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Message-ID: <5893A0CD.5090107@junovagen.se>
Date: Thu, 02 Feb 2017 22:12:45 +0100
From: Thomas Nilefalk <thomas AT junovagen DOT se>
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To: cygwin AT cygwin DOT com
Subject: Providing cygwin1.dll in both 32- and 64-bit versions
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I'm cross-compiling on Cygwin64 to 32-bit. Or rather I'm trying to 
figure out what compiling to 32-bits on a Cygwin64 actually means.

Using 'gcc -m32' causes a lot of "skipping incompatible ..." for a lot 
of libraries, so obviously ld then looks for the 32-bit libraries in the 
"wrong" place.

Using 'i686-pc-cygwin-gcc' creates an executable but that is un-runnable 
in a Cygwin64 environment because the cygwin1.dll is a 64-bit version 
and not compatible with the produced executable. A cygwin32 DLL needs to 
be put first in the path to make the executable run.

Is this by design? At least it seems to me that 'gcc -m32' could be 
taken to mean 'create an executable in the current ABI-environment 
(cygwin64) which uses a 32-bit architecture'. I'm not sure that makes 
sense or is even possible, but if it is not, then the question becomes 
'how can I make a 32-bit compiled cygwin program run under cygwin64'?

One way to answer that is to say "export 
PATH=<path_to_32_bit_cygwin1.dll>:$PATH" but that requires installing 
cygwin32.

I understand that could be the semantics of "cross-compilation" (you 
could compile for any platform, and can't expect that to run in your 
host environment), but then I'd like to propose that "-m32" should/could 
mean compile for host using 32-bit restrictions.

Which part of this is true? Is there a bug with "-m32" or am I just 
misunderstanding everything?

/Thomas

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