Re: $(EXEEXT) in TESTS required?

[Bruno Haible]

Ralf Wildenhues wrote:
> > check_PROGRAMS = test-lock$(EXEEXT)
> >
> > TESTS = test-lock
> ...
> What about @substituted@ values?

TESTS = @substituted@

You could treat it like @substituted@ in check_PROGRAMS, namely
  - assume that $(EXEEXT) is contained in the substituted value,
  - warn if EXTRA_PROGRAMS is not defined. (For
      check_PROGRAMS = @substituted@
    automake gives an error if EXTRA_PROGRAMS is not defined; I think here
    a warning would be appropriate).

> > From 2) I conclude that it would even be acceptable to change the TESTS
> > variable itself, to assume the contents of am_TESTS_DEPENDENCIES.
>
> If I see correctly that would allow Automake to kill this line
> completely (were it not for my question above):
> >     $(MAKE) $(AM_MAKEFLAGS) $(check_PROGRAMS)

Yes, I agree. The current wording in the automake doc doesn't say
that all $(check_PROGRAMS) are build before all tests are executed, therefore
someone who writes

    check_PROGRAMS = foo bar    # foo, when run, calls bar
    TESTS = foo

is already relying on "make" to have built the 'bar' program;
"make check" will not check its status again.

Bruno