Several unary operators in C and C++
Is it standard-conforming to use expressions like
int i = 1;
+-+-+i;
and how the sign of i variable is determined?
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.
你的操作符没有任何副作用， +i对int本身没有做任何事情，你不使用临时生成的值，但删除+什么都不做，并且-(-i)女巫等于i自己。（删除代码中的+将转换运算符，我的意思是在计算中删除它，因为它没有效果）
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.
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