I have two lazarus installations.
I have been having problems compiling the svn branch lately, and this has
caused me to examine further how lazarus determines which compiler installation
to use.
My setup is this:
1) I start with a clean svn directory. "Clean" means that I purged all
generated *.ppu, *.o, etc. and reverted all changes to the .po files (why the
.po files are included in the svn and change upon building is another topic
altogether)
2) I have a separate binutils directory, with ar.exe, as.exe, cp.exe, make.exe,
etc. -- about 32 files in all (no fpc.exe here)
3) I set my path to include only the binutils and the compiler directory (e.g.
PATH=C:\FPCBinutils;C:\FPC_2.6.4\bin\i386-win32)
4) I build lazbuild by running "make lazbuild"
5) I build lazarus by running "make bigide"
My expectation was that only the 2.6.4 compiler would be used in building
lazarus.
I was wrong.
In my cursory examination, it appears that lazarus looks in a hardwired
directory for configuration information. If this is true, it seems very, very
wrong to me. My expectation was that there would be a configuration file for
lazarus in the current search path that would guide its compilation.
It seems that I should be able to change configurations (e.g.
laz_svn/fpc_2.6.4, laz_svn/fpc_2.7.1, laz_1.2.6/fpc_2.6.4, etc.) simply by
changing the search path (and subsequently the configuration file found on that
search path).
I also acknowledge the very real possibility that I am once again confused and
have misunderstood things, and I welcome to be corrected. : -)
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