Mailinglist Archive: opensuse-packaging (266 mails)

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Re: [opensuse-packaging] How to disable no-return-in-nonvoid-function?

Quoting Ruediger Meier <sweet_f_a@xxxxxx>:

On Monday 26 November 2012, Dominique Leuenberger a.k.a DimStar wrote:
Quoting Ilya Chernykh <anixxsus@xxxxxxxxx>:
> On Monday 26 November 2012 13:36:46 Dominique Leuenberger a.k.a
> DimStar wrote:
>> > I have a package that brings about 50 errors
>> > "no-return-in-nonvoid-function". It is very difficult to fix all
>> > these cases (not to mention I do not know C++ and can only guess
>> > where to insert "0" and where "NULL")
>> > and also I am afraid this can break all the existing and future
>> > patches to this package. Previously I could disable it by
>> > setting the badness
>> > to 0 but it seems this does not work any more. What should I do?
>> You should contact upstream and fix the issues...
>> If you really 'must' disable this (there is no chance for the
>> package to enter Factory though!) you can add
>> #!BuildIgnore: brp-check-suse
>> to the .spec file...
> I know I can buildrequire -post-build-checks (was this package
> renamed?), but this
> will disable all ckecks. I want other checks in place, but only
> this one removed (why it was transferred from
> rpmlint to brp by the way?). Currently I added "-Wno-return-type"
> to the compiller command line, but this
> still seems to be too broad. Still looking for more specific
> method.

The most specific one is: patch the source and fix the underlying
issue. The no-return-in-nonvoid has been a brp check for as long as I

Best thing is really: bring it to upstream to get a proper fix to

But don't forget that "no-return-in-nonvoid" is just a warning and not
an error. In many cases you can safely ignore it.

A warning by the compiler does not mean you don't have to care for it.. It means the programmer is responsible to know what is happening in this code and there is NO guarantee between gcc versions that this is not 'just changing'.

The only valid solution is to fix the code.

See for example this code piece (C++)

#include <iostream>

int foo() {
int a = 5;
int b = a + 1;

int main() { std::cout << foo() << std::endl; }

What is the recommended output of this? this example is somewhat simple enough to understand what is going on... BUT the bad thing is, if you ever extend the function foo() with anything AFTER, you migh get a new return value, which is not what you wanted... hence the warning.

Best regards,
Dominique (who wishes upstreams would learn about -Werror -Wall)
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