explain | Explain explains your Ruby code in natural language | Runtime Evironment library

IntelliJ (Android Studio) wants to know what kind of project you have to show the right menus.

1. click left in file tree on the root node (project name)
2. MENU -> FILE -> PROJECT STRUCTURE
3. select left MODULES
4. hit + button to add ANDROID
5. hit OK Button
6. top of the root node (project name) select PROJECT... to see your old view

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.

Code readability, short-circuiting and it is not guaranteed that Ord will always outperform a || operand. Computer systems are more complicated than expected, even though they are man-made.

There was a case where a for loop with a much more complicated condition ran faster on an IBM. The CPU didn't cool and thus instructions were executed faster, that was a possible reason. What I am trying to say, focus on other areas to improve code than fighting small-cases which will differ depending on the CPU and the boolean evaluation (compiler optimizations).

Although I can't provide a formal definition of why/how the rounding modes were chosen as they were, the citation about compatibility with the % operator, which you have included, does make sense when you consider that % is not quite the same thing in C++ and Python.

In C++, it is the remainder operator, whereas, in Python, it is the modulus operator – and, when the two operands have different signs, these aren't necessarily the same thing. There are some fine explanations of the difference between these operators in the answers to: What's the difference between “mod” and “remainder”?

Now, considering this difference, the rounding (truncation) modes for integer division have to be as they are in the two languages, to ensure that the relationship you quoted, (m/n)*n + m%n == m, remains valid.

Here are two short programs that demonstrate this in action (please forgive my somewhat naïve Python code – I'm a beginner in that language):

C++:

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.

When exactly does segmentation fault happen (=when is SIGSEGV sent)?

When you attempt to access memory you don’t have access to, such as accessing an array out of bounds or dereferencing an invalid pointer. The signal SIGSEGV is standardized but different OS might implement it differently. "Segmentation fault" is mainly a term used in *nix systems, Windows calls it "access violation".

Why is the process in undefined behavior state after that point?

Because one or several of the variables in the program didn’t behave as expected. Let’s say you have some array that is supposed to store a number of values, but you didn’t allocate enough room for all them. So only those you allocated room for get written correctly, and the rest written out of bounds of the array can hold any values. How exactly is the OS to know how critical those out of bounds values are for your application to function? It knows nothing of their purpose.

Furthermore, writing outside allowed memory can often corrupt other unrelated variables, which is obviously dangerous and can cause any random behavior. Such bugs are often hard to track down. Stack overflows for example are such segmentation faults prone to overwrite adjacent variables, unless the error was caught by protection mechanisms.

If we look at the behavior of "bare metal" microcontroller systems without any OS and no virtual memory features, just raw physical memory - they will just silently do exactly as told - for example, overwriting unrelated variables and keep on going. Which in turn could cause disastrous behavior in case the application is mission-critical.

Why is it not recoverable?

Because the OS doesn’t know what your program is supposed to be doing.

Though in the "bare metal" scenario above, the system might be smart enough to place itself in a safe mode and keep going. Critical applications such as automotive and med-tech aren’t allowed to just stop or reset, as that in itself might be dangerous. They will rather try to "limp home" with limited functionality.

Why does this solution avoid that unrecoverable state? Does it even?

That solution is just ignoring the error and keeps on going. It doesn’t fix the problem that caused it. It’s a very dirty patch and setjmp/longjmp in general are very dangerous functions that should be avoided for any purpose.

We have to realize that a segmentation fault is a symptom of a bug, not the cause.

Python uses the symbol - as both a unary (-x) and a binary (x-y) operator. These have different operator precedence.

In specific, the ordering wrt // is:

• unary -
• binary //
• binary -

By introducing a 0 as 0--3//2, the first - is a binary - and is applied last. Without a leading 0 as --3//2, both - are unary and applied together.

The corresponding evaluation/syntax tree is roughly like this, evaluating nodes at the bottom first to use them in the parent node:

The following section from this Draft C11 Standard completely refutes the claim made (even with the clarification mentioned in the 'errata', in the comment by GSerg).

6.3.2.3 Pointers

1     A pointer to void may be converted to or from a pointer to any object type. A pointer to any object type may be converted to a pointer to void and back again; the result shall compare equal to the original pointer.

Or, this section from the same draft Standard:

7.20.1.4 Integer types capable of holding object pointers

1    The following type designates a signed integer type with the property that any valid pointer to void can be converted to this type, then converted back to pointer to void, and the result will compare equal to the original pointer:

      intptr_t