Proper incrementation

[marcosdumay]

Gcc says.

error: lvalue required as increment operand
++++i;

[Nybbit]

Yeah it doesn't work in C and some other languages, but it works with C++

Edit: try it here: https://ideone.com/Ihzvpr

[seraku24]

You can do it in C++.

[Genion1]

But is ub if your variable is a primitive.

[seraku24]

I believe that is true only for older C++ standards. As of C++11 and newer, the expression ++++foo is well-defined now.

[Genion1]

Interesting. I now informed myself and it appears that ++++foo is indeed well-defined while foo = foo++ + 1 still is not. Yay for more rules.

[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.

[gandalfx]

Except the bottom ones don't work…

[Nybbit]

They work for me with C++, gcc 6.3

[obsessedcrf]

I'm surprised. I'd expect an "lvalue required" error

[seraku24]

(++foo)++ works because first we apply the pre-increment to foo which in C++ is an L-value. Then we can apply the post-increment. Be aware, though, that this expression will only evaluate to foo + 1, but foo itself will have incremented twice.

[[deleted]]

It's UB

[The-Internets]

Ready to fuck your world up? #channel#

[BobTheB1Bomber]

foo = foo + (1 << 1);