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POMac OSX 10.6 compiler: a puzzling experience with 32 vs 64 bit
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copied!<p>I have trouble understanding the gcc compiler provided by OSX 10.6 snow leopard, mainly because of my lack of experience with 64 bits environments.</p> <pre><code>$ cat >foo.c main() {} $ gcc foo.c -o foo $ file foo foo: Mach-O 64-bit executable x86_64 $ lipo -detailed_info foo input file foo is not a fat file Non-fat file: foo is architecture: x86_64 </code></pre> <p>However, my architecture is seen as an intel i386 type (I have one of the latest Intel Core2 duo MacBook) </p> <pre><code>$ arch i386 </code></pre> <p>and the compiler targets i686-apple-darwin10</p> <pre><code>$ gcc --version i686-apple-darwin10-gcc-4.2.1 (GCC) 4.2.1 (Apple Inc. build 5646) </code></pre> <p>Of course, if I compile 32 bits I get a 32 bit executable.</p> <pre><code>$ gcc -m32 foo.c -o foo $ file foo foo: Mach-O executable i386 </code></pre> <p>but I don't get the big picture. The default setup for the compiler is to produce x86_64 executables, even if I have arch saying I have a 32 bit machine (why? Core2 is 64); even if (I guess) I am running a 32 bit kernel; even if I have a compiler targeting the i686-apple-darwin platform. Why? How can they run ? Should I compile 64 or 32 ?</p> <p>This question is due to my attempt to compile gcc 4.2.3 on the mac, but I am having a bunch of issues with gmp, mpfr and libiberty getting (in some cases) compiled for x86_64. Should I compile everything x86_64 ? If so, what's the target (not i686-apple-darwin10 I guess)?</p> <p>Thanks for the help</p>

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