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<body class='hmmessage'><div dir='ltr'>I have two lazarus installations.<br><br>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.<br><br>My setup is this:<br><br>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)<br>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)<br>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)<br>4) I build lazbuild by running "make lazbuild"<br>5) I build lazarus by running "make bigide"<br><br>My expectation was that only the 2.6.4 compiler would be used in building lazarus.<br><br>I was wrong.<br><br>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.<br><br>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).<br><br>I also acknowledge the very real possibility that I am once again confused and have misunderstood things, and I welcome to be corrected. : -)<br> </div></body>
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