Re: export does not export to $(shell ...

[John Graham-Cumming]

Martin d'Anjou wrote:
> l:=$(shell echo $$VAR)
> $(warning VAR=$(VAR))
> export VAR=10
> l:=$(shell echo $$VAR)
> $(warning VAR=$(VAR))
>
> make
>
> Makefile:2: VAR=
> Makefile:5: VAR=10
>
> Why would VAR be defined for $(shell) just like it is for $(warning) for 
> example? Without export, ok, but with export, why not?

The output you show makes sense to me.  The first $(warning) get 
evaluated before VAR has been set to anything and hence you get the VAR= 
output; the second $(warning) is after VAR has been set to 10 and hence 
you get the VAR=10 output.

I'd also assume that l is always empty in this Makefile since VAR will 
not be exported into the environment of the $(shell).

I think the authors of GNU Make want to avoid the following situation:

    export FOO=$(shell do_something)
    export BAR=$(shell do_something_else)

What should the values of FOO and BAR be in the environment for each of 
those shell commands?  It's hard to answer because in order to run the 
shell you'd have to figure out the value of FOO and BAR to put into the 
environment, which requires a $(shell) which requires... a loop.

John.
--
John Graham-Cumming
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