sc | Common libraries and data structures for C | Hashing library

If you have two images that are similar (or the same) and you want to align them, you can do it using both functions rotate and shift :

I think actually a more efficient way would be to sort by Brand and then Year, and then use interpolate:

I rewrote your code with some better ways to create table. My idea was to pick out the buttons that fell onto a range of type and then loop through those buttons and change its color to those type.

You can try this:

We'll use gtools::permutations to calculate the ... permutations of inputs, and then use expand.grid to show all combinations within them.

First, we can do it easily on one order of the inputs with:

You are better off avoiding serializable isolation level. The way the serializable guarantee is provided is often deadlock prone.

If you can't alter your stored procs to use more targeted locking hints that guarantee the results you require at a lesser isolation level then you can prevent this particular deadlock scenario shown by ensuring that all locks are taken out on ServiceChange first before any are taken out on ServiceChangeParameter.

One way of doing this would be to introduce a table variable in spGetManageServicesRequest and materialize the results of

Multiple threads may compete to do this (see "initialization race" comment).

To paraphrase the code:

I have such a solution. And I'm surprised myself that I used two loops for!! Incredibly, I did it. First things first.

My proposal is based on a simplification. However, the mistake you will make at short distances will be relatively small. But the time gain is huge!

Well, I propose to count the distance in Cartesian coordinates, not spherical.

So we're going to need a simple function that computes the Cartesian coordinates based on the two arguments latitude and longitude. Here is our LatLong2Cart feature.

Yes, I think we can prove that a == 1 || b == 1 is always true. Most of the ideas here were worked out in comments by zwhconst and Peter Cordes, so I just thought I would write it up as an exercise.

(Note that X, Y, A, B below are used as the dummy variables in the standard's axioms, and may change from line to line. They do not coincide with the labels in your code.)

Suppose b = x.load() in thread2 yields 0.

We do have the coherence ordering that you asked about. Specifically, if b = x.load yields 0, then I claim that x.load() in thread2 is coherence ordered before x.store(1) in thread1, thanks to the third bullet in the definition of coherence ordering. For let A be x.load(), B be x.store(1), and X be the initialization x{0} (see below for quibble). Clearly X precedes B in the modification order of x, since X happens-before B (synchronization occurs when the thread is started), and if b == 0 then A has read the value stored by X.

(There is possibly a gap here: initialization of an atomic object is not an atomic operation (3.18.1p3), so as worded, the coherence ordering does not apply to it. I have to believe it was intended to apply here, though. Anyway, we could dodge the issue by putting x.store(0, std::memory_order_relaxed); in main before starting the threads, which would still address the spirit of your question.)

Now in the definition of the ordering S, apply the second bullet with A = x.load() and B = x.store(1) as before, and Y being the atomic_thread_fence in thread1. Then A is coherence-ordered before B, as we just showed; A is seq_cst; and B happens-before Y by sequencing. So therefore A = x.load() precedes Y = fence in the order S.

Now suppose a = y.load() in thread1 also yields 0.

By a similar argument to before, y.load() is coherence ordered before y.store(1), and they are both seq_cst, so y.load() precedes y.store(1) in S. Also, y.store(1) precedes x.load() in S by sequencing, and likewise atomic_thread_fence precedes y.load() in S. We therefore have in S:

• x.load precedes fence precedes y.load precedes y.store precedes x.load

which is a cycle, contradicting the strict ordering of S.