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Home/ Questions/Q 3487154
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T11:06:56+00:00 2026-05-18T11:06:56+00:00

Suppose I have a Makefile: all: $(BINARY) $(BINARY): $(OBJS) $(DEBUG_OBJS) #Link objects here $(OBJS):

  • 0

Suppose I have a Makefile:

all: $(BINARY)

$(BINARY): $(OBJS) $(DEBUG_OBJS)
    #Link objects here

$(OBJS): headers
    #Compile code into objects without debug option

$(DEBUG_OBJS): headers
    #Compile code into objects with debug option

headers:
    #Create on-the-fly header files

As you can see, the target headers is required by both $(OBJS) and $(DEBUG_OBJS). The question is, will headers be called twice? Also, would the below code be equal/equivalent to the above:

all: $(BINARY)

$(BINARY): headers $(OBJS) $(DEBUG_OBJS)
    #Link objects here

$(OBJS): 
    #Compile code into objects without debug option

$(DEBUG_OBJS): 
    #Compile code into objects with debug option

headers:
    #Create on-the-fly header files

in that, would headers get called before $(OBJS) and $(DEBUG_OBJS) by $(BINARY)?

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  1. Editorial Team
    Editorial Team
    2026-05-18T11:06:56+00:00Added an answer on May 18, 2026 at 11:06 am

    No, headers will be done just once.

    You can write a simple makefile to test it:

    all: foo bar
    
    foo: baz
    
    bar: baz
    
    baz:
            echo 'hi'
    

    On doing make, hi will be echoed just once.

    And in your 2nd case make sees that $(BINARY) depends on headers first, so it goes and does headers before other dependencies.

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