Simplify Makefile without changing its logic - makefile

Question

I am wondering is there a way to refactor these 3 lines without changing the logic?
scalac -cp ${SCALA_FLAGS} GrammarUtil.scala
scalac -cp ${SCALA_FLAGS} FirstSpec.scala
scalac -cp ${SCALA_FLAGS} FollowSpec.scala
Makefile
SPECS=FirstSpec FollowSpec
EXAMPLES_PATH=../..
ROOT_PATH=../${EXAMPLES_PATH}
SCALAV=2.12
APSLIB=${ROOT_PATH}/lib/aps-library-${SCALAV}.jar
SCALA_FLAGS=.:${APSLIB}
APS2SCALA=${ROOT_PATH}/bin/aps2scala
.PHONY: all
all: $(addsuffix .run, $(SPECS))
.PHONY: clean
clean:
rm -f *.class grammar.scala first.scala follow.scala
# for anything.scala it compiles it to create class file
%.class: %.scala
scalac -cp ${SCALA_FLAGS} $<
# for anything.scala that does not already exist, it generates it
%.scala:
${APS2SCALA} -DCOT -p ${EXAMPLES_PATH}:${ROOT_PATH}/base $*
# for anything.run it needs to be compiled first before running
%.run: %.class
scala -cp ${SCALA_FLAGS} $(basename $<)
# for GrammarUtil.class it needs to generate and compile grammar.class and then compile itself
GrammarUtil.class: grammar.class
scalac -cp ${SCALA_FLAGS} GrammarUtil.scala
# How to avoid above line: scalac -cp ${SCALA_FLAGS} GrammarUtil.scala
# for FirstSpec.class it needs to compile Spec.class, GrammarUtil.class and compile and then generate first.class
FirstSpec.class: Spec.class GrammarUtil.class first.class
scalac -cp ${SCALA_FLAGS} FirstSpec.scala
# How to avoid above line: scalac -cp ${SCALA_FLAGS} FirstSpec.scala
# for FollowSpec.class it needs to compile Spec.class, GrammarUtil.class and compile and then generate follow.class
FollowSpec.class: Spec.class GrammarUtil.class follow.class
scalac -cp ${SCALA_FLAGS} FollowSpec.scala
# How to avoid above line: scalac -cp ${SCALA_FLAGS} FollowSpec.scala

Answer 1

I hope, that the answer could look like that (put it to some clean directory and just run make, then look at those files with text editor/wiever):
SPECS=FirstSpec FollowSpec
EXAMPLES_PATH=../..
ROOT_PATH=../${EXAMPLES_PATH}
SCALAV=2.12
APSLIB=${ROOT_PATH}/lib/aps-library-${SCALAV}.jar
SCALA_FLAGS=.:${APSLIB}
APS2SCALA=${ROOT_PATH}/bin/aps2scala
.PHONY: all
all: $(addsuffix .run, $(SPECS))
.PHONY: clean
clean:
rm -f *.class grammar.scala first.scala follow.scala
# for anything.scala it compiles it to create class file
# Lets use ALL prequisities ($^) and create target by name (--output $#)
%.class: %.scala
echo "scalac -cp ${SCALA_FLAGS} $^ --output $#" > $#
# for anything.scala that does not already exist, it generates it
%.scala:
echo "${APS2SCALA} -DCOT -p ${EXAMPLES_PATH}:${ROOT_PATH}/base $*" >$#
# for anything.run it needs to be compiled first before running
%.run: %.class
echo "scala -cp ${SCALA_FLAGS} $(basename $<)" >$#
# for GrammarUtil.class it needs to generate and compile grammar.class and then compile itself
# for FirstSpec.class it needs to compile Spec.class, GrammarUtil.class and compile and then generate first.class
# for FollowSpec.class it needs to compile Spec.class, GrammarUtil.class and compile and then generate follow.class
GrammarUtil.class: grammar.class
FirstSpec.class: Spec.class GrammarUtil.class first.class
FollowSpec.class: Spec.class GrammarUtil.class follow.class
I have no scala, so I could not try it, but I let it put commands to create each file to that file for inspection - so it may be enought just to delete those echo " and " >$# to get the desired result.
Also I hope, that the scalacp command have some way to place output to explicitly named file - I used hypotetical --output filename switch for that. Or that is somehow get it from its first file parametr right (so the --outpu filename is not needed)
There are used 3 "tricks"
to use all from dependencies there is $^ variable
to get name of target file is variable $#
there can be more dependencies lines (like those 3 at bottom) and they use the same rule for the target, if it meets the same pattern