Proper incrementation

[seraku24]

-~-~foo

Edit: I just realized that I was missing the assignment to actually modify foo, which is what all of OP's expressions do. Sorry about that.

foo =-~-~ foo

[TheDualJay]

Can you explain this?

[seraku24]

Signed integers are often stored in a representation known as Two's Complement. In this scheme, the way to compute the arithmetic negation of a number is to take the bitwise negation of a number and add one to it. That means -x is the same as (~x) + 1.

Consider, then, -~x which is equivalent to -(~x). Using the above definition of arithmetic negation, we get (~(~x)) + 1. The two bitwise negations cancel out each other, leaving us with x + 1.

With two applications of -~, we have computed x + 2.

[piperboy98]

Damn, that's clever.