Several unary operators in C and C++

Question

Is it standard-conforming to use expressions like

int i = 1;
+-+-+i;

and how the sign of i variable is determined?

Answer 1

Yes it is. Unary + and - associate right-to-left, so the expression is parsed as

+(-(+(-(+i))));

Which results in 1.

Note that these can be overloaded, so for a user-defined type the answer may differ.

Answer 2

Your operators has no side effect, +i do nothing with int itself and you do not use the temporary generated value but remove + that do nothing and you have -(-i) witch is equal to i itself.(removing + in the code will convert the operator, I mean remove it in computation because it has no effect)

Answer 3

i isn't modified (C: without intervening sequence points|C++: in an unsequenced manner) so it's legal. You're just creating a new temporary with each operator.

The unary + doesn't even do anything, so all you have is two negations which just give 1 for that expression. The variable i itself is never changed.