Integer overflow

What do you think of this piece of C code?

  void foo(long v) {
    unsigned long u;
    unsigned sign;
    if (v < 0) {
      u = -v;
      sign = 1;
    } else {
      u = v;
      sign = 0;
    }
    ...

Seems pretty simple, right? Then what do you think of this output from MySQL:

  mysql> create table t1 (a bigint) as select '-9223372036854775807.5' as a;
  mysql> select * from t1;
  +----------------------+
  | a                    |
  +----------------------+
  | -'..--).0-*(+,))+(0( | 
  +----------------------+

Yes, that is authentic output from older versions of MySQL. Not just the wrong number, the output is complete garbage! This is my all-time favorite MySQL bug#31799. It was caused by code like the above C snippet.

So can you spot what is wrong with the code? Looks pretty simple, does it not? But the title of this post may give a hint…

It is a little known fact that signed integer overflow is undefined in C! The code above contains such undefined behaviour. The expression -v overflows when v contains the smallest negative integer of the long type (-2⁶³ on 64-bit) – the absolute value of this cannot be represented in the type. The correct way to put the absolute value of signed v into unsigned u is u = (unsigned long)0 - (unsigned long)v. Unsigned overflow is well-defined in C, in contrast to signed overflow.

And yes, GCC will generate unexpected (but technically valid) assembler for such code, as seen in the Bug#31799. If you do not like this, then use -fno-strict-overflow like I believe Postgresql and the Linux kernel do.

(But better write correct C code from the start).