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Home/ Questions/Q 4060004
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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T15:16:58+00:00 2026-05-20T15:16:58+00:00

i have a small java project i want to build using a makefile, the

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i have a small java project i want to build using a makefile, the code is in src/package... /*.java, the bytecode should go to bin/package.../*.class.
My current file looks like this (simplified):

JC = javac
SRCDIR = src
BINDIR = bin
JCFLAGS = -d $(BINDIR)/

CLASSES = $(SRCDIR)/package/class1.java $(SRCDIR)/package/class2.java $(SRCDIR)/package/class3.java 

default:
    $(JC) $(JCFLAGS) $(CLASSES)

It works and does what it should, but there has to be a more elegant way to do this.
For example, is there a way to apply the path ($(SRCDIR) and the package name) as a prefix to all class filenames, so i do not have to put the path seperately in front of every class?

All classes have to be compiled in one javac-call, as there are circular dependencies in them, so using an own target for each class does not work:

default: $(CLASSES)
%.java:
    $(JC) $(JCFLAGS) $(SRCDIR)/$@

Thanks for your help.

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  1. Editorial Team
    Editorial Team
    2026-05-20T15:16:59+00:00Added an answer on May 20, 2026 at 3:16 pm

    From the GNU make manual:

    $(addprefix prefix,names...)
    

    The argument names is regarded as a series of names, separated by whitespace; prefix is used as a unit. The value of prefix is prepended to the front of each individual name and the resulting larger names are concatenated with single spaces between them. For example,

    $(addprefix src/,foo bar)
    

    produces the result ‘src/foo src/bar’.

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