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Makefile strange variable substitution

2

My Makefile looks like this:

%.foo: %.bar

    cp $< $@



test: *.foo

    echo *.foo

I have 2 files in the directory: a.bar and b.bar. When I run make test it outputs:

cp b.bar *.foo

echo *.foo

*.foo

It also creates a file *.foo in the current directory.

I am actually expecting to see this:

cp a.bar a.foo

cp b.bar b.foo

echo *.foo

a.foo b.foo

And also creating a.foo and b.foo. How to achieve that?

make

share|improve this question

asked 3 hours ago

Martin ŽdilaMartin Ždila

155115

add a comment | 

2

My Makefile looks like this:

%.foo: %.bar

    cp $< $@



test: *.foo

    echo *.foo

I have 2 files in the directory: a.bar and b.bar. When I run make test it outputs:

cp b.bar *.foo

echo *.foo

*.foo

It also creates a file *.foo in the current directory.

I am actually expecting to see this:

cp a.bar a.foo

cp b.bar b.foo

echo *.foo

a.foo b.foo

And also creating a.foo and b.foo. How to achieve that?

make

share|improve this question

asked 3 hours ago

Martin ŽdilaMartin Ždila

155115

add a comment | 

My Makefile looks like this:

%.foo: %.bar

    cp $< $@



test: *.foo

    echo *.foo

I have 2 files in the directory: a.bar and b.bar. When I run make test it outputs:

cp b.bar *.foo

echo *.foo

*.foo

It also creates a file *.foo in the current directory.

I am actually expecting to see this:

cp a.bar a.foo

cp b.bar b.foo

echo *.foo

a.foo b.foo

And also creating a.foo and b.foo. How to achieve that?

make

share|improve this question

asked 3 hours ago

Martin ŽdilaMartin Ždila

155115

%.foo: %.bar

    cp $< $@



test: *.foo

    echo *.foo

cp b.bar *.foo

echo *.foo

*.foo

cp a.bar a.foo

cp b.bar b.foo

echo *.foo

a.foo b.foo

share|improve this question

asked 3 hours ago

Martin ŽdilaMartin Ždila

155115

share|improve this question

asked 3 hours ago

Martin ŽdilaMartin Ždila

155115

asked 3 hours ago

Martin ŽdilaMartin Ždila

155115

asked 3 hours ago

Martin ŽdilaMartin Ždila

155115

2 Answers
2

active

oldest

votes

3

In this case you need to handle wildcards explicitly, with the wildcard function (at least in GNU Make):

%.foo: %.bar

    cp $< $@



foos = $(patsubst %.bar,%.foo,$(wildcard *.bar))



test: $(foos)

    echo $(foos)

$(wildcard *.bar) expands to all the files ending in .bar, the patsubst call replaces .bar with .foo, and all the targets are then processed as you’d expect.

share|improve this answer

answered 3 hours ago

Stephen KittStephen Kitt

175k24400478

add a comment | 

3

There is no *.foo file to begin with. So what make does is look for how to make *.foo literaly and the first rule does this. Make expands $< to the first pre-requisite (*.bar, which happens to be b.bar in this case). Make then runs the shell command cp b.bar *.foo. Since there is no *.foo, shell expands it to cp b.bar *.foo literally. That's how you get a *.foo file.

You can verify this by running make -d test.

You can get the effect you want by generating the list of targets based on list of prerequisites.

TARGETS = $(patsubst %.bar,%.foo,$(wildcard *.bar))

%.foo: %.bar

    @cp $< $@

test: $(TARGETS)

    @echo $(TARGETS)

    echo *.foo

share|improve this answer

answered 3 hours ago

Satya MishraSatya Mishra

35615

add a comment | 

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3

In this case you need to handle wildcards explicitly, with the wildcard function (at least in GNU Make):

%.foo: %.bar

    cp $< $@



foos = $(patsubst %.bar,%.foo,$(wildcard *.bar))



test: $(foos)

    echo $(foos)

$(wildcard *.bar) expands to all the files ending in .bar, the patsubst call replaces .bar with .foo, and all the targets are then processed as you’d expect.

share|improve this answer

answered 3 hours ago

Stephen KittStephen Kitt

175k24400478

add a comment | 

3

In this case you need to handle wildcards explicitly, with the wildcard function (at least in GNU Make):

%.foo: %.bar

    cp $< $@



foos = $(patsubst %.bar,%.foo,$(wildcard *.bar))



test: $(foos)

    echo $(foos)

$(wildcard *.bar) expands to all the files ending in .bar, the patsubst call replaces .bar with .foo, and all the targets are then processed as you’d expect.

share|improve this answer

answered 3 hours ago

Stephen KittStephen Kitt

175k24400478

add a comment | 

In this case you need to handle wildcards explicitly, with the wildcard function (at least in GNU Make):

%.foo: %.bar

    cp $< $@



foos = $(patsubst %.bar,%.foo,$(wildcard *.bar))



test: $(foos)

    echo $(foos)

$(wildcard *.bar) expands to all the files ending in .bar, the patsubst call replaces .bar with .foo, and all the targets are then processed as you’d expect.

share|improve this answer

answered 3 hours ago

Stephen KittStephen Kitt

175k24400478

%.foo: %.bar

    cp $< $@



foos = $(patsubst %.bar,%.foo,$(wildcard *.bar))



test: $(foos)

    echo $(foos)

share|improve this answer

answered 3 hours ago

Stephen KittStephen Kitt

175k24400478

answered 3 hours ago

Stephen KittStephen Kitt

175k24400478

answered 3 hours ago

Stephen KittStephen Kitt

175k24400478

3

There is no *.foo file to begin with. So what make does is look for how to make *.foo literaly and the first rule does this. Make expands $< to the first pre-requisite (*.bar, which happens to be b.bar in this case). Make then runs the shell command cp b.bar *.foo. Since there is no *.foo, shell expands it to cp b.bar *.foo literally. That's how you get a *.foo file.

You can verify this by running make -d test.

You can get the effect you want by generating the list of targets based on list of prerequisites.

TARGETS = $(patsubst %.bar,%.foo,$(wildcard *.bar))

%.foo: %.bar

    @cp $< $@

test: $(TARGETS)

    @echo $(TARGETS)

    echo *.foo

share|improve this answer

answered 3 hours ago

Satya MishraSatya Mishra

35615

add a comment | 

3

There is no *.foo file to begin with. So what make does is look for how to make *.foo literaly and the first rule does this. Make expands $< to the first pre-requisite (*.bar, which happens to be b.bar in this case). Make then runs the shell command cp b.bar *.foo. Since there is no *.foo, shell expands it to cp b.bar *.foo literally. That's how you get a *.foo file.

You can verify this by running make -d test.

You can get the effect you want by generating the list of targets based on list of prerequisites.

TARGETS = $(patsubst %.bar,%.foo,$(wildcard *.bar))

%.foo: %.bar

    @cp $< $@

test: $(TARGETS)

    @echo $(TARGETS)

    echo *.foo

share|improve this answer

answered 3 hours ago

Satya MishraSatya Mishra

35615

add a comment | 

There is no *.foo file to begin with. So what make does is look for how to make *.foo literaly and the first rule does this. Make expands $< to the first pre-requisite (*.bar, which happens to be b.bar in this case). Make then runs the shell command cp b.bar *.foo. Since there is no *.foo, shell expands it to cp b.bar *.foo literally. That's how you get a *.foo file.

You can verify this by running make -d test.

You can get the effect you want by generating the list of targets based on list of prerequisites.

TARGETS = $(patsubst %.bar,%.foo,$(wildcard *.bar))

%.foo: %.bar

    @cp $< $@

test: $(TARGETS)

    @echo $(TARGETS)

    echo *.foo

share|improve this answer

answered 3 hours ago

Satya MishraSatya Mishra

35615

TARGETS = $(patsubst %.bar,%.foo,$(wildcard *.bar))

%.foo: %.bar

    @cp $< $@

test: $(TARGETS)

    @echo $(TARGETS)

    echo *.foo

share|improve this answer

answered 3 hours ago

Satya MishraSatya Mishra

35615

answered 3 hours ago

Satya MishraSatya Mishra

35615

answered 3 hours ago

Satya MishraSatya Mishra

35615