Re: How to preserve $-containing terms in an expanded variable (with call)?

[Paul D. Smith]

%% "Dimitry Golubovsky" <address@hidden> writes:

  dg> I am puzzled at the following problem I cannot solve.
  dg> I need to create a variable to use with "call", which after
  dg> substitution of all parameters must still contain something with
  dg> dollar-sign.

  dg> Namely, the $(call myvar,foo) must expand into:
  dg> awk '$0 != foo { ... }'
  dg> where $0 is in the awk sense (i. e. whole line read from the input).

  dg> I tried to define myvar like this:
  dg> myvar = awk '$$0 != $(0) { .... }'
  dg> this does not work: $$0 expands to /bin/sh i. e. shell's $0, not awk's

If $0 expands to /bin/sh, then that expansion is being done by the
shell, not by make.  Make will never expand $0 to /bin/sh, unless, I
suppose, you do something like:

    0 = /bin/sh

in your makefile.

So, your quoting to make is correct (make is sending $0 to the shell)
but you haven't properly quoted this to the shell so the shell is
expanding it.

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