could be overloaded such that the two versions are expected to produce different net results even when used standalone.Originally Posted by thescratchy
Maybe your lecturer thinks otherwise. After all, we can expect that questions will be of varying degrees of difficulty. Besides, you told us what the exam is not about, but did not tell us what the exam is about, i.e., what you are actually being taught and tested.
There is the micro-optimisation, which might be implemented by an optimising compiler, that changes this:
to:Code:list[i] = list[i] ∗ extra + list[i] ∗ 5;
which can then be simplified to:Code:list[i] = list[i] ∗ (extra + 5);
Observe that (extra + 5) is constant in the loop body. Observe also that in the previous line, list[i] is multiplied by another constant, 36. Therefore, the loop can be written as:Code:list[i] *= (extra + 5);
Code:unsigned int i; extra = 36 * (extra + 5); for (i = 0; i < size; i++) { list[i] *= extra; }
Look up a C++ Reference and learn How To Ask Questions The Smart Way
Thank you for your clear answer. Don't know why I didn't see that myself.
Really appreciate your effort!
Thank you very much
@laserlight: Great optimization for both size and speed.
@thescratchy: Remember when optimizing what your are optimizing for, either speed or size. Most times you have to pick one over the other.
Tim S.
