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Re: How does make determine which shells to invoke when executing external commands?
- From: Warren Young <warren at etr-usa dot com>
- To: Cygwin-L <cygwin at cygwin dot com>
- Date: Thu, 06 Jun 2013 10:25:12 -0600
- Subject: Re: How does make determine which shells to invoke when executing external commands?
- References: <CAPzB_RzA3LBR8QKZA0NT6UJmkt6dr=qHj-s85RmY2xAd4K3KAQ at mail dot gmail dot com> <51AF6A55 dot 2090203 at etr-usa dot com> <CAPzB_RxkcD1ypzHe8j9EZBfV96VSkd3=ce-dywSdzby+mWt6_A at mail dot gmail dot com>
On 6/5/2013 20:43, Hua Ai wrote:
I thought this is a cygwin issue since windows
commands were invoked.
Cygwin make should *always* be using /bin/sh to interpret commands,
unless you've overridden it with SHELL.
Type this into a Makefile:
all:
@echo Shell is $(SHELL)
Then say "make" in that directory. It should confirm my assertion.
If it tells you something different on a problem machine, say "make
--version" and verify that it says "Built for i686-pc-cygwin".
We don't have MinGW, but we do have multiple versions of a software
(development tools from Altera) installed on these computers, which
all contains a copy of cygwin (different versions but all with make).
Are you *certain* Altera is shipping Cygwin GNU make and not MinGW make?
Either way, you're kind of in a bind.
If Altera is shipping MinGW GNU make, you have the conflict I originally
proposed.
If they are in fact shipping Cygwin, you have this problem:
http://www.cygwin.com/faq.html#faq.using.multiple-copies
(Items 4.19 through 4.23.)
This makefile however was run from a standalone cygwin.
What does it say when you say "which make"?
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