This afternoon, I really don't know what I was doing with Operators and C. Eventually, I wrote some code which I was thinking wouldn't compile, But I don't know how it worked.

The code is:

`#include <stdio.h> int main() { int n=2; int sum = n + - + - + - + n; /* This line */ printf("%d\n", sum); return 0; } `

And the output is:

0

I am completely confused how the code compiled and what is happening behind the scene.

How does the line `int sum = n + - + - + - + n;`

work?

--------------Solutions-------------

`int sum = n + - + - + - + n;`

/* b u u u u u u */

/* Order: 7 6 5 4 3 2 1 */

is equivalent to:

`n + (-(+(-(+(-(+n))))));`

or simply `n + (-n)`

Note that unary operators bind more tightly than binary operators in `C opeartor precedance table`

and associativity of unary operator `+-`

is from right to left while of binary `+-`

operators in from left to right.

All but the first are just *unary operators*.

`n + - + - + - + n`

is equivalent to

`n + (-(+(-(+(-(+n))))))`

which is in turn simply equal to

`n + (-n)`

after resolving all the unary operators.

`-n`

is, of course, ordinary negation; `+n`

does essentially nothing (though it has the side effect of forcing integral promotion).

Both of `+`

and `-`

are also unary operators. The result of `+n`

is the (promoted) value of `n`

. The result of `-n`

is the negative of (promoted) `n`

.

`n + - + - + - + n;`

is equivalent to:

`n + (-(+(-(+(-(+n))))))`

which is basically `n + (-n)`

assuming no overflow happens.