Re: Pass a C option based on command-line output?
Re: Pass a C option based on command-line output?
- Subject: Re: Pass a C option based on command-line output?
- From: Eduard de Jong <email@hidden>
- Date: Wed, 11 Jan 2006 11:05:42 -0800
Title: Re: Pass a C option based on command-line
output?
this is how i just did it:
1 select the source file that needs this variable set
2 open its inspector
3 in the "Build" tab enter the compiler flag :
-DHOST=`hostname`
4 close the inspector
5 build and run
Yet another problem may still exist if you want to use the
defined macro as a string literal in the source file: it will need
quotes as part of the definition, and i couldn't get Xcode to include
quotes in the buildsetting that worked. I either got the compiler
choking or the literal string: `hostname`..
Cheers
Eduard
At 17:46 -0500 07-01-2006, Daniel Jalkut wrote:
On Jan 7, 2006, at 4:34 PM, Chris
Espinosa wrote:
Invoking
random shell commands at any given step in an Xcode compilation would
introduce unbounded complexity, and things are bad enough as it is.
Besides, you really don't need to backtick-execute a shell command to
get the host name, it's already set in the shell's $HOSTNAME
variable.
Yeah - but executing random shell scripts
in a build phase also introduces unbounded complexity, right? I
don't know if you're gonna get away from that problem in a highly
customizable IDE :)
So in
Preprocessor Definitions just set HOSTNAME=$HOSTNAME and be done with
it.
On my machine it looks like I have $HOST.
But in any case I don't have any luck trying to set a build setting
with the expanded value. I've tried:
HOSTNAME=$HOST
HOSTNAME=$(HOST)
HOSTNAME=${HOST}
(This is on the Preprocessor Macros build
setting)
Is there a particular build setting where
I can "promote" shell variables into Xcode variables? Or is
there a different format I'm supposed to use to expand them in a build
setting's definition?
Thanks!
Daniel
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Eduard
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possible worlds.
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fears this is true.
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