On Mon, Jun 08, 2015 at 10:43:39 +0200, Michal Privoznik wrote:
> Not every architecture out there has 'char' signed by default.
> For instance, my arm box has it unsigned by default:
>
> $ gcc -dM -E - < /dev/null | grep __CHAR_UNSIGNED__
> #define __CHAR_UNSIGNED__ 1
>
> Therefore, after 65c61e50 the test if failing for me. Problem is,
> we are trying to assign couple of negative values into char
> assuming some will overflow and some don't. That can't be the
> case if 'char' is unsigned by default.
>
> Signed-off-by: Michal Privoznik <mprivozn@xxxxxxxxxx>
> ---
> tests/utiltest.c | 2 +-
> 1 file changed, 1 insertion(+), 1 deletion(-)
>
> diff --git a/tests/utiltest.c b/tests/utiltest.c
> index 3a1f8eb..9eb7fb6 100644
> --- a/tests/utiltest.c
> +++ b/tests/utiltest.c
> @@ -185,7 +185,7 @@ testOverflowCheckMacro(const void *data ATTRIBUTE_UNUSED)
> {
> long long tmp;
> unsigned char luchar;
> - char lchar;
> + signed char lchar;
I actually did not know that without explicit specification char may be
signed or unsigned. Anyways, since char is not defined I'd rather change
the type to uint8_t and int8_t.
Peter
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