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Re: Remove USE_GCC_RUNTIME?
nia <nia%NetBSD.org@localhost> writes:
> Hey, since the recent talks where people expressed a distate for
> GCC-specific hacks in pkgsrc, I wonder whether it's worth bringing
> this variable up:
>
> # USE_GCC_RUNTIME
> # Packages which build shared libraries but do not use libtool to
> # do so should define this variable. It is used to determine whether
> # the gcc runtime should be depended upon when a user has enabled
> # USE_PKGSRC_GCC_RUNTIME.
>
> In my opinion this is a historical wart - not using libtool (while
> unfortunate) is increasingly common with more packages using cmake/meson
> instead. In my eyes the gcc runtime should be depended upon whenever
> USE_PKGSRC_GCC_RUNTIME is set.
Do you mean
If the user has set USE_PKGSRC_GCC_RUNTIME, and if the chosen compiler
is gcc, then always add a pkgsrc dependency on the gccN-libs package
corresponding to the gccN being used, and
Remove the code and docs for this (except that the main part is
hoisted above to non-conditional).
Remove this declaration from all packges.
Or is this about forcing packges to link with pkgsrc's libstdc++ instead
of native, by putting it in buildlink?
Does it make any sense for someone to want to use pkgsrc gcc (always) by
seting USE_PKGSRC_GCC, but to use the native gcc libs? Might it make
sense for C, and not make sense for C++? (If so, is that case
sufficiently odd to be not supported?)
How does this interact with use of pkgsrc gcc due to GCC_REQD or
USE_*_FEATURES? Or is it only about the USE_PKGSRC_GCC?
Why is this about libtool? What is it doing, that the other build
systems don't? Is libtool, or the rest, buggy? Or is it that when a
package is declared to use libtool this dependency processing happens?
Practically, what will change in built packages, under what
circumstances? I think you are suggesting that pretty much every
package that uses C++ should have defined USE_GCC_RUNTIME, unless it
somehow used libtool which did this implicitly, but I don't see how that
added a dependency. But it is hard to tell.
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