我想知道为什么在执行 .jar 文件时scala不需要-jar, while java?
例如:
scala target/some.jar
java -jar target/someother.jar
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我查看了scalashell 脚本,它调用java如下:
execCommand \
"${JAVACMD:=java}" \
$JAVA_OPTS \
"${java_args[@]}" \
"${classpath_args[@]}" \
-Dscala.home="$SCALA_HOME" \
$OVERRIDE_USEJAVACP \
$WINDOWS_OPT \
scala.tools.nsc.MainGenericRunner "$@"
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难道纠正scala通行证target/some.jar作为一部分"$@"来java,因此java命令没有前缀 target/some.jar用-jar?
谢谢。
分析MainGenericRunner我们看到scalarunner的源代码试图猜测用户的意图
def runTarget(): Option[Throwable] = howToRun match {
case AsObject =>
ObjectRunner.runAndCatch(settings.classpathURLs, thingToRun, command.arguments)
case AsScript if isE =>
ScriptRunner(settings).runScriptText(combinedCode, thingToRun +: command.arguments)
case AsScript =>
ScriptRunner(settings).runScript(thingToRun, command.arguments)
case AsJar =>
JarRunner.runJar(settings, thingToRun, command.arguments)
case Error =>
None
case _ =>
...
}
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其中howToRun潜在的呼吁guessHowToRun
private def guessHowToRun(target: String): GenericRunnerCommand.HowToRun = {
if (!ok) Error
else if (io.Jar.isJarOrZip(target)) AsJar
else if (ScalaClassLoader.classExists(settings.classpathURLs, target)) AsObject
else {
val f = io.File(target)
if (!f.hasExtension("class", "jar", "zip") && f.canRead) AsScript
else {
Console.err.println("No such file or class on classpath: " + target)
Error
}
}
}
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我们可以通过提供-howtorun标志来明确:
scala -howtorun:jar some.jar
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Scala 应用程序也可以通过 运行java -jar,例如,如果我们在build.sbt
Compile / mainClass := Some("example.MyMain")
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并执行sbt assembly,那么下面生成的fat jartarget/就可以这样执行了
java -jar myapp-assembly.jar
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