array | Supports events | Functional Programming library

From the cpython source code for sum sum initially seems to attempt a fast path that assumes all inputs are the same type. If that fails it will just iterate:

This problem's caused by a bug in Springfox. It's making an assumption about how Spring MVC is set up that doesn't always hold true. Specifically, it's assuming that MVC's path matching will use the Ant-based path matcher and not the PathPattern-based matcher. PathPattern-based matching has been an option for some time now and is the default as of Spring Boot 2.6.

As described in Spring Boot 2.6's release notes, you can restore the configuration that Springfox assumes will be used by setting spring.mvc.pathmatch.matching-strategy to ant-path-matcher in your application.properties file. Note that this will only work if you are not using Spring Boot's Actuator. The Actuator always uses PathPattern-based parsing, irrespective of the configured matching-strategy. A change to Springfox will be required if you want to use it with the Actuator in Spring Boot 2.6 and later.

This problem has a fun O(n) solution.

If you draw a graph of cumulative sum vs index, then:

The average value in the subarray between any two indexes is the slope of the line between those points on the graph.

The first highest-average-prefix will end at the point that makes the highest angle from 0. The next highest-average-prefix must then have a smaller average, and it will end at the point that makes the highest angle from the first ending. Continuing to the end of the array, we find that...

These segments of highest average are exactly the segments in the upper convex hull of the cumulative sum graph.

Find these segments using the monotone chain algorithm. Since the points are already sorted, it takes O(n) time.

there could be at least one difference - case when you need to pass variable to some other function, for example:

It looks like GCC's naïveté about store-forwarding stalls is hurting its auto-vectorization strategy here. See also Store forwarding by example for some practical benchmarks on Intel with hardware performance counters, and What are the costs of failed store-to-load forwarding on x86? Also Agner Fog's x86 optimization guides.

(gcc -O3 enables -ftree-vectorize and a few other options not included by -O2, e.g. if-conversion to branchless cmov, which is another way -O3 can hurt with data patterns GCC didn't expect. By comparison, Clang enables auto-vectorization even at -O2, although some of its optimizations are still only on at -O3.)

It's doing 64-bit loads (and branching to store or not) on pairs of ints. This means, if we swapped the last iteration, this load comes half from that store, half from fresh memory, so we get a store-forwarding stall after every swap. But bubble sort often has long chains of swapping every iteration as an element bubbles far, so this is really bad.

(Bubble sort is bad in general, especially if implemented naively without keeping the previous iteration's second element around in a register. It can be interesting to analyze the asm details of exactly why it sucks, so it is fair enough for wanting to try.)

Anyway, this is pretty clearly an anti-optimization you should report on GCC Bugzilla with the "missed-optimization" keyword. Scalar loads are cheap, and store-forwarding stalls are costly. (Can modern x86 implementations store-forward from more than one prior store? no, nor can microarchitectures other than in-order Atom efficiently load when it partially overlaps with one previous store, and partially from data that has to come from the L1d cache.)

Even better would be to keep buf[x+1] in a register and use it as buf[x] in the next iteration, avoiding a store and load. (Like good hand-written asm bubble sort examples, a few of which exist on Stack Overflow.)

If it wasn't for the store-forwarding stalls (which AFAIK GCC doesn't know about in its cost model), this strategy might be about break-even. SSE 4.1 for a branchless pmind / pmaxd comparator might be interesting, but that would mean always storing and the C source doesn't do that.

If this strategy of double-width load had any merit, it would be better implemented with pure integer on a 64-bit machine like x86-64, where you can operate on just the low 32 bits with garbage (or valuable data) in the upper half. E.g.,

It does not default to zero. The sample answer is wrong. Undefined behaviour is undefined; the value may be 0, it may be 100. Accessing it may cause a seg fault, or cause your computer to be formatted.

As to why it's not an error, it's because C++ is not required to do bounds checking on arrays. You could use a vector and use the at function, which throws exceptions if you go outside the bounds, but arrays do not.

The question is not why this fails for int[], but why it works for all the other types! Unfortunately, you have fallen prey to ADL which is actually calling std::size instead of the size function you have written. This is because all overloads of your function fail, and so it looks in the namespace of the first argument for a matching function, where it finds std::size. Rerun your program with the function renamed to something else:

The answer is in the PEP of the generator expressions, in particular the session Early Binding vs Late biding:

After much discussion, it was decided that the first (outermost) for-expression should be evaluated immediately and that the remaining expressions be evaluated when the generator is executed.

So basically the array in:

To prove that it's correct, you have to find some sort of invariant. Something that's true during every pass of the loop.

Looking at it, after the very first pass of the inner loop, the largest element of the list will actually be in the first position.

Now in the second pass of the inner loop, i = 1, and the very first comparison is between i = 1 and j = 0. So, the largest element was in position 0, and after this comparison, it will be swapped to position 1.

In general, then it's not hard to see that after each step of the outer loop, the largest element will have moved one to the right. So after the full steps, we know at least the largest element will be in the correct position.

What about all the rest? Let's say the second-largest element sits at position i of the current loop. We know that the largest element sits at position i-1 as per the previous discussion. Counter j starts at 0. So now we're looking for the first A[j] such that it's A[j] > A[i]. Well, the A[i] is the second largest element, so the first time that happens is when j = i-1, at the first largest element. Thus, they're adjacent and get swapped, and are now in the "right" order. Now A[i] again points to the largest element, and hence for the rest of the inner loop no more swaps are performed.

So we can say: Once the outer loop index has moved past the location of the second largest element, the second and first largest elements will be in the right order. They will now slide up together, in every iteration of the outer loop, so we know that at the end of the algorithm both the first and second-largest elements will be in the right position.

What about the third-largest element? Well, we can use the same logic again: Once the outer loop counter i is at the position of the third-largest element, it'll be swapped such that it'll be just below the second largest element (if we have found that one already!) or otherwise just below the first largest element.

Ah. And here we now have our invariant: After k iterations of the outer loop, the k-length sequence of elements, ending at position k-1, will be in sorted order:

After the 1st iteration, the 1-length sequence, at position 0, will be in the correct order. That's trivial.

After the 2nd iteration, we know the largest element is at position 1, so obviously the sequence A[0], A[1] is in the correct order.

Now let's assume we're at step k, so all the elements up to position k-1 will be in order. Now i = k and we iterate over j. What this does is basically find the position at which the new element needs to be slotted into the existing sorted sequence so that it'll be properly sorted. Once that happens, the rest of the elements "bubble one up" until now the largest element sits at position i = k and no further swaps happen.

Thus finally at the end of step N, all the elements up to position N-1 are in the correct order, QED.