Re: a couple questions about variable assignment and wildcards

[Paul D. Smith]

%% "Robert P. J. Day" <address@hidden> writes:

  rpjd>   objects = *.o
  rpjd>   objects := $(wildcard *.o)

    vs.

  rpjd>   objects = $(wildcard *.o)
  rpjd>   objects := *.o

  rpjd> the first again, i would guess, is a deferred assignment, with
  rpjd> the actual value of '$(wildcard *.o)' which will be evaluated
  rpjd> when used.  how can i print the actual value of a variable
  rpjd> without having it evaluated?

If you have GNU make 3.80 you can use the $(value ...) function.  If you
don't, you can't do it.

Note this is only relevant for the $(wildcard) line.  The *.o doesn't
contain any variable or function values, so it always returns *.o.

  rpjd> and that last statement is an immediate assignment, right?  so
  rpjd> that the wildcard expansion is done before assignment, as long
  rpjd> as something matches that pattern.

GNU make doesn't do wildcarding when expanding a variable that contains
a "*".  The result of evaluating that value is always the literal string
'*.o'.

The wildcarding may be done by the user of that string, but it isn't
handled during the expansion of the variable.

-- 
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