Re: [PATCH 4/4] config: avoid calling `labs()` on too-large data type

["René Scharfe" <l.s.r@web.de>]

Am 16.06.19 um 08:48 schrieb René Scharfe:
> Am 13.06.19 um 13:49 schrieb Johannes Schindelin via GitGitGadget:
>> From: Johannes Schindelin <johannes.schindelin@gmx.de>
>>
>> The `labs()` function operates, as the initial `l` suggests, on `long`
>> parameters. However, in `config.c` we tried to use it on values of type
>> `intmax_t`.
>>
>> This problem was found by GCC v9.x.
>>
>> To fix it, let's just "unroll" the function (i.e. negate the value if it
>> is negative).
>
> There's also imaxabs(3).
>
>>
>> Signed-off-by: Johannes Schindelin <johannes.schindelin@gmx.de>
>> ---
>>  config.c | 4 ++--
>>  1 file changed, 2 insertions(+), 2 deletions(-)
>>
>> diff --git a/config.c b/config.c
>> index 296a6d9cc4..01c6e9df23 100644
>> --- a/config.c
>> +++ b/config.c
>> @@ -869,9 +869,9 @@ static int git_parse_signed(const char *value, intmax_t *ret, intmax_t max)
>>  			errno = EINVAL;
>>  			return 0;
>>  		}
>> -		uval = labs(val);
>> +		uval = val < 0 ? -val : val;
>>  		uval *= factor;
>> -		if (uval > max || labs(val) > uval) {
>> +		if (uval > max || (val < 0 ? -val : val) > uval) {
>>  			errno = ERANGE;
>>  			return 0;
>>  		}
>
> So this check uses unsigned arithmetic to find out if the multiplication
> overflows, right?  Let's say value is "4G", then val will be 4 and
> factor will be 2^30.  Multiplying the two yields 2^32.  On a 32-bit
> system this will wrap around to 0, so that's what we get for uval there.
> The range check will then pass unless max is negative, which is wrong.

No, this example is wrong.  (I need to remember to take baby steps while
carrying numbers. o_O)

So value = "5G", then val = 5 and factor = 2^30.  After multiplication
uval = 2^32 + 2^30, on 32-bit machines this is 2^30 due to wrap-around.
Correct so far?

If uval is 2^30, then it's smaller than 2^31, so will pass a check
against INT_MAX.  val is 5, which is smaller than 2^30, so will pass the
second check as well.  Makes sense?

That would mean "5G" will overflow on a 32-bit machine, but we won't
detect it.

René