Makefile strange variable substitution












1















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



























    1















    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

























      1












      1








      1








      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














      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




      make






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 1 hour ago









      Martin ŽdilaMartin Ždila

      150115




      150115






















          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































            2














            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























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              2 Answers
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              2 Answers
              2






              active

              oldest

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              active

              oldest

              votes






              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




























                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


























                  3












                  3








                  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













                  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












                  share|improve this answer



                  share|improve this answer










                  answered 59 mins ago









                  Stephen KittStephen Kitt

                  175k24400478




                  175k24400478

























                      2














                      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




























                        2














                        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


























                          2












                          2








                          2







                          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













                          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












                          share|improve this answer



                          share|improve this answer










                          answered 45 mins ago









                          Satya MishraSatya Mishra

                          34615




                          34615






























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