Re: [Qemu-devel] [PULL 2/9] QEMU does not care about left shifts of signed negative values

[Paolo Bonzini]

On 25/11/2015 20:54, Peter Maydell wrote:
> > > Your latest patch at 
> > > https://gcc.gnu.org/ml/gcc-patches/2015-11/msg03055.html
> > > doesn't seem to touch the documentation of -fwrapv at all, so I
> > > don't think it is sufficient to allow users of the compiler
> > > to say "-fwrapv means signed behaviour for shifts".
> >
> > GCC *always* does signed behavior for shifts, even without -fwrapv.
> > I'll commit tomorrow the patch that promises that for the future.
> >
> > GCC does not need -fwrapv at all.
>
> Yes it does, because without -fwrapv it still wants to warn
> about them. We need to tell the compiler that we really do
> want a C dialect where they have specific behaviour and so no
> warnings are ever correct. By default (as documented, even
> with your patch) GCC just promises that it won't actually
> do undefined behaviour for signed negative shifts. (It doesn't
> even actually say what impdef semantics it does provide,

It says it above the text I changed:

     GCC supports only two's complement integer types, and all bit
     patterns are ordinary values.
     [...]
     Bitwise operators act on the representation of the value including
     both the sign and value bits, where the sign bit is considered
     immediately above the highest-value value bit.

> and in practice the impdef semantics include "warn about
> this", which we don't want.)

No, it doesn't warn with the commonly used options.  Only with 
-pedantic, which is documented as

    Issue all the warnings demanded by strict ISO C and ISO C++;
    reject all programs that use forbidden extensions, and some
    other programs that do not follow ISO C and ISO C++.

So the combination of -fwrapv and -pedantic is not particularly 
interesting.  Even then the warning is "initializer element is not
a constant expression"; nothing to do with overflow.  For example:

    #define INT_MIN ((int)-0x80000000)
    int y = INT_MIN - 1;
    int z = -1 << 2;
    int w = INT_MIN << 1;
    int u = 1 << 31;

$ gcc f.c -std=c11 -Wall -pedantic
f.c:2:1: warning: overflow in constant expression [-Woverflow]
f.c:3:9: warning: initializer element is not a constant expression [-Wpedantic]
f.c:4:9: warning: initializer element is not a constant expression [-Wpedantic]
f.c:5:9: warning: initializer element is not a constant expression [-Wpedantic]

The first warning is activated by -Wall, the others aren't.

Paolo