Date: Thu, 7 Jun 2012 19:03:16 -1000 From: parv@pair.com To: Tim Daneliuk <tundra@tundraware.com> Cc: FreeBSD Mailing List <freebsd-questions@freebsd.org> Subject: Re: Somewhat OT - A Makefile Question Message-ID: <20120608050316.GA2190@holstein.holy.cow> In-Reply-To: <4FD0ADFB.8030508@tundraware.com> References: <4FCF48AF.307@tundraware.com> <20120607051901.GA2205@holstein.holy.cow> <4FD0ADFB.8030508@tundraware.com>
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in message <4FD0ADFB.8030508@tundraware.com>, wrote Tim Daneliuk thusly... > > On 06/07/2012 12:19 AM, Parv wrote: > > in message<4FCF48AF.307@tundraware.com>, > > wrote Tim Daneliuk thusly... > >> > > ... > >> Within a makefile, I need to assign the name of a program as > >> in: > >> > >> FOO = "bar". > >> > >> The problem is that 'bar' may also be know as, say, "bar.sh". > > ... > >> Is there a simple way to determine which form "bar" or "bar.sh" > >> on on a given system *at the time the make is run*? If both > >> exist, I will pick one arbitrarily, > > ... > >> For example I don't think this works when both are there: > >> > >> FOO = $(shell `which bar bar.sh) > > > > Modify the subshell command to ... > > > > which bar bar.sh | head -n 1 > > > > > > ... as in (for FreeBSD make) ... > > > > shell=`which zsh sh tcsh csh 2>/dev/null | fgrep -v 'not found' | head -n 3` ... > Thanks. Happy to help. > I came up with something similar, but I think your recipe is a bit > more elegant ... It was "If both exist, I will pick one arbitrarily ... " that helped much in modification of the original. - parv --
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