Recently I came across a query while reading THE STRUCTURE OF THE JAVA VIRTUAL MACHINE
On page no-5 ch. 2,
Compiled code to be executed by the Java Virtual Machine is represented using a hardware- and operating system-independent binary format, typically (but not necessarily) stored in a file, known as the class file format.
That bracket But not necessarily is the reason.
Question is,
In which case compiled code will not be stored in class file ? and if it will not be stored in class file, then where and how ?
Edit : Please note that question is not about ClassLoader.
You seem more interest in how to generate a java class file in memory so here goes:
public class CompileSourceInMemory {
public static void main(String args[]) throws IOException {
JavaCompiler compiler = ToolProvider.getSystemJavaCompiler();
DiagnosticCollector<JavaFileObject> diagnostics = new DiagnosticCollector<JavaFileObject>();
StringWriter writer = new StringWriter();
PrintWriter out = new PrintWriter(writer);
out.println("public class HelloWorld {");
out.println(" public static void main(String args[]) {");
out.println(" System.out.println(\"This is in another java file\");");
out.println(" }");
out.println("}");
out.close();
JavaFileObject file = new JavaSourceFromString("HelloWorld", writer.toString());
Iterable<? extends JavaFileObject> compilationUnits = Arrays.asList(file);
CompilationTask task = compiler.getTask(null, null, diagnostics, null, null, compilationUnits);
boolean success = task.call();
if (success) {
try {
Class.forName("HelloWorld").getDeclaredMethod("main", new Class[] { String[].class }).invoke(null, new Object[] { null });
} catch (ClassNotFoundException e) {
System.err.println("Class not found: " + e);
} catch (NoSuchMethodException e) {
System.err.println("No such method: " + e);
} catch (IllegalAccessException e) {
System.err.println("Illegal access: " + e);
} catch (InvocationTargetException e) {
System.err.println("Invocation target: " + e);
}
}
}
}
class JavaSourceFromString extends SimpleJavaFileObject {
final String code;
JavaSourceFromString(String name, String code) {
super(URI.create("string:///" + name.replace('.','/') + Kind.SOURCE.extension),Kind.SOURCE);
this.code = code;
}
@Override
public CharSequence getCharContent(boolean ignoreEncodingErrors) {
return code;
}
}
Source: In memory compilation
This generates a class in memory without any external representation of the java source or compiled class.