building the same file for different targets with different options

Question

I have one "helper" file included in two "main" files which are built into two executables with the same makefile. I have debug print statements in the helper file. I want the print statements to actually be printed in one executable, but not the other. Is there a way to do it? Right now I have the following, and I was hoping to compile with DEBUG_PRINT defined for one executable but not the other, but I don't see how.

main1.cpp:

    #include "helper.h"
    ...

main2.cpp:

    #include "helper.h"
    ...

helper.cpp:

    #ifdef DEBUG_PRINT
    cout << "here is a debug message" << endl;
    #endif

Makefile:

    build: main1 main2
    main1: main1.o helper.o
        g++ -g -o main1 main1.o helper.o
    main2: main2.o helper.o
        g++ -g -o main2 main2.o helper.o
    %.o: %.cpp
        gcc -g -c $<

thiton · Accepted Answer · 2011-10-21 16:57:48Z

4

You will need two different object files (main1-helper.o and main2-helper.o) and target-specific compiler flags:

 main1: CXXFLAGS=-DDEBUG_PRINT
 %.o: %.cpp
      gcc $(CXXFLAGS) -g -o $@ -c $<

Note: This leaves you with the problem of generating main1-helper.o from helper.o. There are a few ways to solve this; however, you might be more comfortable using automake from the start instead of rolling your own solutions.

edited Oct 21, 2011 at 16:57

answered Oct 21, 2011 at 16:42

thiton

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1 Comment

user786653 Over a year ago

+1 I misread the Makefile at first. There's no way around having two different object files aside from moving $(CXXFLAGS) helper.cpp to the g++ link command (which seems kind of dirty).

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