My project structure looks like this:
.
├── Cargo.lock
├── Cargo.toml
└── src
├── bin
│ └── other.rs
├── main.rs
└── util.rs
(code: https://gitlab.com/msrd0/cargo-bin-import)
In my other.rs, I'm trying to reuse code from the util mod, which is declared as a public mod in my main.rs file. I tried the following:
use util::do_sth
use crate::util::do_sth
use cargo_bin_import::util::do_sth (with and without extern crate)
mod util; use util::do_sth
extern crate util; use util::do_sth (suggested by rustc)
None of the above worked and gave me error messages similar to this one:
error[E0432]: unresolved import `crate::util`
--> src/bin/other.rs:1:12
|
1 | use crate::util::do_sth;
| ^^^^ maybe a missing `extern crate util;`?
error: aborting due to previous error
Use a library and two binaries, then reuse the lib's code in the two binaries. Example:
Cargo.toml
[lib]
name = "utils"
path = "src/utils.rs"
# cargo build --bin other
[[bin]]
name = "other"
path = "src/bin/other.rs"
# cargo build --bin main
[[bin]]
name = "main"
path = "src/main.rs"
Then use utils::{...}. The path are taken from your question, but putting main inside bin/ and renaming utils.rs to lib.rs could be a more standard way to do it.
If the library is generic enough, you could even release it on crates.io for others to benefit from it.
Related
This is a much asked question, but none of the solutions mentioned on SO have worked so far.
The folder structure is as follows:
project/
└── tests/
├── conftest.py
├── __init__.py
└── int_tests/
└── test_device.py
└── project_core/
└── tests/
├── conftest.py
├── __init__.py
└── int_tests/
└── test_device.py
import file mismatch:
imported module 'test_device' has this __file__ attribute:
/home/.../project/project_core/tests/int_tests/test_device.py
which is not the same as the test file we want to collect:
/home/.../project/tests/int_tests/test_device.py
HINT: remove __pycache__ / .pyc files and/or use a unique basename for your test file modules
Steps tried so far:
Removing pycache and pyc files.
Adding _init to each folder. (As is stated in pytest GIP)
Removing _init from each folder.
Do i need init files in each tests/subfolder?
The same error occurs with conftest.py as well. This error is not limited to vscode-pytest plugin, also occurs on the terminal.
PS : For CI purposes, the system is configured with docker & tox. Development is done in venv.
I am learning Scala now and I want to write some silly little app like a console Twitter client, or whatever. The question is, how to structure application on disk and logically. I know python, and there I would just create some files with classes and then import them in the main module like import util.ssh or from tweets import Retweet (strongly hoping you wouldn't mind that names, they are just for reference). But how should I do this stuff using Scala? Also, I have not much experience with JVM and Java, so I am a complete newbie here.
I'm going to disagree with Jens, here, though not all that much.
Project Layout
My own suggestion is that you model your efforts on Maven's standard directory layout.
Previous versions of SBT (before SBT 0.9.x) would create it automatically for you:
dcs#ayanami:~$ mkdir myproject
dcs#ayanami:~$ cd myproject
dcs#ayanami:~/myproject$ sbt
Project does not exist, create new project? (y/N/s) y
Name: myproject
Organization: org.dcsobral
Version [1.0]:
Scala version [2.7.7]: 2.8.1
sbt version [0.7.4]:
Getting Scala 2.7.7 ...
:: retrieving :: org.scala-tools.sbt#boot-scala
confs: [default]
2 artifacts copied, 0 already retrieved (9911kB/134ms)
Getting org.scala-tools.sbt sbt_2.7.7 0.7.4 ...
:: retrieving :: org.scala-tools.sbt#boot-app
confs: [default]
15 artifacts copied, 0 already retrieved (4096kB/91ms)
[success] Successfully initialized directory structure.
Getting Scala 2.8.1 ...
:: retrieving :: org.scala-tools.sbt#boot-scala
confs: [default]
2 artifacts copied, 0 already retrieved (15118kB/160ms)
[info] Building project myproject 1.0 against Scala 2.8.1
[info] using sbt.DefaultProject with sbt 0.7.4 and Scala 2.7.7
> quit
[info]
[info] Total session time: 8 s, completed May 6, 2011 12:31:43 PM
[success] Build completed successfully.
dcs#ayanami:~/myproject$ find . -type d -print
.
./project
./project/boot
./project/boot/scala-2.7.7
./project/boot/scala-2.7.7/lib
./project/boot/scala-2.7.7/org.scala-tools.sbt
./project/boot/scala-2.7.7/org.scala-tools.sbt/sbt
./project/boot/scala-2.7.7/org.scala-tools.sbt/sbt/0.7.4
./project/boot/scala-2.7.7/org.scala-tools.sbt/sbt/0.7.4/compiler-interface-bin_2.7.7.final
./project/boot/scala-2.7.7/org.scala-tools.sbt/sbt/0.7.4/compiler-interface-src
./project/boot/scala-2.7.7/org.scala-tools.sbt/sbt/0.7.4/compiler-interface-bin_2.8.0.RC2
./project/boot/scala-2.7.7/org.scala-tools.sbt/sbt/0.7.4/xsbti
./project/boot/scala-2.8.1
./project/boot/scala-2.8.1/lib
./target
./lib
./src
./src/main
./src/main/resources
./src/main/scala
./src/test
./src/test/resources
./src/test/scala
So you'll put your source files inside myproject/src/main/scala, for the main program, or myproject/src/test/scala, for the tests.
Since that doesn't work anymore, there are some alternatives:
giter8 and sbt.g8
Install giter8, clone ymasory's sbt.g8 template and adapt it to your necessities, and use that. See below, for example, this use of unmodified ymasory's sbt.g8 template. I think this is one of the best alternatives to starting new projects when you have a good notion of what you want in all your projects.
$ g8 ymasory/sbt
project_license_url [http://www.gnu.org/licenses/gpl-3.0.txt]:
name [myproj]:
project_group_id [com.example]:
developer_email [john.doe#example.com]:
developer_full_name [John Doe]:
project_license_name [GPLv3]:
github_username [johndoe]:
Template applied in ./myproj
$ tree myproj
myproj
├── build.sbt
├── LICENSE
├── project
│ ├── build.properties
│ ├── build.scala
│ └── plugins.sbt
├── README.md
├── sbt
└── src
└── main
└── scala
└── Main.scala
4 directories, 8 files
np plugin
Use softprops's np plugin for sbt. In the example below, the plugin is configured on ~/.sbt/plugins/build.sbt, and its settings on ~/.sbt/np.sbt, with standard sbt script. If you use paulp's sbt-extras, you'd need to install these things under the right Scala version subdirectory in ~/.sbt, as it uses separate configurations for each Scala version. In practice, this is the one I use most often.
$ mkdir myproj; cd myproj
$ sbt 'np name:myproj org:com.example'
[info] Loading global plugins from /home/dcsobral/.sbt/plugins
[warn] Multiple resolvers having different access mechanism configured with same name 'sbt-plugin-releases'. To avoid conflict, Remove duplicate project resolvers (`resolvers`) or rename publishing resolver (`publishTo`).
[info] Set current project to default-c642a2 (in build file:/home/dcsobral/myproj/)
[info] Generated build file
[info] Generated source directories
[success] Total time: 0 s, completed Apr 12, 2013 12:08:31 PM
$ tree
.
├── build.sbt
├── src
│ ├── main
│ │ ├── resources
│ │ └── scala
│ └── test
│ ├── resources
│ └── scala
└── target
└── streams
└── compile
└── np
└── $global
└── out
12 directories, 2 files
mkdir
You could simply create it with mkdir:
$ mkdir -p myproj/src/{main,test}/{resource,scala,java}
$ tree myproj
myproj
└── src
├── main
│ ├── java
│ ├── resource
│ └── scala
└── test
├── java
├── resource
└── scala
9 directories, 0 files
Source Layout
Now, about the source layout. Jens recommends following Java style. Well, the Java directory layout is a requirement -- in Java. Scala does not have the same requirement, so you have the option of following it or not.
If you do follow it, assuming the base package is org.dcsobral.myproject, then source code for that package would be put inside myproject/src/main/scala/org/dcsobral/myproject/, and so on for sub-packages.
Two common ways of diverging from that standard are:
Omitting the base package directory, and only creating subdirectories for the sub-packages.
For instance, let's say I have the packages org.dcsobral.myproject.model, org.dcsobral.myproject.view and org.dcsobral.myproject.controller, then the directories would be myproject/src/main/scala/model, myproject/src/main/scala/view and myproject/src/main/scala/controller.
Putting everything together. In this case, all source files would be inside myproject/src/main/scala. This is good enough for small projects. In fact, if you have no sub-projects, it is the same as above.
And this deals with directory layout.
File Names
Next, let's talk about files. In Java, the practice is separating each class in its own file, whose name will follow the name of the class. This is good enough in Scala too, but you have to pay attention to some exceptions.
First, Scala has object, which Java does not have. A class and object of the same name are considered companions, which has some practical implications, but only if they are in the same file. So, place companion classes and objects in the same file.
Second, Scala has a concept known as sealed class (or trait), which limits subclasses (or implementing objects) to those declared in the same file. This is mostly done to create algebraic data types with pattern matching with exhaustiveness check. For example:
sealed abstract class Tree
case class Node(left: Tree, right: Tree) extends Tree
case class Leaf(n: Int) extends Tree
scala> def isLeaf(t: Tree) = t match {
| case Leaf(n: Int) => println("Leaf "+n)
| }
<console>:11: warning: match is not exhaustive!
missing combination Node
def isLeaf(t: Tree) = t match {
^
isLeaf: (t: Tree)Unit
If Tree was not sealed, then anyone could extend it, making it impossible for the compiler to know whether the match was exhaustive or not. Anyway, sealed classes go together in the same file.
Another naming convention is to name the files containing a package object (for that package) package.scala.
Importing Stuff
The most basic rule is that stuff in the same package see each other. So, put everything in the same package, and you don't need to concern yourself with what sees what.
But Scala also have relative references and imports. This requires a bit of an explanation. Say I have the following declarations at the top of my file:
package org.dcsobral.myproject
package model
Everything following will be put in the package org.dcsobral.myproject.model. Also, not only everything inside that package will be visible, but everything inside org.dcsobral.myproject will be visible as well. If I just declared package org.dcsobral.myproject.model instead, then org.dcsobral.myproject would not be visible.
The rule is pretty simple, but it can confuse people a bit at first. The reason for this rule is relative imports. Consider now the following statement in that file:
import view._
This import may be relative -- all imports can be relative unless you prefix it with _root_.. It can refer to the following packages: org.dcsobral.myproject.model.view, org.dcsobral.myproject.view, scala.view and java.lang.view. It could also refer to an object named view inside scala.Predef. Or it could be an absolute import refering to a package named view.
If more than one such package exists, it will pick one according to some precedence rules. If you needed to import something else, you can turn the import into an absolute one.
This import makes everything inside the view package (wherever it is) visible in its scope. If it happens inside a class, and object or a def, then the visibility will be restricted to that. It imports everything because of the ._, which is a wildcard.
An alternative might look like this:
package org.dcsobral.myproject.model
import org.dcsobral.myproject.view
import org.dcsobral.myproject.controller
In that case, the packages view and controller would be visible, but you'd have to name them explicitly when using them:
def post(view: view.User): Node =
Or you could use further relative imports:
import view.User
The import statement also enable you to rename stuff, or import everything but something. Refer to relevant documentation about it for more details.
So, I hope this answer all your questions.
Scala supports and encourages the package structure of Java /JVM and pretty much the same recommendation apply:
mirror the package structure in the directory structure. This isn't necessary in Scala, but it helps to find your way around
use your inverse domain as a package prefix. For me that means everything starts with de.schauderhaft. Use something that makes sense for you, if you don't have you own domain
only put top level classes in one file if they are small and closely related. Otherwise stick with one class/object per file. Exceptions: companion objects go in the same file as the class. Implementations of a sealed class go into the same file.
if you app grows you might want to have something like layers and modules and mirror those in the package structure, so you might have a package structure like this: <domain>.<module>.<layer>.<optional subpackage>.
don't have cyclic dependencies on a package, module or layer level
TLDR: What is the simplest way to create an importable file in Scala?
This is a beginner's question.
I've been learning Scala for a few weeks and now have some code that I would like to share between a couple of files/different projects. I have tried a lot of import structures but none of them worked. The requirements are:
File to be imported should reside in a totally different directory.
File to be imported should be importable by independent projects.
File to be imported is a single .scala file.
(Optional) file to be imported should contain defs, objects and case classes.
Example:
File to be imported location: /some/path/to_be_imported.scala.
File using project (1) location: /abc/def/will_import01.scala.
File using project (2) location: /xyz/rst/will_import02.scala.
I'm not trying to create a package or distribute it.
See how I would address this considering the programming language I already know:
Since I'm versed in Python I'll give an expected version of the answer should this problem refer o Python:
In that case you could:
Put your file on the same directory of your executed file then just run: python3 ./your_file.py. For instance:
➜ another_path|$ python3 ./main_module/main_file.py
1
self printing
➜ another_path|$ tree .
=======================================================================
.
└── main_module
├── main_file.py
├── __pycache__
│ └── sample_file_to_be_imported.cpython-36.pyc
└── sample_file_to_be_imported.py
Notice that they are in the exact same directory (this contradicts point 2 above nevertheless it solves the problem).
Add the directory of your file to the PYTHONPATH environment variable then run your module (best answer):
➜ random_path|$ PYTHONPATH=$PYTHONPATH:./sample_module python3 ./main_module/main_file.py
1
self printing
=======================================================================
➜ random_path|$ tree .
.
├── main_module
│ └── main_file.py
└── sample_module
├── __pycache__
│ └── sample_file_to_be_imported.cpython-36.pyc
└── sample_file_to_be_imported.py
3 directories, 3 files
Content of the files:
➜ random_path|$ cat ./main_module/main_file.py
from sample_file_to_be_imported import func1, Class01
print(func1())
x = Class01()
x.cprint()
=======================================================================
➜ random_path|$ cat ./sample_module/sample_file_to_be_imported.py
def func1():
return 1
class Class01():
def cprint(self):
print('self printing')
Edit 01: #felipe-rubin answer does not work:
$ scala -cp /tmp/scala_stack_exchange/ myprogram.scala
/tmp/scala_stack_exchange/path01/myprogram.scala:3: error: not found: value Timer
val x = Timer(1)
^
one error found
=======================================================================
➜ path01 tree /tmp/scala_stack_exchange
/tmp/scala_stack_exchange
├── anotherpath
│ ├── Timer.class
│ └── timer.scala
└── path01
└── myprogram.scala
2 directories, 3 files
=======================================================================
$ cat /tmp/scala_stack_exchange/anotherpath/timer.scala
class Timer(a: Int) {
def time(): Unit = println("time this")
}
=======================================================================
$ cat /tmp/scala_stack_exchange/path01/myprogram.scala
import anotherpath.Timer
val x = Timer(1)
x.time()
The simplest way would be to compile a .scala file with scalac:
Linux/OSX: scalac mypackage/Example.scala
Windows: scalac mypackage\Example.scala
The above should generate a .class file (or more).
Assuming the file contains a class called Example you can import it somewhere else like this:
import mypackage.Example
When compiling another file which does the above import, you will need to have 'mypackage' in the classpath. You can add directories to the classpath when calling scalac by using the -cp flag like:
Linux/OSX: scalac -cp .:path/to/folder/where/mypackage/is/located AnotherExample.scala
Windows: scalac -cp .;path\to\folder\where\mypackage\is\located AnotherExample.scala
Doing this for bigger projects gets complicated, in which case you might resort to a build tool (e.g. SBT) or an IDE (e.g. IntelliJ Idea) to do the complicated work for you.
Other notes:
If you don't have scalac, you can get it from the scala website ('download binaries' option)
the -cp flag stand for "classpath". There is also a -classpath flag which does the same thing
Welcome to Scala :)
I finally got this working. Thanks for the valuable input from the other answers.
I have diversified the name of every path, file and object to be as general as possible. This probably does not follow the guidelines of the scala community but is the most explicit, illustrative help I could find. Project layout:
File Layout
$ tree /tmp/scala_stack_exchange
/tmp/scala_stack_exchange
├── anotherpath
│ ├── file_defines_class.scala
│ └── some_package_name
│ ├── MyObj.class
│ └── MyObj$.class
└── path01
└── myprogram.scala
3 directories, 4 files
Where I want to run myprogram.scala which should import classes defined in file_defines_class.scala.
Preparation
Compile the file you want to be imported by other modules:
cd /tmp/scala_stack_exchange/anotherpath && scalac ./file_defines_class.scala
Execution
cd /tmp/scala_stack_exchange/path01 && scala -cp /tmp/scala_stack_exchange/anotherpath/ ./myprogram.scala
Results
myobj time
Contents of the files
// /tmp/scala_stack_exchange/path01/myprogram.scala
import some_package_name.MyObj
val x = new MyObj(10)
x.time()
// /tmp/scala_stack_exchange/anotherpath/file_defines_class.scala
package some_package_name
object MyObj
case class MyObj(i: Int) {
def time(): Unit = println("myobj time")
}
Feels like magic. However this whole process is rather cumbersome :/
I'm using kotlin on command line and I'm getting import errors error: unresolved reference: ConnectionHandler
Following is my directory tree:
$ tree .
.
├── LICENSE
├── main.jar
├── main.kt
└── server
├── ConnectionHandler.class
├── ConnectionHandler.kt
├── HttpRequest.class
└── HttpRequest.kt
1 directory, 7 files
When I run kotlinc main.kt -include-runtime -d main.jar I get
main.kt:2:8: error: unresolved reference: server
import server.*
I have declared package server in both of server/ConnectionHandler.kt and server/HttpRequest.kt
Note: The META-INF folder is missing. It is not regenerated either on subsequent compilations.
What am I doing wrong? If it has anything to do with META-INF folder, how can I regenerate it?
You need to include all your source files when using kotlinc (or at least include the compiled classes on the classpath). e.g.
kotlinc main.kt server/ConnectionHandler.kt server/HttpRequest.kt -include-runtime -d main.jar
I am learning Scala now and I want to write some silly little app like a console Twitter client, or whatever. The question is, how to structure application on disk and logically. I know python, and there I would just create some files with classes and then import them in the main module like import util.ssh or from tweets import Retweet (strongly hoping you wouldn't mind that names, they are just for reference). But how should I do this stuff using Scala? Also, I have not much experience with JVM and Java, so I am a complete newbie here.
I'm going to disagree with Jens, here, though not all that much.
Project Layout
My own suggestion is that you model your efforts on Maven's standard directory layout.
Previous versions of SBT (before SBT 0.9.x) would create it automatically for you:
dcs#ayanami:~$ mkdir myproject
dcs#ayanami:~$ cd myproject
dcs#ayanami:~/myproject$ sbt
Project does not exist, create new project? (y/N/s) y
Name: myproject
Organization: org.dcsobral
Version [1.0]:
Scala version [2.7.7]: 2.8.1
sbt version [0.7.4]:
Getting Scala 2.7.7 ...
:: retrieving :: org.scala-tools.sbt#boot-scala
confs: [default]
2 artifacts copied, 0 already retrieved (9911kB/134ms)
Getting org.scala-tools.sbt sbt_2.7.7 0.7.4 ...
:: retrieving :: org.scala-tools.sbt#boot-app
confs: [default]
15 artifacts copied, 0 already retrieved (4096kB/91ms)
[success] Successfully initialized directory structure.
Getting Scala 2.8.1 ...
:: retrieving :: org.scala-tools.sbt#boot-scala
confs: [default]
2 artifacts copied, 0 already retrieved (15118kB/160ms)
[info] Building project myproject 1.0 against Scala 2.8.1
[info] using sbt.DefaultProject with sbt 0.7.4 and Scala 2.7.7
> quit
[info]
[info] Total session time: 8 s, completed May 6, 2011 12:31:43 PM
[success] Build completed successfully.
dcs#ayanami:~/myproject$ find . -type d -print
.
./project
./project/boot
./project/boot/scala-2.7.7
./project/boot/scala-2.7.7/lib
./project/boot/scala-2.7.7/org.scala-tools.sbt
./project/boot/scala-2.7.7/org.scala-tools.sbt/sbt
./project/boot/scala-2.7.7/org.scala-tools.sbt/sbt/0.7.4
./project/boot/scala-2.7.7/org.scala-tools.sbt/sbt/0.7.4/compiler-interface-bin_2.7.7.final
./project/boot/scala-2.7.7/org.scala-tools.sbt/sbt/0.7.4/compiler-interface-src
./project/boot/scala-2.7.7/org.scala-tools.sbt/sbt/0.7.4/compiler-interface-bin_2.8.0.RC2
./project/boot/scala-2.7.7/org.scala-tools.sbt/sbt/0.7.4/xsbti
./project/boot/scala-2.8.1
./project/boot/scala-2.8.1/lib
./target
./lib
./src
./src/main
./src/main/resources
./src/main/scala
./src/test
./src/test/resources
./src/test/scala
So you'll put your source files inside myproject/src/main/scala, for the main program, or myproject/src/test/scala, for the tests.
Since that doesn't work anymore, there are some alternatives:
giter8 and sbt.g8
Install giter8, clone ymasory's sbt.g8 template and adapt it to your necessities, and use that. See below, for example, this use of unmodified ymasory's sbt.g8 template. I think this is one of the best alternatives to starting new projects when you have a good notion of what you want in all your projects.
$ g8 ymasory/sbt
project_license_url [http://www.gnu.org/licenses/gpl-3.0.txt]:
name [myproj]:
project_group_id [com.example]:
developer_email [john.doe#example.com]:
developer_full_name [John Doe]:
project_license_name [GPLv3]:
github_username [johndoe]:
Template applied in ./myproj
$ tree myproj
myproj
├── build.sbt
├── LICENSE
├── project
│ ├── build.properties
│ ├── build.scala
│ └── plugins.sbt
├── README.md
├── sbt
└── src
└── main
└── scala
└── Main.scala
4 directories, 8 files
np plugin
Use softprops's np plugin for sbt. In the example below, the plugin is configured on ~/.sbt/plugins/build.sbt, and its settings on ~/.sbt/np.sbt, with standard sbt script. If you use paulp's sbt-extras, you'd need to install these things under the right Scala version subdirectory in ~/.sbt, as it uses separate configurations for each Scala version. In practice, this is the one I use most often.
$ mkdir myproj; cd myproj
$ sbt 'np name:myproj org:com.example'
[info] Loading global plugins from /home/dcsobral/.sbt/plugins
[warn] Multiple resolvers having different access mechanism configured with same name 'sbt-plugin-releases'. To avoid conflict, Remove duplicate project resolvers (`resolvers`) or rename publishing resolver (`publishTo`).
[info] Set current project to default-c642a2 (in build file:/home/dcsobral/myproj/)
[info] Generated build file
[info] Generated source directories
[success] Total time: 0 s, completed Apr 12, 2013 12:08:31 PM
$ tree
.
├── build.sbt
├── src
│ ├── main
│ │ ├── resources
│ │ └── scala
│ └── test
│ ├── resources
│ └── scala
└── target
└── streams
└── compile
└── np
└── $global
└── out
12 directories, 2 files
mkdir
You could simply create it with mkdir:
$ mkdir -p myproj/src/{main,test}/{resource,scala,java}
$ tree myproj
myproj
└── src
├── main
│ ├── java
│ ├── resource
│ └── scala
└── test
├── java
├── resource
└── scala
9 directories, 0 files
Source Layout
Now, about the source layout. Jens recommends following Java style. Well, the Java directory layout is a requirement -- in Java. Scala does not have the same requirement, so you have the option of following it or not.
If you do follow it, assuming the base package is org.dcsobral.myproject, then source code for that package would be put inside myproject/src/main/scala/org/dcsobral/myproject/, and so on for sub-packages.
Two common ways of diverging from that standard are:
Omitting the base package directory, and only creating subdirectories for the sub-packages.
For instance, let's say I have the packages org.dcsobral.myproject.model, org.dcsobral.myproject.view and org.dcsobral.myproject.controller, then the directories would be myproject/src/main/scala/model, myproject/src/main/scala/view and myproject/src/main/scala/controller.
Putting everything together. In this case, all source files would be inside myproject/src/main/scala. This is good enough for small projects. In fact, if you have no sub-projects, it is the same as above.
And this deals with directory layout.
File Names
Next, let's talk about files. In Java, the practice is separating each class in its own file, whose name will follow the name of the class. This is good enough in Scala too, but you have to pay attention to some exceptions.
First, Scala has object, which Java does not have. A class and object of the same name are considered companions, which has some practical implications, but only if they are in the same file. So, place companion classes and objects in the same file.
Second, Scala has a concept known as sealed class (or trait), which limits subclasses (or implementing objects) to those declared in the same file. This is mostly done to create algebraic data types with pattern matching with exhaustiveness check. For example:
sealed abstract class Tree
case class Node(left: Tree, right: Tree) extends Tree
case class Leaf(n: Int) extends Tree
scala> def isLeaf(t: Tree) = t match {
| case Leaf(n: Int) => println("Leaf "+n)
| }
<console>:11: warning: match is not exhaustive!
missing combination Node
def isLeaf(t: Tree) = t match {
^
isLeaf: (t: Tree)Unit
If Tree was not sealed, then anyone could extend it, making it impossible for the compiler to know whether the match was exhaustive or not. Anyway, sealed classes go together in the same file.
Another naming convention is to name the files containing a package object (for that package) package.scala.
Importing Stuff
The most basic rule is that stuff in the same package see each other. So, put everything in the same package, and you don't need to concern yourself with what sees what.
But Scala also have relative references and imports. This requires a bit of an explanation. Say I have the following declarations at the top of my file:
package org.dcsobral.myproject
package model
Everything following will be put in the package org.dcsobral.myproject.model. Also, not only everything inside that package will be visible, but everything inside org.dcsobral.myproject will be visible as well. If I just declared package org.dcsobral.myproject.model instead, then org.dcsobral.myproject would not be visible.
The rule is pretty simple, but it can confuse people a bit at first. The reason for this rule is relative imports. Consider now the following statement in that file:
import view._
This import may be relative -- all imports can be relative unless you prefix it with _root_.. It can refer to the following packages: org.dcsobral.myproject.model.view, org.dcsobral.myproject.view, scala.view and java.lang.view. It could also refer to an object named view inside scala.Predef. Or it could be an absolute import refering to a package named view.
If more than one such package exists, it will pick one according to some precedence rules. If you needed to import something else, you can turn the import into an absolute one.
This import makes everything inside the view package (wherever it is) visible in its scope. If it happens inside a class, and object or a def, then the visibility will be restricted to that. It imports everything because of the ._, which is a wildcard.
An alternative might look like this:
package org.dcsobral.myproject.model
import org.dcsobral.myproject.view
import org.dcsobral.myproject.controller
In that case, the packages view and controller would be visible, but you'd have to name them explicitly when using them:
def post(view: view.User): Node =
Or you could use further relative imports:
import view.User
The import statement also enable you to rename stuff, or import everything but something. Refer to relevant documentation about it for more details.
So, I hope this answer all your questions.
Scala supports and encourages the package structure of Java /JVM and pretty much the same recommendation apply:
mirror the package structure in the directory structure. This isn't necessary in Scala, but it helps to find your way around
use your inverse domain as a package prefix. For me that means everything starts with de.schauderhaft. Use something that makes sense for you, if you don't have you own domain
only put top level classes in one file if they are small and closely related. Otherwise stick with one class/object per file. Exceptions: companion objects go in the same file as the class. Implementations of a sealed class go into the same file.
if you app grows you might want to have something like layers and modules and mirror those in the package structure, so you might have a package structure like this: <domain>.<module>.<layer>.<optional subpackage>.
don't have cyclic dependencies on a package, module or layer level