CODEWITHSUNDEEP
makefilegnu-make
mpen
mpen

Reputation: 282895

Make from *either* source file

I have this make rule:

dist/node/%.js: src/%.node.js yarn.lock .babelrc | $(NODE_DIST_DIRS)
    $(BIN)/babel $< -o $@

It works fine when my source file ends with .node.js. e.g.,

$ make dist/node/fs.js
mkdir -p dist/node/
node_modules/.bin/babel src/fs.node.js -o dist/node/fs.js

However, I want to compile dist/node/fs.js from either src/fs.node.js or src/fs.js, whichever exists.

The reason for this is that I have some shared files with just the .js extension (work in both browser and node), and then more specific files with the .node.js extension. If there's a more specific version, I want to use that.

I don't know how to do conditional dependencies in combination with %. Is this possible? Can I give precedence to dependencies and take the best match?

Upvotes: 1

Views: 52

Answers (2)

MadScientist
MadScientist

Reputation: 100856

There is no way to do that in a single rule. You'll have to write two rules with different prerequisites but otherwise the same:

dist/node/%.js: src/%.node.js yarn.lock .babelrc | $(NODE_DIST_DIRS)
        $(BIN)/babel $< -o $@

dist/node/%.js: src/%.js yarn.lock .babelrc | $(NODE_DIST_DIRS)
        $(BIN)/babel $< -o $@

Upvotes: 2

reinierpost
reinierpost

Reputation: 8591

I suppose you could try using a double-colon rule, but those cannot be pattern rules.

As an alternative, consider generating and including a makefile containing just those dependencies.

Upvotes: 1

Related Questions