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[SUGGESTION] Having 'make check' use AM_CPPFLAGS
From: |
Stephen Torri |
Subject: |
[SUGGESTION] Having 'make check' use AM_CPPFLAGS |
Date: |
Sat, 11 Oct 2003 22:23:47 -0500 |
When I do the normal build the Makefile variable AM_CPPFLAGS is honored
but it is not when I doing 'make check'. I am trying to compile a
testing file that uses the same include headers as the files which it is
suppose to test. For example:
------------ BEGIN EXAMPLE --------------
LT_LIBRARIES = libFoo.la
libFoo_la_SOURCES = foo.cpp
noinst_HEADERS = foo.h
TESTS = test_Foo
test_Foo_SOURCES = test_Foo.cpp
AM_CPPFLAGS =-I$(top_srcdir)/src/Bar
------------ END EXAMPLE ---------------
So when I do 'make' you will see that the include for Bar's directory is
there in the compile argument. If you do 'make check' you will find that
it is not and therefore test_Foo does not. The reason behind is in
foo.h.
------------ foo.h -----------------
#include <bar.h>
class Foo : public Bar {
};
------------------------------------
The location of bar.h is in the directory $(top_srcdir)/src/Bar. The
attempt to compile fails because bar.h cannot be found in path for basic
includes.
The fix for this was to do:
------------ BEGIN EXAMPLE --------------
LT_LIBRARIES = libFoo.la
noinst_PROGRAMS = test_Foo
libFoo_la_SOURCES = foo.cpp
noinst_HEADERS = foo.h
TESTS = test_Foo
test_Foo_SOURCES = test_Foo.cpp
AM_CPPFLAGS =-I$(top_srcdir)/src/Bar
------------ END EXAMPLE ---------------
By adding noinst_PROGRAMS the AM_CPPFLAGS was honored during 'make' and
because the binary test_Foo was there when doing 'make check' the test
was able to go ahead. I am not sure this is the best semantics to get
the job doing. I am curious to hear opinions on this?
Stephen
--
Stephen Torri
GPG Key: http://www.cs.wustl.edu/~storri/storri.asc
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- [SUGGESTION] Having 'make check' use AM_CPPFLAGS,
Stephen Torri <=