Re: May be bug in export PATH and $(shell ...).

[Philip Guenther]

On Fri, Jun 11, 2010 at 1:29 AM, Oleksandr Gavenko <address@hidden> wrote:
> Just do:
>
>  $ mkdir test
>  $ cat <<EOF >test/test.sh
> #!/bin/sh
> echo I am HERE
> EOF

(Missing a "chmod +x test/test.sh here, but that doesn't really affect
your point...)


>  $ cat <<EOF >Makefile

(Should quote EOF or else the shell will try to expand $(PATH) as a
command substitution in the here-doc, but again, that doesn't really
affect your point...)


> export PATH := test:$(PATH)
> $(shell test.sh)
> all:
>        test.sh
> EOF
>  $ make
> make: temp.sh: Command not found    # from $(shell ...)
> temp.sh
> I am HERE      # from target command
>
>
> So export PATH work for spawned process under target command and don't work
> for $(shell ...).

I can't fix it in the C bits, but you can work around this at the
Makefile level: replace the $(shell test.sh) line with this:

shll = $(shell PATH=${PATH} $1)
$(call shll,test.sh)

You can use $(shll) over and over again.


Philip Guenther