Linux – How to setup path for maven

NOTE:

The mentioned LATEST and RELEASE metaversions have been dropped for plugin dependencies in Maven 3 "for the sake of reproducible builds", over 6 years ago. (They still work perfectly fine for regular dependencies.) For plugin dependencies please refer to this Maven 3 compliant solution.

If you always want to use the newest version, Maven has two keywords you can use as an alternative to version ranges. You should use these options with care as you are no longer in control of the plugins/dependencies you are using.

When you depend on a plugin or a dependency, you can use the a version value of LATEST or RELEASE. LATEST refers to the latest released or snapshot version of a particular artifact, the most recently deployed artifact in a particular repository. RELEASE refers to the last non-snapshot release in the repository. In general, it is not a best practice to design software which depends on a non-specific version of an artifact. If you are developing software, you might want to use RELEASE or LATEST as a convenience so that you don't have to update version numbers when a new release of a third-party library is released. When you release software, you should always make sure that your project depends on specific versions to reduce the chances of your build or your project being affected by a software release not under your control. Use LATEST and RELEASE with caution, if at all.

See the POM Syntax section of the Maven book for more details. Or see this doc on Dependency Version Ranges, where:

• A square bracket ( [ & ] ) means "closed" (inclusive).
• A parenthesis ( ( & ) ) means "open" (exclusive).

Here's an example illustrating the various options. In the Maven repository, com.foo:my-foo has the following metadata:

<?xml version="1.0" encoding="UTF-8"?><metadata>
  <groupId>com.foo</groupId>
  <artifactId>my-foo</artifactId>
  <version>2.0.0</version>
  <versioning>
    <release>1.1.1</release>
    <versions>
      <version>1.0</version>
      <version>1.0.1</version>
      <version>1.1</version>
      <version>1.1.1</version>
      <version>2.0.0</version>
    </versions>
    <lastUpdated>20090722140000</lastUpdated>
  </versioning>
</metadata>

If a dependency on that artifact is required, you have the following options (other version ranges can be specified of course, just showing the relevant ones here):

Declare an exact version (will always resolve to 1.0.1):

<version>[1.0.1]</version>

Declare an explicit version (will always resolve to 1.0.1 unless a collision occurs, when Maven will select a matching version):

<version>1.0.1</version>

Declare a version range for all 1.x (will currently resolve to 1.1.1):

<version>[1.0.0,2.0.0)</version>

Declare an open-ended version range (will resolve to 2.0.0):

<version>[1.0.0,)</version>

Declare the version as LATEST (will resolve to 2.0.0) (removed from maven 3.x)

<version>LATEST</version>

Declare the version as RELEASE (will resolve to 1.1.1) (removed from maven 3.x):

<version>RELEASE</version>

Note that by default your own deployments will update the "latest" entry in the Maven metadata, but to update the "release" entry, you need to activate the "release-profile" from the Maven super POM. You can do this with either "-Prelease-profile" or "-DperformRelease=true"

It's worth emphasising that any approach that allows Maven to pick the dependency versions (LATEST, RELEASE, and version ranges) can leave you open to build time issues, as later versions can have different behaviour (for example the dependency plugin has previously switched a default value from true to false, with confusing results).

It is therefore generally a good idea to define exact versions in releases. As Tim's answer points out, the maven-versions-plugin is a handy tool for updating dependency versions, particularly the versions:use-latest-versions and versions:use-latest-releases goals.

If your goal is to use a profiler, use one of the suggested ones.

However, if you're in a hurry and you can manually interrupt your program under the debugger while it's being subjectively slow, there's a simple way to find performance problems.

Just halt it several times, and each time look at the call stack. If there is some code that is wasting some percentage of the time, 20% or 50% or whatever, that is the probability that you will catch it in the act on each sample. So, that is roughly the percentage of samples on which you will see it. There is no educated guesswork required. If you do have a guess as to what the problem is, this will prove or disprove it.

You may have multiple performance problems of different sizes. If you clean out any one of them, the remaining ones will take a larger percentage, and be easier to spot, on subsequent passes. This magnification effect, when compounded over multiple problems, can lead to truly massive speedup factors.

Caveat: Programmers tend to be skeptical of this technique unless they've used it themselves. They will say that profilers give you this information, but that is only true if they sample the entire call stack, and then let you examine a random set of samples. (The summaries are where the insight is lost.) Call graphs don't give you the same information, because

1. They don't summarize at the instruction level, and
2. They give confusing summaries in the presence of recursion.

They will also say it only works on toy programs, when actually it works on any program, and it seems to work better on bigger programs, because they tend to have more problems to find. They will say it sometimes finds things that aren't problems, but that is only true if you see something once. If you see a problem on more than one sample, it is real.

P.S. This can also be done on multi-thread programs if there is a way to collect call-stack samples of the thread pool at a point in time, as there is in Java.

P.P.S As a rough generality, the more layers of abstraction you have in your software, the more likely you are to find that that is the cause of performance problems (and the opportunity to get speedup).

Added: It might not be obvious, but the stack sampling technique works equally well in the presence of recursion. The reason is that the time that would be saved by removal of an instruction is approximated by the fraction of samples containing it, regardless of the number of times it may occur within a sample.

Another objection I often hear is: "It will stop someplace random, and it will miss the real problem". This comes from having a prior concept of what the real problem is. A key property of performance problems is that they defy expectations. Sampling tells you something is a problem, and your first reaction is disbelief. That is natural, but you can be sure if it finds a problem it is real, and vice-versa.

Added: Let me make a Bayesian explanation of how it works. Suppose there is some instruction I (call or otherwise) which is on the call stack some fraction f of the time (and thus costs that much). For simplicity, suppose we don't know what f is, but assume it is either 0.1, 0.2, 0.3, ... 0.9, 1.0, and the prior probability of each of these possibilities is 0.1, so all of these costs are equally likely a-priori.

Then suppose we take just 2 stack samples, and we see instruction I on both samples, designated observation o=2/2. This gives us new estimates of the frequency f of I, according to this:

Prior                                    
P(f=x) x  P(o=2/2|f=x) P(o=2/2&&f=x)  P(o=2/2&&f >= x)  P(f >= x | o=2/2)

0.1    1     1             0.1          0.1            0.25974026
0.1    0.9   0.81          0.081        0.181          0.47012987
0.1    0.8   0.64          0.064        0.245          0.636363636
0.1    0.7   0.49          0.049        0.294          0.763636364
0.1    0.6   0.36          0.036        0.33           0.857142857
0.1    0.5   0.25          0.025        0.355          0.922077922
0.1    0.4   0.16          0.016        0.371          0.963636364
0.1    0.3   0.09          0.009        0.38           0.987012987
0.1    0.2   0.04          0.004        0.384          0.997402597
0.1    0.1   0.01          0.001        0.385          1

                  P(o=2/2) 0.385                

The last column says that, for example, the probability that f >= 0.5 is 92%, up from the prior assumption of 60%.

Suppose the prior assumptions are different. Suppose we assume P(f=0.1) is .991 (nearly certain), and all the other possibilities are almost impossible (0.001). In other words, our prior certainty is that I is cheap. Then we get:

Prior                                    
P(f=x) x  P(o=2/2|f=x) P(o=2/2&& f=x)  P(o=2/2&&f >= x)  P(f >= x | o=2/2)

0.001  1    1              0.001        0.001          0.072727273
0.001  0.9  0.81           0.00081      0.00181        0.131636364
0.001  0.8  0.64           0.00064      0.00245        0.178181818
0.001  0.7  0.49           0.00049      0.00294        0.213818182
0.001  0.6  0.36           0.00036      0.0033         0.24
0.001  0.5  0.25           0.00025      0.00355        0.258181818
0.001  0.4  0.16           0.00016      0.00371        0.269818182
0.001  0.3  0.09           0.00009      0.0038         0.276363636
0.001  0.2  0.04           0.00004      0.00384        0.279272727
0.991  0.1  0.01           0.00991      0.01375        1

                  P(o=2/2) 0.01375                

Now it says P(f >= 0.5) is 26%, up from the prior assumption of 0.6%. So Bayes allows us to update our estimate of the probable cost of I. If the amount of data is small, it doesn't tell us accurately what the cost is, only that it is big enough to be worth fixing.

Yet another way to look at it is called the Rule Of Succession. If you flip a coin 2 times, and it comes up heads both times, what does that tell you about the probable weighting of the coin? The respected way to answer is to say that it's a Beta distribution, with average value (number of hits + 1) / (number of tries + 2) = (2+1)/(2+2) = 75%.

(The key is that we see I more than once. If we only see it once, that doesn't tell us much except that f > 0.)

So, even a very small number of samples can tell us a lot about the cost of instructions that it sees. (And it will see them with a frequency, on average, proportional to their cost. If n samples are taken, and f is the cost, then I will appear on nf+/-sqrt(nf(1-f)) samples. Example, n=10, f=0.3, that is 3+/-1.4 samples.)

Added: To give an intuitive feel for the difference between measuring and random stack sampling:
There are profilers now that sample the stack, even on wall-clock time, but what comes out is measurements (or hot path, or hot spot, from which a "bottleneck" can easily hide). What they don't show you (and they easily could) is the actual samples themselves. And if your goal is to find the bottleneck, the number of them you need to see is, on average, 2 divided by the fraction of time it takes. So if it takes 30% of time, 2/.3 = 6.7 samples, on average, will show it, and the chance that 20 samples will show it is 99.2%.

Here is an off-the-cuff illustration of the difference between examining measurements and examining stack samples. The bottleneck could be one big blob like this, or numerous small ones, it makes no difference.

Measurement is horizontal; it tells you what fraction of time specific routines take. Sampling is vertical. If there is any way to avoid what the whole program is doing at that moment, and if you see it on a second sample, you've found the bottleneck. That's what makes the difference - seeing the whole reason for the time being spent, not just how much.