Re: need help understanding $(eval ...) variable expansion

[gk]

I have partially answered my question:
        if I substitute '$$(a)' for '$(a)' I get the result I expected.

I think my problem was in understanding how $(call ...) works.
It seems that 'call' does not do any evaluation or variable assignment; 
only expansion.
When 'call' sees:
a-copy:=$(a)
'$(a)' and '$(1)' are expanded:
$(1) -> hello
$(a) ->
Then finally 'eval' makes the assignments:
a:=hello
a-copy:=
etc.

At 06:48 PM 11/25/2002 -0800, gk wrote:
> Can anyone explain how to assign the value of $(a) in the following makefile?
> This is a simplified version of what I am trying to do: I need to perform 
> filtering on the input $(1) so I cannot simply substitue $(1) for $(a) in 
> the following example.
>
> # makefile
> define macro
>
> a:=$(1)
> a-copy:=$(a)
> a-reference=$(a)
> $(eval a-copy-eval:=$(a))
> $(eval a-reference-eval=$(a))
>
> foo:
>         @echo 'a=$$(a)'
>         @echo 'a-copy=$$(a-copy)'
>         @echo 'a-reference=$$(a-reference)'
>         @echo 'a-copy-eval=$$(a-copy-eval)'
>         @echo 'a-reference-eval=$$(a-reference-eval)'
> endef # XMake_outfileRule
>
> $(eval $(call macro,hello))
> # eof
>
> - Greg Keraunen
>
>
>
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- Greg Keraunen