Linking with Libraries

Usually, you build a library because you want to link it with one or more programs. You link libraries with a program by specifying the libraries in the $LIBS construction variable, and by specifying the directory in which the library will be found in the $LIBPATH construction variable:

      Library('foo', ['f1.c', 'f2.c', 'f3.c'])
      Program('prog.c', LIBS=['foo', 'bar'], LIBPATH='.')
    

Notice, of course, that you don't need to specify a library prefix (like lib) or suffix (like .a or .lib). SCons uses the correct prefix or suffix for the current system.

On a POSIX or Linux system, a build of the above example would look like:

      % scons -Q
      cc -o f1.o -c f1.c
      cc -o f2.o -c f2.c
      cc -o f3.o -c f3.c
      ar rc libfoo.a f1.o f2.o f3.o
      ranlib libfoo.a
      cc -o prog.o -c prog.c
      cc -o prog prog.o -L. -lfoo -lbar
    

On a Windows system, a build of the above example would look like:

      C:\>scons -Q
      cl /nologo /c f1.c /Fof1.obj
      cl /nologo /c f2.c /Fof2.obj
      cl /nologo /c f3.c /Fof3.obj
      lib /nologo /OUT:foo.lib f1.obj f2.obj f3.obj
      cl /nologo /c prog.c /Foprog.obj
      link /nologo /OUT:prog.exe /LIBPATH:. foo.lib bar.lib prog.obj
    

As usual, notice that SCons has taken care of constructing the correct command lines to link with the specified library on each system.

Note also that, if you only have a single library to link with, you can specify the library name in single string, instead of a Python list, so that:

      Program('prog.c', LIBS='foo', LIBPATH='.')
    

is equivalent to:

      Program('prog.c', LIBS=['foo'], LIBPATH='.')
    

This is similar to the way that SCons handles either a string or a list to specify a single source file.