Hi,
If the <then> and <else> expression in the ?: ternary operator have
different signedness they will both be implicitely casted to unsigned.
When the result is stored in a variable with a storage capable of
holding both values, this is very unexpected. Consider this example:
int rc = -1;
unsigned int foo = 123;
long x = y ? foo : rc;
If one of the branch of the ?: is unsigned, then the compiler will cast
both branch to unsigned _before_ storing it in x. Despite long being
able to store INT_MIN, INT_MAX, UINT_MAX (assuming 64b long/32b int).
So if y is 0, it's basically doing
long x = (long)((unsigned int)-1);
Which will result in storing 0x00000000ffffffff (4294967295) instead of
expected 0xffffffffffffffff (-1).
I thought we hit some sort of weird compiler bug but after reducing the
problem to the simple example above and trying it GCC, clang, ICC and
MSVC they all do the same thing: https://godbolt.org/z/P5Ts7o1df
So it is most likely a C quirk. Standard reads 6.5.15. 5)
> If both the second and third operands have arithmetic type, the result
> type that would be determined by the usual arithmetic conversions, were
> they applied to those two operands, is the type of the result.
Cheers,
--
Aurélien Aptel / SUSE Labs Samba Team
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