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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:

I suspect this may have been an accident, though I prefer the new behavior.

The new behavior is a consequence of a change to how the bytecode for * arguments works. The change is in the changelog under Python 3.9.0 alpha 3:

bpo-39320: Replace four complex bytecodes for building sequences with three simpler ones.

The following four bytecodes have been removed:

• BUILD_LIST_UNPACK
• BUILD_TUPLE_UNPACK
• BUILD_SET_UNPACK
• BUILD_TUPLE_UNPACK_WITH_CALL

The following three bytecodes have been added:

• LIST_TO_TUPLE
• LIST_EXTEND
• SET_UPDATE

On Python 3.8, the bytecode for f(*a, a.pop()) looks like this:

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.

Cache misses. When N int objects are allocated back-to-back, the memory reserved to hold them tends to be in a contiguous chunk. So crawling over the list in allocation order tends to access the memory holding the ints' values in sequential, contiguous, increasing order too.

Shuffle it, and the access pattern when crawling over the list is randomized too. Cache misses abound, provided there are enough different int objects that they don't all fit in cache.

At r==1, and r==2, CPython happens to treat such small ints as singletons, so, e.g., despite that you have 10 million elements in the list, at r==2 it contains only (at most) 100 distinct int objects. All the data for those fit in cache simultaneously.

Beyond that, though, you're likely to get more, and more, and more distinct int objects. Hardware caches become increasingly useless then when the access pattern is random.

Illustrating:

Looks like a bug in Visual Studio, according to the standard [except.handle]:

A handler is a match for an exception object of type E if

[...]

• the handler is of type cv T or const T& where T is a pointer or pointer-to->member type and E is std​::​nullptr_t.

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.

std::memcpy(arr_copy, arr, sizeof arr); (your example) is well-defined.

std::memcpy(arr_copy, arr[0], sizeof arr);, on the other hand, causes undefined behavior (at least in C++; not entirely sure about C).

Multidimensional arrays are 1D arrays of arrays. As far as I know, they don't get much (if any) special treatment compared to true 1D arrays (i.e. arrays with elements of non-array type).

Consider an example with a 1D array:

The __enter__ method should return the context object. with ... as ... uses the return value of __enter__ to determine what object to give you. Since your __enter__ returns nothing, it implicitly returns None, so test is None.