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bio website gustedt.wordpress.com
location Strasbourg, France
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visits member for 3 years, 7 months
seen Jan 2 '13 at 8:30

Over the years I mostly programmed in C++ on POSIX systems, namely linux. But then, quite suddenly I became allergic to C++, to its useless bloat, crude syntax and its tendency to send apples three times around the earth just to be delivered to your neighbor.

I now mostly program in C, C99 to be precise, starting with C11, and occasionally in perl when I just need to do some scripting.

A first result of that is P99 a set of preprocessor macros for C99. People will say that P99 is a useless bloat, has crude syntax and has tendency to send apples once around the earth just to be delivered to your neighbor. But if it is already two orbits that we gain, it is worth it, isn't it?

As an additional feature, on some platforms P99 now also emulates a lot of C11, atomics, threads, type generics, be surprised!


Sep
24
awarded  Autobiographer
Jul
26
comment Big-O of this algorithm?
@Fallen, concerning big-O notation you are completely on the wrong track. For sure are 3 nested loops of length bounded by n enough to prove that there is a C such that C x N is an upper bound. You are arguing about a lower bound.
Jul
26
comment Big-O of this algorithm?
I have the impression that you are trying to prove a lower bound. Big-O is looking for an upper bound. Stating that the algorithm is O(n^3) is trivial. Omega(n^3) is the difficult part.
Jul
26
comment Big-O of this algorithm?
You should perhaps first read up about the differences between big-O, Omega, Theta and little-o. Big-O asks if a function is a upper bound. First there is no the bounding function, but the trick of big-O is that this is a whole family of functions. Then you should much more interested in a lower bound of the complexity (Omega) then in an upper bound big-O.
Jul
26
comment Big-O of this algorithm?
who is voting this down? this answer is correct. It is O(n^3). That it is maybe also O(n^2) doesn't contradict this, O(n^2) is included in O(n^3).
Sep
16
comment Schemes to resolve deadlocks
If an algorithmic scheme allows by some trick to recover when a deadlock occurs, by definition such a situation must be detectable beforehand and can thus entirely avoided. Or said otherwise, if you have an algorithm that potentially goes into deadlock, change your design.
Feb
27
comment If you need more than 3 levels of indentation, you're screwed?
@jokon, @Steve, @larsmans: You should really see this quote in its context. This gives you all explanations that you are looking for: computing.llnl.gov/linux/slurm/coding_style.pdf