A way to overflow or underflow on purpose about wuffs HOT 1 CLOSED

This is already implemented. x ~mod+ y will add while ignoring overflow. There's also the x ~mod+= y form that's analogous to the += operator. An example of its use is in std/adler32/common_adler32.wuffs. For that particular case, it's not as though the computation will actually overflow, but using ~mod+ instead of a plain + makes the proof checking easier.

It's a little verbose, but breaking it down:

• the ~ means an overflow-concious operator.
• the mod says overflow is treated as modular arithmetic (the sat form chooses saturating arithmetic).
• the + means, y'know, plus.

from wuffs.